2 pens and 1 eraser cost $35 and 3 pens and 4 erasers cost $65. X represents the cost of 1 pen and Y represents the cost of 1 eraser. Write the 2 simultaneous equations and solve. Set up our 2 equations where cost = price * quantity: [LIST=1] [*]2x + y = 35 [*]3x + 4y = 65 [/LIST] We can solve this one of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+%2B+y+%3D+35&term2=3x+%2B+4y+%3D+65&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+%2B+y+%3D+35&term2=3x+%2B+4y+%3D+65&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+%2B+y+%3D+35&term2=3x+%2B+4y+%3D+65&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]x (cost of 1 pen) = 15[/B] [*][B]y (cost of 1 eraser) = 5[/B] [/LIST]

2 times a number minus 4 times another number is 6. The sum of 2 numbers is 8. Find the 2 numbers. Let the first number be x, and the second number be y. We're given two equations: [LIST=1] [*]2x - 4y = 6 [*]x + y = 8 [/LIST] Using our simultaneous equation calculator, there are 3 ways to solve this: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+4y+%3D+6&term2=x+%2B+y+%3D+8&pl=Substitution']Substitution[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+4y+%3D+6&term2=x+%2B+y+%3D+8&pl=Elimination']Elimination[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+4y+%3D+6&term2=x+%2B+y+%3D+8&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] They all give the same answers: (x, y) = [B](6.3333333, 1.6666667)[/B]

4 adults and 3 children cost $40. Two adults and 6 children cost $38 Givens and Assumptions: [LIST] [*]Let the number of adults be a [*]Let the number of children be c [*]Cost = Price * Quantity [/LIST] We're given 2 equations: [LIST=1] [*]4a + 3c = 40 [*]2a + 6c = 38 [/LIST] We can solve this system of equations 3 ways [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4a+%2B+3c+%3D+40&term2=2a+%2B+6c+%3D+38&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4a+%2B+3c+%3D+40&term2=2a+%2B+6c+%3D+38&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4a+%2B+3c+%3D+40&term2=2a+%2B+6c+%3D+38&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we use, we get: [LIST] [*][B]a = 7[/B] [*][B]c = 4[/B] [/LIST]

414 people used public pool. Daily prices are $1.75 for children and $2.00 for adults. Total cost was $755.25. How many adults and children used the pool Let the number of children who used the pool be c, and the number of adults who used the pool be a. We're given two equations: [LIST=1] [*]a + c = 414 [*]2a + 1.75c = 755.25 [/LIST] We have a simultaneous equations. You can solve this any of 3 ways below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+414&term2=2a+%2B+1.75c+%3D+755.25&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+414&term2=2a+%2B+1.75c+%3D+755.25&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+414&term2=2a+%2B+1.75c+%3D+755.25&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] Whichever method you choose, you get the same answer: [LIST] [*][B]a = 123[/B] [*][B]c = 291[/B] [/LIST]

A 100 point test contains a total of 20 questions. The multiple choice questions are worth 3 points each and short response questions are worth 8 points each. Write a system of linear equations that represents this situation Assumptions: [LIST] [*]Let m be the number of multiple choice questions [*]Let s be the number of short response questions [/LIST] Since total points = points per problem * number of problems, we're given 2 equations: [LIST=1] [*][B]m + s = 20[/B] [*][B]3m + 8s = 100[/B] [/LIST] We can solve this system of equations 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+s+%3D+20&term2=3m+%2B+8s+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get: [B]m = 12, s = 8[/B]

A 12% acid solution is made by mixing 8% and 20% solutions. If the 450 ml of the 12% solution is required, how much of each solution is required? Component Unit Amount 8% Solution: 0.08 * x = 0.08x 20% Solution: 0.2 * y = 0.2y 12% Solution: 0.12 * 450 = 54 We add up the 8% solution and 20% solution to get two equations: [LIST=1] [*]0.08x + 0.2y = 54 [*]x + y = 450 [/LIST] We have a simultaneous set of equations. We can solve it using three methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.08x+%2B+0.2y+%3D+54&term2=x+%2B+y+%3D+450&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.08x+%2B+0.2y+%3D+54&term2=x+%2B+y+%3D+450&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.08x+%2B+0.2y+%3D+54&term2=x+%2B+y+%3D+450&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]x = 300 ml[/B] [*][B]y = 150 ml[/B] [/LIST]

A Bouquet of lillies and tulips has 12 flowers. Lillies cost $3 each, and tulips cost $2 each. The bouquet costs $32. Write and solve a system of linear equations to find the number of lillies and tulips in the bouquet. Let l be the number of lillies and t be the number of tulips. We're given 2 equations: [LIST=1] [*]l + t = 12 [*]3l + 2t = 32 [/LIST] With this system of equations, we can solve it 3 ways. [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=l+%2B+t+%3D+12&term2=3l+%2B+2t+%3D+32&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=l+%2B+t+%3D+12&term2=3l+%2B+2t+%3D+32&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=l+%2B+t+%3D+12&term2=3l+%2B+2t+%3D+32&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter which method we choose, we get: [LIST] [*][B]l = 8[/B] [*][B]t = 4[/B] [/LIST] [B]Now Check Your Work For Equation 1[/B] l + t = 12 8 + 4 ? 12 12 = 12 [B]Now Check Your Work For Equation 2[/B] 3l + 2t = 32 3(8) + 2(4) ? 32 24 + 8 ? 32 32 = 32

A cash register contains $5 bills and $20 bills with a total value of $180 . If there are 15 bills total, then how many of each does the register contain? Let f be the number of $5 dollar bills and t be the number of $20 bills. We're given the following equations: [LIST=1] [*]f + t = 15 [*]5f + 20t = 180 [/LIST] We can solve this system of equations 3 ways. We get [B]t = 7[/B] and [B]f = 8[/B]. [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+15&term2=5f+%2B+20t+%3D+180&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+15&term2=5f+%2B+20t+%3D+180&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+15&term2=5f+%2B+20t+%3D+180&pl=Cramers+Method']Cramers Method[/URL] [/LIST]

A cashier has a total of 52 bills in her cash drawer. There are only $10 bills and $5 bills in her drawer. The value of the bills is $320. How many $10 bills are in the drawer? Let f be the amount of $5 bills in her drawer. Let t be the amount of $10 bills in her drawer. We're given two equations: [LIST=1] [*]f + t = 52 [*]5f + 10t = 320 [/LIST] We have a system of equations. We can solve this 3 ways below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+52&term2=5f+%2B+10t+%3D+320&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+52&term2=5f+%2B+10t+%3D+320&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+52&term2=5f+%2B+10t+%3D+320&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter what method we choose, we get: f = 40 and t = 12 So the answer for how many $10 bills are in the drawer is [B]12[/B]. Let's check our work for equation 1: 40 + 12 ? 52 52 = 52 <-- Confirmed Let's check our work for equation 2: 5(40) + 10(12) ? 320 200 + 120 ? 320 320 = 320 <-- Confirmed

A crate contains 300 coins and stamps. The coins cost $3 each and the stamps cost $1.5 each. The total value of the items is $825. How many coins are there? Let c be the number of coins, and s be the number of stamps. We're given: [LIST=1] [*]c + s = 300 [*]3c + 1.5s = 825 [/LIST] We have a set of simultaneous equations, or a system of equations. We can solve this 3 ways: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Cramers+Method']Cramers Method[/URL] [/LIST] No matter which way we pick, we get: s = 50 c = [B]250[/B]

A first number plus twice a second number is 10. Twice the first number plus the second totals 29. Find the numbers. Let the first number be x. Let the second number be y. We are given the following two equations: [LIST=1] [*]x + 2y = 10 [*]2x + y = 29 [/LIST] We can solve this 3 ways using: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+10&term2=2x+%2By+%3D+29&pl=Substitution']Substitution[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+10&term2=2x+%2By+%3D+29&pl=Elimination']Elimination[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+10&term2=2x+%2By+%3D+29&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] Using any of the 3 methods, we get the same answers of [B](x, y) = (16, -3)[/B]

A first number plus twice a second number is 11. Twice the first number plus the second totals 34. Find the numbers. Let the first number be x and the second number be y. We're given: [LIST=1] [*]x + 2y = 11 [*]2x + y = 34 [/LIST] Using our simultaneous equations calculator, we have 3 methods to solve this: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+11&term2=2x+%2B+y+%3D+34&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+11&term2=2x+%2B+y+%3D+34&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+11&term2=2x+%2B+y+%3D+34&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] All 3 methods give the same solution: [LIST] [*][B]x = 19[/B] [*][B]y = -4[/B] [/LIST]

A first number plus twice a second number is 22. Twice the first number plus the second totals 28. Find the numbers. Let the first number be x. Let the second number be y. We're given two equations: [LIST=1] [*]x + 2y = 22 <-- Since twice means multiply by 2 [*]2x + y = 28 <-- Since twice means multiply by 2 [/LIST] We have a set of simultaneous equations. We can solve this three ways [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+22&term2=2x+%2B+y+%3D+28+&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+22&term2=2x+%2B+y+%3D+28&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+22&term2=2x+%2B+y+%3D+28&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter which method we use, we get the same answer: [LIST] [*][B]x = 11 & 1/3[/B] [*][B]y = 5 & 1/3[/B] [/LIST]

A first number plus twice a second number is 3. Twice the first number plus the second totals 24. Let the first number be x. Let the second number be y. We're given: [LIST=1] [*]x + 2y = 3 <-- Because [I]twice[/I] means multiply by 2 [*]2x + y = 24 <-- Because [I]twice[/I] means multiply by 2 [/LIST] We have a system of equations. We can solve it any one of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+3&term2=2x+%2B+y+%3D+24&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+3&term2=2x+%2B+y+%3D+24&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+2y+%3D+3&term2=2x+%2B+y+%3D+24&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which way we choose, we get: [LIST] [*]x = [B]15[/B] [*]y = [B]-6[/B] [/LIST]

A food truck sells salads for $6.50 each and drinks for $2.00 each. The food trucks revenue from selling a total of 209 salads and drinks in one day was $836.50. How many salads were sold that day? Let the number of drinks be d. Let the number of salads be s. We're given two equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d + s = 209 [/LIST] We can use substitution to solve this system of equations quickly. The question asks for the number of salads (s). Therefore, we want all expressions in terms of s. Rearrange Equation 2 by subtracting s from both sides: d + s - s = 209 - s Cancel the s's, we get: d = 209 - s So we have the following system of equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d = 209 - s [/LIST] Substitute equation (2) into equation (1) for d: 2(209 - s) + 6.50s = 836.50 Multiply through to remove the parentheses: 418 - 2s + 6.50s = 836.50 To solve this equation for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=418-2s%2B6.50s%3D836.50&pl=Solve']type it into our search engine and we get[/URL]: s = [B]93[/B]

A movie theater charges $7 for adults and $3 for seniors on a particular day when 324 people paid an admission the total receipts were 1228 how many were seniors and how many were adults? Let the number of adult tickets be a. Let the number of senior tickets be s. We're given two equations: [LIST=1] [*]a + s = 324 [*]7a + 3s = 1228 [/LIST] We have a set of simultaneous equations we can solve using 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+324&term2=7a+%2B+3s+%3D+1228&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+324&term2=7a+%2B+3s+%3D+1228&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+324&term2=7a+%2B+3s+%3D+1228&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [LIST] [*][B]a = 64[/B] [*][B]s = 260[/B] [/LIST]

A movie theater charges 7.00 for adults and 2.00 for seniors citizens. On a day when 304 people paid for admission, the total receipt were 1118. How many who paid were adults ? How many were senior citizens? Let a be the number of adult tickets. Let s be the number of senior citizen tickets. We're given two equations: [LIST=1] [*]a + s = 304 [*]7a + 2s = 1118 [/LIST] We can solve this system of equations three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+304&term2=7a+%2B+2s+%3D+1118&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+304&term2=7a+%2B+2s+%3D+1118&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+304&term2=7a+%2B+2s+%3D+1118&pl=Cramers+Method']Cramer's Method[/URL] [/LIST] No matter which way we choose, we end up with the same answer: [LIST] [*]a = [B]102[/B] [*]s = [B]202[/B] [/LIST]

A party rental company has chairs and tables for rent. The total cost to rent 5 chairs and 3 tables is $37. The total cost to rent 2 chairs and 6 tables is $64. What is the cost to rent each chair and each table? Let c be the cost of renting one chair and t be the cost of renting one table. We're given two equations: [LIST=1] [*]5c + 3t = 37 [*]2c + 6t =64 [/LIST] We have a system of equations. Using our system of equations calculator, we can solve this problem any of 3 ways below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5c+%2B+3t+%3D+37&term2=2c+%2B+6t+%3D+64&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5c+%2B+3t+%3D+37&term2=2c+%2B+6t+%3D+64&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5c+%2B+3t+%3D+37&term2=2c+%2B+6t+%3D+64&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] All 3 methods give the same answer: [LIST] [*][B]Chairs (c) cost $1.25[/B] [*][B]Tables (t) cost $10.25[/B] [/LIST]

A person has $13,000 invested in stock A and stock B. Stock A currently sells for $20 a share and stock B sells for $90 a share. If stock B triples in value and stock A goes up 50%, his stock will be worth $33,000. How many shares of each stock does he own? Set up the given equations, where A is the number of shares for Stock A, and B is the number of shares for Stock B [LIST=1] [*]90A + 20B = 13000 [*]3(90A) + 1.5(20B) = 33000 <-- [I]Triple means multiply by 3, and 50% gain means multiply by 1.5[/I] [/LIST] Rewrite (2) by multiplying through: 270A + 30B = 33000 Using our simultaneous equations calculator, we get [B]A = 100 and B = 200[/B]. Click the links below to solve using each method: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=90A+%2B+20B+%3D+13000&term2=270A+%2B+30B+%3D+33000&pl=Cramers+Method']Cramers Method[/URL] [/LIST] Check our work using equation (1) 90(100) + 20(200) ? 13,000 9000 + 4000 ? 13,000 13000 = 13000

a son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. Declare variables: [LIST] [*]Let f be the father's age [*]Let s be the son's age [/LIST] We're given two equations: [LIST=1] [*]s = f/4 [*]f - s = 30. [I]The reason why we subtract s from f is the father is older[/I] [/LIST] Using substitution, we substitute equaiton (1) into equation (2) for s: f - f/4 = 30 To remove the denominator/fraction, we multiply both sides of the equation by 4: 4f - 4f/4 = 30 *4 4f - f = 120 3f = 120 To solve for f, we divide each side of the equation by 3: 3f/3 = 120/3 Cancel the 3's on the left side and we get: f = [B]40[/B]

A straight line has the equation ax + by=23. The points (5,-2) and (1,-5) lie on the line. Find the values of a and b. plug in both points and form 2 equations: [LIST=1] [*]5a - 2b = 23 [*]1x - 5b = 23 [/LIST] We can solve this simultaneous equations any one of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5a+-+2b+%3D+23&term2=1a+-+5b+%3D+23&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5a+-+2b+%3D+23&term2=1a+-+5b+%3D+23&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=5a+-+2b+%3D+23&term2=1a+-+5b+%3D+23&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]a = 3[/B] [*][B]b = -4[/B] [/LIST]

A test has twenty questions worth 100 points total. the test consists of true/false questions worth 3 points each and multiple choice questions worth 11 points each. How many true/false questions are on the test? Let m be the number of multiple choice questions and t be the number of true/false questions. We're given: [LIST=1] [*]m + t = 20 [*]11m + 3t = 100 [/LIST] We can solve this system of equations 3 ways below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the following answers: [LIST] [*][B]m = 5[/B] [*][B]t = 15[/B] [/LIST] Check our work in equation 1: 5 + 15 ? 20 [I]20 = 20[/I] Check our work in equation 2: 11(5) + 3(15) ? 100 55 + 45 ? 100 [I]100 = 100[/I]

A test has twenty questions worth 100 points. The test consists of True/False questions worth 3 points each and multiple choice questions worth 11 points each. How many multiple choice questions are on the test? Let the number of true/false questions be t. Let the number of multiple choice questions be m. We're given two equations: [LIST=1] [*]m + t = 20 [*]11m + 3t = 100 [/LIST] We have a set of simultaneous equations. We can solve this using 3 methods: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1m+%2B+t+%3D+20&term2=11m+%2B+3t+%3D+100&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we pick, we get the same answer: [LIST] [*][B]m = 5[/B] [*][B]t = 15[/B] [/LIST]

A used book store also started selling used CDs and videos. In the first week, the store sold a combination of 40 CDs and videos. They charged $4 per CD and $6 per video and the total sales were $180. Determine the total number of CDs and videos sold Let c be the number of CDs sold, and v be the number of videos sold. We're given 2 equations: [LIST=1] [*]c + v = 40 [*]4c + 6v = 180 [/LIST] You can solve this system of equations three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter what method we choose, we get [B]c = 30, v = 10[/B]. Now let's check our work for both given equations for c = 30 and v = 10: [LIST=1] [*]30 + 10 = 40 <-- This checks out [*]4c + 6v = 180 --> 4(30) + 6(10) --> 120 + 60 = 180 <-- This checks out [/LIST]

A used book store also started selling used CDs and videos. In the first week, the store sold a combination of 40 CDs and videos. They charged $4 per CD and $6 per video and the total sales were $180. Determine the total number of CDs and videos sold. Let the number of cd's be c and number of videos be v. We're given two equations: [LIST=1] [*]c + v = 40 [*]4c + 6v = 180 [/LIST] We can solve this system of equations using 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+v+%3D+40&term2=4c+%2B+6v+%3D+180&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [B]c = 30 v = 10[/B]

Admission to a baseball game is $2.00 for general admission and $5.50 for reserved seats. The receipts were $3577.00 for 1197 paid admissions. How many of each ticket were sold? Let g be the number of tickets for general admission Let r be the number of tickets for reserved seats We have two equations: [LIST=1] [*]g + r = 1197 [*]2g + 5.50r = 3577 [/LIST] We can solve this a few ways, but let's use substitution using our [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=g+%2B+r+%3D+1197&term2=2g+%2B+5.50r+%3D+3577&pl=Substitution']simultaneous equations calculator[/URL]: [LIST] [*][B]r = 338[/B] [*][B]g = 859[/B] [/LIST]

admission to the school fair is $2.50 for students and $3.75 for others. if 2848 admissions were collected for a total of 10,078.75, how many students attended the fair Let the number of students be s and the others be o. We're given two equations: [LIST=1] [*]o + s = 2848 [*]3.75o + 2.50s = 10078.75 [/LIST] Since we have no coefficients for equation 1, let's solve this the fast way using substitution. Rearrange equation 1 by subtracting o from each side to isolate s [LIST=1] [*]o = 2848 - s [*]3.75o + 2.50s = 10078.75 [/LIST] Now substitute equation 1 into equation 2: 3.75(2848 - s) + 2.50s =10078.75 To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=3.75%282848-s%29%2B2.50s%3D10078.75&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]481[/B]

Ahmad has a jar containing only 5-cent and 20-cent coins. In total there are 31 coins with a total value of $3.50. How many of each type of coin does Ahmad have? Let the number of 5-cent coins be f. Let the number of 20-cent coins be t. We're given two equations: [LIST=1] [*]f + t = 31 [*]0.05f + 0.2t = 3.50 [/LIST] We can solve this simultaneous system of equations 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+31&term2=0.05f+%2B+0.2t+%3D+3.50&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter which method we choose, we get: [LIST] [*][B]f = 18[/B] [*][B]t = 13[/B] [/LIST]

Alberto and Willie each improved their yards by planting daylilies and ivy. They bought their supplies from the same store. Alberto spent $64 on 3 daylilies and 8 pots of ivy. Willie spent $107 on 9 daylilies and 7 pots of ivy. What is the cost of one daylily and the cost of one pot of ivy? Assumptions: [LIST] [*]Let d be the cost of one daylily [*]Let i be the cost of one pot of ivy [/LIST] Givens: [LIST=1] [*]3d + 8i = 64 [*]9d + 7i = 107 [/LIST] To solve this system of equations, you can use any of three methods below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3d+%2B+8i+%3D+64&term2=9d+%2B+7i+%3D+107&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3d+%2B+8i+%3D+64&term2=9d+%2B+7i+%3D+107&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3d+%2B+8i+%3D+64&term2=9d+%2B+7i+%3D+107&pl=Cramers+Method']Cramer's Method[/URL] [/LIST] No matter what method we use, we get the same answer: [LIST] [*][B]d = 8[/B] [*][B]i = 5[/B] [/LIST] [B][MEDIA=youtube]K1n3niERg-U[/MEDIA][/B]

Alexis is working at her schools bake sale. Each mini cherry pie sells for $4 and each mini peach pie sells for $3. Alexis sells 25 pies and collects $84. How many pies of each kind does she sell? Let each cherry pie be c and each peach pie be p. We have the following equations: [LIST=1] [*]c + p = 25 [*]4c + 3p = 84 [/LIST] You can solve this system of equations 3 ways. [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c%2Bp%3D25&term2=4c+%2B+3p+%3D+84&pl=Substitution']Substitution Rule[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c%2Bp%3D25&term2=4c+%2B+3p+%3D+84&pl=Elimination']Elimination Rule[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c%2Bp%3D25&term2=4c+%2B+3p+%3D+84&pl=Cramers+Method']Cramers Rule[/URL] No matter which way you choose, you get [B]c = 9 and p = 16[/B].

Free Algebraic Substitutions Calculator - Given an algebraic statement with variables [a-z], this calculator takes a set of given substitution values, i.e., x=2,y=3,z=4, and evaluates your statement using the substitution values.

at a bakery the cost of one cupcake and 2 slices of pie is $12.40. the cost of 2 cupcakes and 3 slices of pie costs $20.20. what is the cost of one cupcake? Let the number of cupcakes be c Let the number of pie slices be p Total Cost = Unit cost * quantity So we're given two equations: [LIST=1] [*]1c + 2p = 12.40 [*]2c + 3p = 20.20 [/LIST] We can solve this system of equations any one of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1c+%2B+2p+%3D+12.40&term2=2c+%2B+3p+%3D+20.20&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1c+%2B+2p+%3D+12.40&term2=2c+%2B+3p+%3D+20.20&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=1c+%2B+2p+%3D+12.40&term2=2c+%2B+3p+%3D+20.20&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]c = 3.2[/B] [*]p = 4.6 [/LIST]

Consider the following 15 numbers 1, 2, y - 4, 4, 5, x, 6, 7, 8, y, 9, 10, 12, 3x, 20 - The mean of the last 10 numbers is TWICE the mean of the first 10 numbers - The sum of the last 2 numbers is FIVE times the sum of the first 3 numbers (i) Calculate the values of x and y We're given two equations: [LIST=1] [*](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10 = 2(1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/10 [*]3x - 20 = 5(1 + 2 + y - 4) [/LIST] Let's evaluate and simplify: [LIST=1] [*](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10 = (1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/5 [*]3x - 20 = 5(y - 1) [/LIST] Simplify some more: [URL='https://www.mathcelebrity.com/polynomial.php?num=x%2B6%2B7%2B8%2By%2B9%2B10%2B12%2B3x%2B20&pl=Evaluate'](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10[/URL] = (4x + y + 72)/10 [URL='https://www.mathcelebrity.com/polynomial.php?num=1%2B2%2By-4%2B4%2B5%2Bx%2B6%2B7%2B8%2By&pl=Evaluate'](1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/5[/URL] = (2y + x + 29)/5 5(y - 1) = 5y - 5 So we're left with: [LIST=1] [*](4x + y + 72)/10 = (2y + x + 29)/5 [*]3x - 20 = 5y - 5 [/LIST] Cross multiply equations in 1, we have: 5(4x + y + 72) = 10(2y + x + 29) 20x + 5y + 360 = 20y + 10x + 290 We have: [LIST=1] [*]20x + 5y + 360 = 20y + 10x + 290 [*]3x - 20 = 5y - 5 [/LIST] Combining like terms: [LIST=1] [*]10x - 15y = -70 [*]3x - 5y = 15 [/LIST] Now we have a system of equations which we can solve any of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: (x, y) = [B](-115, -72)[/B]

Given: BC = EF AC = EG AB = 10 BC = 3 Prove FG = 10 [LIST] [*]AC = AB + BC (Segment Addition Postulate) [*]AB = 10, BC = 3 (Given) [*]AC = 10 + 3 (Substitution Property of Equality) [*]AC = 13 (Simplify) [*]AC = EG, BC = EF (Given) [*]EG = 13, EF = 3 (Segment Equivalence) [*]EG = EF + FG (Segment Addition Postulate) [*]13 = 3 + FG (Substitution Property of Equality) [*]FG = 10 (Subtraction Property) [/LIST]

Half of a pepperoni pizza plus 3/4ths of a ham and pineapple pizza has 765 calories. 1/4th of a pepperoni pizza and a whole ham and pineapple pizza contains 745 calories. How many calories are each of the 2 kinds of pizzas individually? Let p be the pepperoni pizza calories and h be the ham and pineapple pizza calories. We're given [LIST=1] [*]0.5p + 0.75h = 765 [*]0.25p + h = 745 [/LIST] With this system of equations, we can solve using 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.5p+%2B+0.75h+%3D+765&term2=0.25p+%2B+h+%3D+745&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.5p+%2B+0.75h+%3D+765&term2=0.25p+%2B+h+%3D+745&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.5p+%2B+0.75h+%3D+765&term2=0.25p+%2B+h+%3D+745&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [B]h = 580 p = 660[/B]

if a=3, then find the value of 30-5a. Using [URL='https://www.mathcelebrity.com/algsubs.php?num=30-5a&varlist=a%3D3&pl=Evaluate']our algebraic substitution calculator[/URL], we get: a = [B]15[/B]

If JK = PQ and PQ = ST, then JK=ST JK = PQ | Given Substitute ST for PQ since PQ = ST | Substitution [B]JK = ST[/B]

if sc = hr and hr=ab then sc=ab sc = hr (given) Since hr = ab, we can substitute ab for hr by substitution: [B]sc = ab[/B]

In a bike shop they sell bicycles & tricycles. I counted 80 wheels & 34 seats. How many bicycles & tricycles were in the bike shop? Let b be the number or bicycles and t be the number of tricycles. Since each bicycle has 2 wheels and 1 seat and each tricycle has 3 wheels and 1 seat, we have the following equations: [LIST=1] [*]2b + 3t = 80 [*]b + t = 34 [/LIST] We can solve this set of simultaneous equations 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2b+%2B+3t+%3D+80&term2=b+%2B+t+%3D+34&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2b+%2B+3t+%3D+80&term2=b+%2B+t+%3D+34&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2b+%2B+3t+%3D+80&term2=b+%2B+t+%3D+34&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]b = 22[/B] [*][B]t = 12[/B] [/LIST]

Jack has 34 bills and coins in 5’s and 2’s. The total value is $116. How many 5 dollar bills does he have? Let the number of 5 dollar bills be f. Let the number of 2 dollar bills be t. We're given two equations: [LIST=1] [*]f + t = 34 [*]5f + 2t = 116 [/LIST] We have a system of equations, which we can solve 3 ways: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=f+%2B+t+%3D+34&term2=5f+%2B+2t+%3D+116&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answers: [LIST] [*][B]f = 16[/B] [*][B]t = 18[/B] [/LIST]

Jill and Jack are getting vegetables from the Farmer's Market. Jill buys 12 carrots and 8 tomatoes for $34. Jack buys 10 carrots and 7 tomatoes for $29. How much does each carrot and each tomato cost? Let the cost of carrots be c and the cost of tomatoes be t. Since the total cost is price times quantity, We're given two equations: [LIST=1] [*]12c + 8t = 34 <-- Jill [*]10c + 7t = 29 <-- Jack [/LIST] We have a system of equations. We can solve this one of three ways: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12c+%2B+8t+%3D+34&term2=10c+%2B+7t+%3D+29&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12c+%2B+8t+%3D+34&term2=10c+%2B+7t+%3D+29&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12c+%2B+8t+%3D+34&term2=10c+%2B+7t+%3D+29&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [LIST] [*][B]t = 2[/B] [*][B]c = 1.5[/B] [/LIST]

Joel bought 88 books. Some books cost $13 each and some cost $17 each. In all, he spent $128. Which system of linear equations represents the given situation? Let a be the number of the $13 book, and b equal the number of $17 books. We have the following system of linear equations: [LIST=1] [*][B]a + b = 88[/B] [*][B]13a + 17b = 128[/B] [/LIST] To solve this system, use our calculator for the following methods: [LIST] [*][URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+b+%3D+88&term2=13a+%2B+17b+%3D+128&pl=Substitution']Substitution[/URL] [*][URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+b+%3D+88&term2=13a+%2B+17b+%3D+128&pl=Elimination']Elimination[/URL] [*][URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+b+%3D+88&term2=13a+%2B+17b+%3D+128&pl=Cramers+Method']Cramers Method[/URL] [/LIST]

Kevin and Randy Muise have a jar containing 52 coins, all of which are either quarters or nickels. The total value of the coins in the jar is $6.20. How many of each type of coin do they have? Let q be the number of quarters, and n be the number of nickels. We have: [LIST=1] [*]n + q = 52 [*]0.05n + 0.25q = 6.20 [/LIST] We can solve this system of equations three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=n+%2B+q+%3D+52&term2=0.05n+%2B+0.25q+%3D+6.20&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter what method we choose, we get the same answer: [LIST] [*][B]n = 34[/B] [*][B]q = 18[/B] [/LIST]

M is the midpoint of AB. Prove AB = 2AM M is the midpoint of AB (Given) AM = MB (Definition of Congruent Segments) AM + MB = AB (Segment Addition Postulate) AM + AM = AB (Substitution Property of Equality) 2AM = AB (Distributive property) [MEDIA=youtube]8BNo_4kvBzw[/MEDIA]

On Monday the office staff at your school paid 8.77 for 4 cups of coffee and 7 bagels. On Wednesday they paid 15.80 for 8 cups of coffee and 14 bagels. Can you determine the cost of a bagel Let the number of cups of coffee be c Let the number of bagels be b. Since cost = Price * Quantity, we're given two equations: [LIST=1] [*]7b + 4c = 8.77 [*]14b + 8c = 15.80 [/LIST] We have a system of equations. We can solve this 3 ways: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=7b+%2B+4c+%3D+8.77&term2=14b+%2B+8c+%3D+15.80&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=7b+%2B+4c+%3D+8.77&term2=14b+%2B+8c+%3D+15.80&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=7b+%2B+4c+%3D+8.77&term2=14b+%2B+8c+%3D+15.80&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we use, we get the same answer [LIST] [*]The system is inconsistent. Therefore, we have no answer. [/LIST]

On the first day of ticket sales the school sold 3 senior citizen tickets and 10 child tickets for a total of $82. The school took in $67 on the second day by selling 8 senior citizen tickets And 5 child tickets. What is the price of each ticket? Let the number of child tickets be c Let the number of senior citizen tickets be s We're given two equations: [LIST=1] [*]10c + 3s = 82 [*]5c + 8s = 67 [/LIST] We have a system of simultaneous equations. We can solve it using any one of 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10c+%2B+3s+%3D+82&term2=5c+%2B+8s+%3D+67&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10c+%2B+3s+%3D+82&term2=5c+%2B+8s+%3D+67&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10c+%2B+3s+%3D+82&term2=5c+%2B+8s+%3D+67&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [LIST] [*][B]c = 7[/B] [*][B]s = 4[/B] [/LIST]

Pam has two part-time jobs. At one job, she works as a cashier and makes $8 per hour. At the second job, she works as a tutor and makes$12 per hour. One week she worked 30 hours and made$268 . How many hours did she spend at each job? Let the cashier hours be c. Let the tutor hours be t. We're given 2 equations: [LIST=1] [*]c + t = 30 [*]8c + 12t = 268 [/LIST] To solve this system of equations, we can use 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+t+%3D+30&term2=8c+%2B+12t+%3D+268&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+t+%3D+30&term2=8c+%2B+12t+%3D+268&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+t+%3D+30&term2=8c+%2B+12t+%3D+268&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we use, we get the same answer: [LIST] [*]c = [B]23[/B] [*]t = [B]7[/B] [/LIST]

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40, or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost? Let the cost of paper packages be p and the cost of staplers be s. We're given two equations: [LIST=1] [*]3p + 4s = 40 [*]5p + 6s = 62 [/LIST] We have a system of equations. We can solve this three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3p+%2B+4s+%3D+40&term2=5p+%2B+6s+%3D+62&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3p+%2B+4s+%3D+40&term2=5p+%2B+6s+%3D+62&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3p+%2B+4s+%3D+40&term2=5p+%2B+6s+%3D+62&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get the same answer: [LIST] [*][B]p = 4[/B] [*][B]s = 7[/B] [/LIST]

Sam has $2.25 in quarters and dimes, and the total number of coins is 12. How many quarters and how many dimes? Let d be the number of dimes. Let q be the number of quarters. We're given two equations: [LIST=1] [*]0.1d + 0.25q = 2.25 [*]d + q = 12 [/LIST] We have a simultaneous system of equations. We can solve this 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=0.1d+%2B+0.25q+%3D+2.25&term2=d+%2B+q+%3D+12&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]d = 5[/B] [*][B]q = 7[/B] [/LIST]

Free Simultaneous Equations Calculator - Solves a system of simultaneous equations with 2 unknowns using the following 3 methods:
1) Substitution Method (Direct Substitution)
2) Elimination Method
3) Cramers Method or Cramers Rule Pick any 3 of the methods to solve the systems of equations 2 equations 2 unknowns

Some History teachers at Richmond High School are purchasing tickets for students and their adult chaperones to go on a field trip to a nearby museum. For her class, Mrs. Yang bought 30 student tickets and 30 adult tickets, which cost a total of $750. Mr. Alexander spent $682, getting 28 student tickets and 27 adult tickets. What is the price for each type of ticket? Let the number of adult tickets be a Let the number of student tickets be s We're given two equations: [LIST=1] [*]30a + 30s = 750 [*]27a + 28s = 682 [/LIST] To solve the simultaneous equations, we can use any of three methods below: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=30a+%2B+30s+%3D+750&term2=27a+%2B+28s+%3D+682&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=30a+%2B+30s+%3D+750&term2=27a+%2B+28s+%3D+682&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=30a+%2B+30s+%3D+750&term2=27a+%2B+28s+%3D+682&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we use, we get the same answers: [LIST] [*][B]a = 18[/B] [*][B]s = 7[/B] [/LIST]

The admission fee at an amusement park is $1.50 for children and $4 for adults. On a certain day, 327 people entered the park , and the admission fee collected totaled 978.00 dollars . How many children and how many adults were admitted? Let the number of children's tickets be c. Let the number of adult tickets be a. We're given two equations: [LIST=1] [*]a + c = 327 [*]4a + 1.50c = 978 [/LIST] We can solve this system of equation 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+327&term2=4a+%2B+1.50c+%3D+978&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+327&term2=4a+%2B+1.50c+%3D+978&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+c+%3D+327&term2=4a+%2B+1.50c+%3D+978&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answers: [LIST] [*][B]a = 195[/B] [*][B]c = 132[/B] [/LIST]

the cost of a buffet at a restaurant is different for adults and kids. the bill for 2 adults and 3 kids is $51. the bill for 3 adults and 1 kid is $45. what is the cost per adult and per kid? Let the cost for each adult be a Let the cost for each kid be k We're given two equations: [LIST=1] [*]2a + 3k = 51 [*]3a + k = 45 [/LIST] To solve this simultaneous set of equations, we can use three methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2a+%2B+3k+%3D+51&term2=3a+%2B+k+%3D+45&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2a+%2B+3k+%3D+51&term2=3a+%2B+k+%3D+45&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2a+%2B+3k+%3D+51&term2=3a+%2B+k+%3D+45&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we use, we get the same answer: [LIST] [*]a = [B]12[/B] [*]k = [B]9[/B] [/LIST]

The first group orders 3 pizzas and 4 drinks for $33.50. The second group orders 5 pizzas and 6 drinks for $54. Find the cost for each pizza and each drink Assumptions: [LIST] [*]Let the cost of each pizza be p [*]Let the cost of each drink be d [/LIST] Givens: [LIST=1] [*]4d + 3p = 33.50 [*]6d + 5p = 54 [/LIST] We have a simultaneous group of equations. To solve this, we can use 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4d+%2B+3p+%3D+33.50&term2=6d+%2B+5p+%3D+54&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4d+%2B+3p+%3D+33.50&term2=6d+%2B+5p+%3D+54&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=4d+%2B+3p+%3D+33.50&term2=6d+%2B+5p+%3D+54&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we use, we get the same answer: [LIST] [*]d = [B]$2.75[/B] [*]p = [B]$7.5[/B] [/LIST]

The Radio City Music Hall is selling tickets to Kayla’s premiere at the Rockettes. On the first day of ticket sales they sold 3 senior citizen tickets and 9 child tickets for a total of $75. It took in $67 on the second day by selling 8 senior citizen tickets and 5 child tickets. What is the price of each senior citizen ticket and each child ticket? Let the cost of child tickets be c Let the cost of senior tickets be s Since revenue = cost * quantity, we're given two equations: [LIST=1] [*]9c + 3s = 75 [*]5c + 8s = 67 [/LIST] To solve this simultaneous group of equations, we can use 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=9c+%2B+3s+%3D+75&term2=5c+%2B+8s+%3D+67&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=9c+%2B+3s+%3D+75&term2=5c+%2B+8s+%3D+67&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=9c+%2B+3s+%3D+75&term2=5c+%2B+8s+%3D+67&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we use, we get the same answer: [LIST] [*][B]c = 7[/B] [*][B]s = 4[/B] [/LIST]

The school is selling potted plants as a fundraiser. Kara sold 12 ferns and 8 ivy plants for 260.00. Paul sold 15 ivy plants and 6 ferns for 240. What’s the selling price of each plant. Let the cost of each fern be f Let the cost of each ivy plant be I We're given: [LIST=1] [*]12f + 8i = 260 [*]15i + 6f = 240 [/LIST] To solve this system of equations, we can use 3 methods: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12f+%2B+8i+%3D+260&term2=15f+%2B+6i+%3D+240&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12f+%2B+8i+%3D+260&term2=15f+%2B+6i+%3D+240&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=12f+%2B+8i+%3D+260&term2=15f+%2B+6i+%3D+240&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: [LIST] [*][B]f = 7.5[/B] [*][B]i= 21.25[/B] [/LIST]

The senior class at high school A and high school B planned separate trips to the state fair. There senior class and high school A rented and filled 10 vans and 6 buses with 276 students. High school B rented and filled 5 vans and 2 buses with 117 students. Every van had the same number of students in them as did the buses. How many students can a van carry?? How many students can a bus carry?? Let b be the number of students a bus can carry. Let v be the number of students a van can carry. We're given: [LIST=1] [*]High School A: 10v + 6b = 276 [*]High School B: 5v + 2b = 117 [/LIST] We have a system of equations. We can solve this 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10v+%2B+6b+%3D+276&term2=5v+%2B+2b+%3D+117&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10v+%2B+6b+%3D+276&term2=5v+%2B+2b+%3D+117&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10v+%2B+6b+%3D+276&term2=5v+%2B+2b+%3D+117&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get: [LIST] [*][B]b = 21[/B] [*][B]v = 15[/B] [/LIST]

the sum of 2 numbers is 5. 5 times the larger number plus 4 times the smaller number is 37. Find the numbers Let the first small number be x. Let the second larger number be y. We're given: [LIST=1] [*]x + y = 5 [*]5y + 4x = 37 [/LIST] We can solve this 3 ways, using the following methods: [LIST=1] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+y%3D5&term2=5y+%2B+4x+%3D+37&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+y%3D5&term2=5y+%2B+4x+%3D+37&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+y%3D5&term2=5y+%2B+4x+%3D+37&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [B]x = -12 y = 17 [/B] Let's check our work using equation 1: -12 + 17 ? 5 5 = 5 <-- Check Let's check our work using equation 2: 5(17) + 4(-12) ? 37 85 - 48 ? 37 37 = 37 <-- Check

There were 175 tickets sold for the upcoming event in the gym. If students tickets cost $5 and adult tickets are $8, tell me how many tickets were sold if gate receipts totaled $1028? Let s be the number of student tickets and a be the number of adult tickets. We are given: a + s = 175 8a + 5s = 1028 There are 3 ways to solve this, all of which give us: [B]a = 51 s = 124 [/B] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Substitution']Substitution Method[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Elimination']Elimination Method[/URL] [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=a+%2B+s+%3D+175&term2=8a+%2B+5s+%3D+1028&pl=Cramers+Method']Cramers Method[/URL]

Tom is 2 years older than Sue and Bill is twice as old as Tom. If you add all their ages and subtract 2, the sum is 20. How old is Bill? Let t be Tom's age., s be Sue's age, and b be Bill's age. We have the following equations: [LIST=1] [*]t = s + 2 [*]b = 2t [*]s + t + b - 2 = 20 [/LIST] Get (2) in terms of s (2) b = 2(s + 2) <-- using (1), substitute for t So we have (3) rewritten with substitution as: s + (s + 2) + 2(s + 2) - 2 = 20 s + (s + 2) + 2s + 4 - 2 = 20 Group like terms: (s + s + 2s) + (2 + 4 - 2) = 20 4s + 4 = 20 Run this through our [URL='https://www.mathcelebrity.com/1unk.php?num=4s%2B4%3D20&pl=Solve']equation calculator [/URL]to get s = 4 Above, we had b = 2(s + 2) Substituting s = 4, we get: 2(4 + 2) = 2(6) = [B]12 Bill is 12 years old[/B]

Twice a first number decreased by a second number is 16. The first number increased by 3 times the second number is 1. Find the numbers. Let the first number be x and the second number be y. We're given: [LIST=1] [*]2x - y = 16 [*]x + 3y = 1 [/LIST] Using our simultaneous equations calculator, you can solve this 3 ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+y+%3D+16&term2=x+%2B+3y+%3D+1&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+y+%3D+16&term2=x+%2B+3y+%3D+1&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=2x+-+y+%3D+16&term2=x+%2B+3y+%3D+1&pl=Cramers+Method']Cramers Rule[/URL] [/LIST] No matter what method we use, we get the same answers: [B]x = 7 y = -2 (x, y) = (7, -2) [/B] Let's check our work in equation 1: 2(7) - -2 ? 16 14 + 2 ? 16 16 = 16 <-- Check Let's check our work in equation 2: 7 + 3(-2) ? 1 7 - 6 ? 1 1 = 1 <-- Check

Tyrone re sells 3 pairs of Yeezys and a pair of Nikes for 250$. Nucci re sells a pair of Yeezys and Nikes for 150$ How much does a pair of Yeezys cost? Let y be the cost of Yeezy's and n be the cost of Nike's. We're given two equations: [LIST=1] [*]3y + n = 250 [*]y + n = 150 [/LIST] We have a system of equations, and we can solve it using one of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3y+%2B+n+%3D+250&term2=y+%2B+n+%3D+150&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3y+%2B+n+%3D+250&term2=y+%2B+n+%3D+150&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=3y+%2B+n+%3D+250&term2=y+%2B+n+%3D+150&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter what method we choose, we get: [LIST] [*][B]n = 100[/B] [*][B]y = 50[/B] [/LIST]