1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers. Let the first integer be x and the second integer be y. We have the following two equations: [LIST=1] [*]x = 7y [*]xy = 448 [/LIST] Substitute (1) into (2), we have: (7y)y = 448 7y^2 = 448 Divide each side by 7 y^2 = 64 y = -8, 8 We use 8, since 8*7 = 56, and 56*8 =448. So the answer is [B](x, y) = (8, 56)[/B]

10 times the first of 2 consecutive even integers is 8 times the second. Find the integers. Let the first integer be x. Let the second integer be y. We're given: [LIST=1] [*]10x = 8y [*]We also know a consecutive even integer means we add 2 to x to get y. y = x + 2 [/LIST] Substitute (1) into (2): 10x = 8(x + 2) Multiply through: 10x = 8x + 16 To solve for x, [URL='https://www.mathcelebrity.com/1unk.php?num=10x%3D8x%2B16&pl=Solve']we type this equation into our search engine[/URL] and we get: [B]x = 8[/B] Since y = x + 2, we plug in x = 8 to get: y = 8 + 2 [B]y = 10 [/B] Now, let's check our work. Does x = 8 and y = 10 make equation 1 hold? 10(8) ? 8(10) 80 = 80 <-- Yes!

Let x be the first even integer. That means the next consecutive even integer must be x + 2. Set up our equation: x + (x + 2) = 118 Group x terms 2x + 2 = 118 Subtract 2 from each side 2x = 116 Divide each side by 2 x = 58 Which means the next consecutive even integer is 58 + 2 = 60 So our two consecutive even integers are [B]58, 60[/B] Check our work: 58 + 60 = 118

2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]

237 what is the place value of 3 Place value for integers with no decimals from right to left is: 7 is the ones digit 3 is the [B]tens digit[/B]

2 consecutive even integers such that the smaller added to 5 times the larger gives a sum of 70. Let the first, smaller integer be x. And the second larger integer be y. Since they are both even, we have: [LIST=1] [*]x = y - 2 <-- Since they're consecutive even integers [*]x + 5y = 70 <-- Smaller added to 5 times the larger gives a sum of 70 [/LIST] Substitute (1) into (2): (y - 2) + 5y = 70 Group like terms: (1 + 5)y - 2 = 70 6y - 2 = 70 [URL='https://www.mathcelebrity.com/1unk.php?num=6y-2%3D70&pl=Solve']Typing 6y - 2 = 70 into our search engine[/URL], we get: [B]y = 12 <-- Larger integer[/B] Plugging this into Equation (1) we get: x = 12 - 2 [B]x = 10 <-- Smaller Integer[/B] So (x, y) = (10, 12)

3 consecutive odd integers such that thrice the middle is 15 more than the sum of the other 2. [LIST] [*]Let the first integer be n [*]The next odd one (middle) is n + 2. [*]The next odd one is n + 4 [/LIST] We are given 3(n + 2) = n + n + 4 + 15. Simplifying, we get: 3n + 6 = 2n + 19 [URL='http://www.mathcelebrity.com/1unk.php?num=3n%2B6%3D2n%2B19&pl=Solve']Plugging that problem[/URL] into our search engine, we get n = 13. So the next odd integer is 13 + 2 = 15 The next odd integer is 15 + 2 = 17

4 consecutive integers such that the sum of the first 3 integers is equal to the 4th Let n be our first consecutive integer. [LIST=1] [*]n [*]n + 1 [*]n + 2 [*]n + 3 [/LIST] The sum of the first 3 integers is equal to the 4th: n + n + 1 + n + 2 = n + 3 Simplify by grouping like terms: (n + n + n) + (1 + 2) = n + 3 3n + 3 = n + 3 3n = n n = 0 n = 0 n + 1 = 1 n + 2 = 2 n + 3 = 3 Check our work: 0 + 1 +2 ? 3 3 = 3 Our final answer is [B](0, 1, 2, 3}[/B]

A is the set of integers greater than or equal to -5 and less than or equal to -2 [B]-5 <= A <= -2[/B]

A is the set of odd integers between 4 and 12 Let A be the set of odd numbers between 4 and 12: [B]A = {5, 7, 9, 11}[/B]

A set of 4 consecutive integers adds up to 314. What is the least of the 4 integers? First integer is x. The next 3 are x + 1, x + 2, and x + 3. Set up our equation: x + (x + 1) + (x + 2) + (x + 3) = 314 Group x terms and group constnats (x + x + x + x) + (1 + 2 + 3) = 314 Simplify and combine 4x + 6 = 314 [URL='http://www.mathcelebrity.com/1unk.php?num=4x%2B6%3D314&pl=Solve']Enter this in the equation solver[/URL] [B]x = 77[/B]

Ages are consecutive integers. The sum of ages are 111. What are the ages In the search engine, we type [I][URL='http://www.mathcelebrity.com/consecintwp.php?num=111&pl=Sum']sum of 2 consecutive integers is 111[/URL][/I]. We get [B]55 and 56[/B].

are all integers whole numbers true or false [B]False [/B] [LIST] [*]All whole numbers are integers but not all integers are whole numbers. [*]Whole numbers are positive integers. Which means negative integers are not whole numbers [*]-1 for instance is an integer, but not a whole number [/LIST]

Before Barry Bonds, Mark McGwire, and Sammy Sosa, Roger Maris held the record for the most home runs in one season. Just behind Maris was Babe Ruth. The numbers of home runs hit by these two athletes in their record-breaking seasons form consecutive integers. Combined, the two athletes hit 121 home runs. Determine the number of home runs hit by Maris and Ruth in their record-breaking seasons. We want [URL='https://www.mathcelebrity.com/consecintwp.php?num=121&pl=Sum']the sum of 2 consecutive integers equals 121[/URL]. [B]We get Maris at 61 and Ruth at 60[/B]

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[INDENT]Consecutive odd integers are odd integers that differ by ______ , such as ______ and 13. Consecutive even integers are even integers that differ by ______ , such as 12 and ______ . Consecutive odd integers are odd integers that differ by ___2___ , such as ___11___ and 13. Consecutive even integers are even integers that differ by ___2___ , such as 12 and ___14___ .[/INDENT]

Find 3 consecutive integers such that the sum of twice the smallest and 3 times the largest is 126. Let the first integer be n, the second integer be n + 1, and the third integer be n + 2. We have: Sum of the smallest and 3 times the largest is 126: n + 3(n + 2) = 126 Multiply through: n + 3n + 6 = 126 Group like terms: 4n + 6 = 126 [URL='https://www.mathcelebrity.com/1unk.php?num=4n%2B6%3D126&pl=Solve']Type 4n + 6 = 126 into our calculator[/URL], we get n = 30. Which means the next two integers are 31 and 32. [B]{30, 31, 32}[/B]

Find 3 Even Integers with a sum of 198 Let x be the first even integer. Then y is the next, and z is the third even integer. [LIST=1] [*]y = x + 2 [*]z = x + 4 [*]x + y + z = 198 [/LIST] Substituting y and z into (3): x + x + 2 + x + 4 = 198 Group x terms 3x + 6 = 198 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B6%3D198&pl=Solve']equation solver[/URL], we get: [B]x = 64[/B] y = 64 + 2 [B]y= 66[/B] z = 64 + 4 [B]z = 68[/B]

Find the largest of three consecutive even integers when six times the first integers is equal to five times the middle integer. Let the first of the 3 consecutive even integers be n. The second consecutive even integer is n + 2. The third (largest) consecutive even integer is n + 4. We are given 6n = 5(n + 2). Multiply through on the right side, and we get: 6n = 5n + 10 [URL='https://www.mathcelebrity.com/1unk.php?num=6n%3D5n%2B10&pl=Solve']Typing 6n = 5n + 10 into the search engine[/URL], we get n = 10. Remember, n was our smallest of 3 consecutive even integers. So the largest is: n + 4 10 + 4 [B]14[/B]

Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29

Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]

Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25

Four consecutive integers beginning with n consecutive meaning one after another. So we have: [LIST] [*][B]n[/B] [*][B]n + 1[/B] [*][B]n + 2[/B] [*][B]n + 3[/B] [/LIST]

Four cousins were born at two-year intervals. The sum of their ages is 36. What are their ages? So the last cousin is n years old. this means consecutive cousins are n + 2 years older than the next. whether their ages are even or odd, we have the sum of 4 consecutive (odd|even) integers equal to 36. We [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof4consecutiveevenintegersis36&pl=Calculate']type this into our search engine[/URL] and we get the ages of: [B]6, 8, 10, 12[/B]

Yes, $40 for both. This was my son's 8th grade problem. They are learning integers with special cases, so this makes sense, I hope. Thank you so much for responding. I did have this written, but I wasn't certain it was correct.

if a and b are odd then a + b is even Let a and b be positive odd integers of the form: [LIST] [*]a = 2n + 1 [*]b = 2m + 1 [/LIST] a + b = 2n + 1 + 2m + 1 a + b = 2n + 2m + 1 + 1 Combing like terms, we get: a + b = 2n + 2m + 2 a + b = 2(n + m) + 2 Let k = n + m a + b = 2k + 2 [B]Therefore a + b is even[/B]

If a, b, and c are positive integers such that a^b = x and c^b = y, then xy = ? A) ac^b B) ac^2b C) (ac)^b D) (ac)^2b E) (ac)^b^2 xy = a^b * c^b We can use the Power of a Product Rule a^b * c^b = (ac)^b Therefore: xy = [B](ac)^b - Answer C[/B]

If first integer is 5y, then the next two consecutive integers are integers increase by 1, so we have: [B]5y + 1[/B] 5y + 1 + 1 = [B]5y + 2[/B]

If x represents the first, or the smaller, of two consecutive odd integers, express the sum of the two integers in terms of x If x is the first of two consecutive odd integers, then we find the next consecutive odd integer by adding 2 to x: x + 2 The sum of the two consecutive odd integers is expressed by x + (x + 2) Simplify by grouping like terms, we get: [B]2x + 2[/B]

Free Integers Calculator - This lesson walks you through what integers are, how to write integers, integer notation, and what's included in integers

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Let n be the middle number of three consecutive integers This means: [LIST] [*]n is the second of three consecutive integers [*]The first consecutive integer is n - 1 [*]The third consecutive integer is n + 1 [/LIST] The sum is found by: n - 1 + n + n + 1 Simplifying, we get: (n + n + n) + 1 - 1 [B]3n[/B]

Let U be the set of all integers between −3 and 3 (including −3 and 3). Let A={−2,0,1,3}. Find Ac. Give your answer in standard set notation Ac is anything not in A, but in U. So we have: Ac = [B]{-3, -1, 2}[/B]

M is the set of integers that are greater than or equal to -1 and less than or equal to 2 We include -1 on the left, and include 2 on the right [B]M = {-1, 0, 1, 1, 2)[/B]

Free Modulus Calculator - Given 2 integers a and b, this modulo calculator determines a mod b or simplifies modular arithmetic such as 7 mod 3 + 5 mod 8 - 32 mod 5

natural numbers that are factors of 16 Natural numbers are positive integers starting at 1. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Of these, [URL='https://www.mathcelebrity.com/factoriz.php?num=16&pl=Show+Factorization']the only factors of 16[/URL] are: {[B]1, 2, 4, 8, 16}[/B]

What do you want to do with this number set? Express it as representation of integers?

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Prove 0! = 1 Let n be a whole number, where n! represents the product of n and all integers below it through 1. The factorial formula for n is: n! = n · (n - 1) * (n - 2) * ... * 3 * 2 * 1 Written in partially expanded form, n! is: n! = n * (n - 1)! [U]Substitute n = 1 into this expression:[/U] n! = n * (n - 1)! 1! = 1 * (1 - 1)! 1! = 1 * (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! <> 1 which contradicts the equation above

[URL='https://www.mathcelebrity.com/proofs.php?num=prove0%21%3D1&pl=Prove']Prove 0! = 1[/URL] Let n be a whole number, where n! represents: The product of n and all integers below it through 1. The factorial formula for n is n! = n · (n - 1) · (n - 2) · ... · 3 · 2 · 1 Written in partially expanded form, n! is: n! = n · (n - 1)! [SIZE=5][B]Substitute n = 1 into this expression:[/B][/SIZE] n! = n · (n - 1)! 1! = 1 · (1 - 1)! 1! = 1 · (0)! For the expression to be true, 0! [U]must[/U] equal 1. Otherwise, 1! ≠ 1 which contradicts the equation above [MEDIA=youtube]wDgRgfj1cIs[/MEDIA]

Use proof by contradiction. Assume sqrt(2) is rational. This means that sqrt(2) = p/q for some integers p and q, with q <>0. We assume p and q are in lowest terms. Square both side and we get: 2 = p^2/q^2 p^2 = 2q^2 This means p^2 must be an even number which means p is also even since the square of an odd number is odd. So we have p = 2k for some integer k. From this, it follows that: 2q^2 = p^2 = (2k)^2 = 4k^2 2q^2 = 4k^2 q^2 = 2k^2 q^2 is also even, therefore q must be even. So both p and q are even. This contradicts are assumption that p and q were in lowest terms. So sqrt(2) [B]cannot be rational. [MEDIA=youtube]tXoo9-8Ewq8[/MEDIA][/B]

Take two consecutive integers: n, n + 1 The difference of their cubes is: (n + 1)^3 - n^3 n^3 + 3n^2 + 3n + 1 - n^3 Cancel the n^3 3n^2 + 3n + 1 Factor out a 3 from the first 2 terms: 3(n^2 + n) + 1 The first two terms are always divisible by 3 but then the + 1 makes this expression not divisible by 3: 3(n^2 + n) + 1 = 1 (mod 3) [MEDIA=youtube]hFvJ3epqmyE[/MEDIA]

Prove the following statement for non-zero integers a, b, c, If a divides b and b divides c, then a divides c. If an integer a divides an integer b, then we have: b = ax for some non-zero integer x If an integer b divides an integer c, then we have: c = by for some non-zero integer y Since b = ax, we substitute this into c = by for b: c = axy We can write this as: c = a(xy) [LIST] [*]Since x and y are integers, then xy is also an integer. [*]Therefore, c is the product of some integer multiplied by a [*]This means a divides c [/LIST] [MEDIA=youtube]VUIUFAFFVU4[/MEDIA]

Take two integers, r and s. We can write r as a/b for integers a and b since a rational number can be written as a quotient of integers We can write s as c/d for integers c and d since a rational number can be written as a quotient of integers Add r and s: r + s = a/b + c/d With a common denominator bd, we have: r + s = (ad + bc)/bd Because a, b, c, and d are integers, ad + bc is an integer since rational numbers are closed under addition and multiplication. Since b and d are non-zero integers, bd is a non-zero integer. Since we have the quotient of 2 integers, r + s is a rational number. [MEDIA=youtube]0ugZSICt_bQ[/MEDIA]

Take two arbitrary integers, x and y We can express the odd integer x as 2a + 1 for some integer a We can express the odd integer y as 2b + 1 for some integer b x + y = 2a + 1 + 2b + 1 x + y = 2a + 2b + 2 Factor out a 2: x + y = 2(a + b + 1) Since 2 times any integer even or odd is always even, then [B]x + y by definition is even[/B]. [MEDIA=youtube]9A-qe4yZXYw[/MEDIA]

Let us take an integer x which is both even [I]and[/I] odd. [LIST] [*]As an even integer, we write x in the form 2m for some integer m [*]As an odd integer, we write x in the form 2n + 1 for some integer n [/LIST] Since both the even and odd integers are the same number, we set them equal to each other 2m = 2n + 1 Subtract 2n from each side: 2m - 2n = 1 Factor out a 2 on the left side: 2(m - n) = 1 By definition of divisibility, this means that 2 divides 1. But we know that the only two numbers which divide 1 are 1 and -1. Therefore, our original assumption that x was both even and odd must be false. [MEDIA=youtube]SMM9ubEVcLE[/MEDIA]

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Roster form of: A = {3x-2/x are integers between 0 and 8} x = 0 = Undefined since we divide by 0 x = 1: 3*1 + 2/1 = 5 x = 2: 3*2 + 2/2 = 7 x = 3: 3*3 + 2/3 = 9.66666666666667 x = 4: 3*4 + 2/4 = 12.5 x = 5: 3*5 + 2/5 = 15.4 x = 6: 3*6 + 2/6 = 18.3333333333333 x = 7: 3*7 + 2/7 = 21.2857142857143 x = 8: 3*8 + 2/8 = 24.25 [B]A = {(0, undefined), (1, 5), (2, 7), (3, 9.6667), (4, 12.5), (5, 15.4), (6, 18.3333), (7, 21.2857142857143), (8, 24.25)}[/B]

Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]

Storm Center 7 was tracking a cold front approaching Cedarburg. Before it rolled in, the temperature was – 7°F. Then, the temperature decreased by 9°F. What was the temperature after the cold front rolled in? Using signed integers, we start with 7 below or -7 -7 The temperature decreased by 9 which means we subtract: -7 - 9 or -7 + (-9) [B]-16°F or 16 below 0 [MEDIA=youtube]oJjEhkdnTxA[/MEDIA][/B]

sum of 3 consecutive odd integers equals 1 hundred 17 The sum of 3 consecutive odd numbers equals 117. What are the 3 odd numbers? 1) Set up an equation where our [I]odd numbers[/I] are n, n + 2, n + 4 2) We increment by 2 for each number since we have [I]odd numbers[/I]. 3) We set this sum of consecutive [I]odd numbers[/I] equal to 117 n + (n + 2) + (n + 4) = 117 [SIZE=5][B]Simplify this equation by grouping variables and constants together:[/B][/SIZE] (n + n + n) + 2 + 4 = 117 3n + 6 = 117 [SIZE=5][B]Subtract 6 from each side to isolate 3n:[/B][/SIZE] 3n + 6 - 6 = 117 - 6 [SIZE=5][B]Cancel the 6 on the left side and we get:[/B][/SIZE] 3n + [S]6[/S] - [S]6[/S] = 117 - 6 3n = 111 [SIZE=5][B]Divide each side of the equation by 3 to isolate n:[/B][/SIZE] 3n/3 = 111/3 [SIZE=5][B]Cancel the 3 on the left side:[/B][/SIZE] [S]3[/S]n/[S]3 [/S]= 111/3 n = 37 Call this n1, so we find our other 2 numbers n2 = n1 + 2 n2 = 37 + 2 n2 = 39 n3 = n2 + 2 n3 = 39 + 2 n3 = 41 [SIZE=5][B]List out the 3 consecutive odd numbers[/B][/SIZE] ([B]37, 39, 41[/B]) 37 ← 1st number, or the Smallest, Minimum, Least Value 39 ← 2nd number 41 ← 3rd or the Largest, Maximum, Highest Value

Free Sum of Consecutive Numbers Calculator - Finds the sum of (n) consecutive integers, even or odd as well. Examples include:
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Free Sum of Three Consecutive Integers Calculator - Finds three consecutive integers, if applicable, who have a sum equal to a number. Sum of 3 consecutive integers

Sum of two consecutive numbers is always odd Definition: [LIST] [*]A number which can be written in the form of 2 m where m is an integer, is called an even integer. [*]A number which can be written in the form of 2 m + 1 where m is an integer, is called an odd integer. [/LIST] Take two consecutive integers, one even, and one odd: 2n and 2n + 1 Now add them 2n + (2n+ 1) = 4n + 1 = 2(2 n) + 1 The sum is of the form 2n + 1 (2n is an integer because the product of two integers is an integer) Therefore, the sum of two consecutive integers is an odd number.

The ages of three siblings are all consecutive integers. The sum of of their ages is 39. Let the age of the youngest sibling be n. This means the second sibling is n + 1. This means the oldest/third sibling is n + 2. So what we want is the[URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutiveintegersequalto39&pl=Calculate'] sum of 3 consecutive integers equal to 39[/URL]. We type this command into our search engine. We get: n = 12. So the youngest sibling is [B]12[/B]. The next sibling is 12 + 1 = [B]13[/B] The oldest/third sibling is 12 + 2 = [B]14[/B]

The domain of a relation is all even negative integers greater than -9. The range y of the relation is the set formed by adding 4 to the numbers in the domain. Write the relation as a table of values and as an equation. The domain is even negative integers greater than -9: {-8, -6, -4, -2} Add 4 to each x for the range: {-8 + 4 = -4, -6 + 4 = -2. -4 + 4 = 0, -2 + 4 = 2} For ordered pairs, we have: (-8, -4) (-6, -2) (-4, 0) (-2, 2) The equation can be written: y = x + 4 on the domain (x | x is an even number where -8 <= x <= -2)

the left and right page numbers of an open book are two consecutive integers whose number is 235 find the page numbers Using our [URL='https://www.mathcelebrity.com/consecintwp.php?pl=Sum&num=+235']consecutive integer calculator[/URL], we get: [B]117, 118[/B]

The left and right page numbers of an open book are two consecutive integers whose sum is 403. Find these page numbers. Page numbers left and right are consecutive integers. So we want to find a number n and n + 1 where: n + n + 1 = 403 Combining like terms, we get: 2n + 1 = 403 Typing that equation into our search engine, we get: [B]n = 201[/B] This is our left hand page. Our right hand page is: 201 + 1 = [B]202[/B]

The number -2.34 can be found between which two integers We want to take the integer of -2.34 which is -2 Since -2.34 is less than 0, we subtract 1: -2 -1 = -3 Therefore, -2.34 lies between [B]-2 and -3[/B] -3 < -2.34 < -2

The product of two consecutive integers is greater than 100 Take an integer x. Next consecutive integer is x + 1 The product of those integers is: x(x + 1) This product is greater than 100 which gives us the algebraic expression of: x(x + 1) > 100 IF we want to solve for x: x^2 + x > 100 Subtract 100 from each side: x^2 + x - 100 > 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-100%3E0&pl=All&hintnum=+0']Solve this quadratic:[/URL]

The square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/B]

The sum of 3 consecutive integers is greater than 30. Let the first consecutive integer be n The second consecutive integer is n + 1 The third consecutive integer is n + 2 The sum is written as: n + n + 1 + n + 2 Combine like terms: (n + n + n) + (1 + 2) 3n + 3 The phrase [I]greater than[/I] means an inequality, which we write as: [B]3n + 3 > 30[/B]

The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].

Let the 3 integers be x, y, and z. y = x + 1 z = y + 1, or x + 2. Set up our equation: x + (x + 1) + (x + 2) = 42 Group our variables and constants: (x + x + x) + (1 + 2) = 42 3x + 3 = 42 Subtract 3 from each side: 3x = 39 Divide each side of the equation by 3: [B]x = 13 So y = x + 1 = 14 z = x + 2 = 15 (x,y,z) = (13,14,15)[/B]

The sum of two consecutive integers if n is the first integer. consecutive means immediately after, so we have: n n + 1 [U]The sum is written as:[/U] n + n + 1 [U]Grouping like terms, we have:[/U] (n + n) + 1 [B]2n + 1[/B]

The sum of two consecutive integers plus 18 is 123. Let our first integer be n and our next integer be n + 1. We have: n + (n + 1) + 18 = 123 Group like terms to get our algebraic expression: 2n + 19 = 123 If we want to solve the algebraic expression using our [URL='http://www.mathcelebrity.com/1unk.php?num=2n%2B19%3D123&pl=Solve']equation solver[/URL], we get n = 52. This means the next integer is 52 + 1 = 53

There are 2 consecutive integers. Twice the first increased by the second yields 16. What are the numbers? Let x be the first integer. y = x + 1 is the next integer. We have the following givens: [LIST=1] [*]2x + y = 16 [*]y = x + 1 [/LIST] Substitute (2) into (1) 2x + (x + 1) = 16 Combine x terms 3x + 1 = 16 Subtract 1 from each side 3x = 15 Divide each side by 3 [B]x = 5[/B] So the other integer is 5 + 1 = [B]6[/B]

Three good friends are in the same algebra class, their scores on a recent test are three consecutive odd integers whose sum is 273. Find the score In our search engine, we type in [URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=3consecutiveintegerswhosesumis273&pl=Calculate']3 consecutive integers whose sum is 273[/URL] and we get: [B]90, 91, 92[/B]

To make an international telephone call, you need the code for the country you are calling. The code for country A, country B, and C are three consecutive integers whose sum is 90. Find the code for each country. If they are three consecutive integers, then we have: [LIST=1] [*]B = A + 1 [*]C = B + 1, which means C = A + 2 [*]A + B + C = 90 [/LIST] Substitute (1) and (2) into (3) A + (A + 1) + (A + 2) = 90 Combine like terms 3A + 3 = 90 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3a%2B3%3D90&pl=Solve']equation calculator[/URL], we get: [B]A = 29[/B] Which means: [LIST] [*]B = A + 1 [*]B = 29 + 1 [*][B]B = 30[/B] [*]C = A + 2 [*]C = 29 + 2 [*][B]C = 31[/B] [/LIST] So we have [B](A, B, C) = (29, 30, 31)[/B]

Two consecutive even integers that equal 126 Let the first integer equal x. So the next even integer must be x + 2. The sum which is equal to 126 is written as x + (x + 2) = 126 Simplify: 2x + 2 = 126 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2x%2B2%3D126&pl=Solve']equation calculator,[/URL] we get: x = 62 This means the next consecutive even integer is 62 = 2 = 64. So our two even consecutive integers with a sum of 126 are [B](62, 64)[/B]

What pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the larger? Let x and y be consecutive integers, where y = x + 1 We have 7x < 6y as our inequality. Substituting x, y = x + 1, we have: 7x < 6(x + 1) 7x < 6x + 6 Subtracting x from each side, we have: x < 6, so y = 6 + 1 = 7 (x, y) = (6, 7)

When 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum? Let the 3 consecutive positive integers be: [LIST=1] [*]x [*]x + 1 [*]x + 2 [/LIST] The product is: x(x + 1)(x + 2) The sum is: x + x + 1 + x + 2 = 3x + 3 We're told the product is equivalent to: x(x + 1)(x + 2) = 16(3x + 3) x(x + 1)(x + 2) = 16 * 3(x + 1) Divide each side by (x + 1) x(x + 2) = 48 x^2 + 2x = 48 x^2 + 2x - 48 = 0 Now subtract the sum from the product: x^2 + 2x - 48 - (3x + 3) [B]x^2 - x - 51[/B]

{x | x is an even integer between -3 and 5} We list even integers out in this range: [B]{-2, 0, 2, 4}[/B]