One integer is nine more than two times another integer. If the product of the integers is 18, how do you find the two integers?

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One integer is nine more than two times another integer. If the product of the integers is 18, how do you find the two integers?

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Answer 1 · 2018-05-19T00:03:36

Solutions integers: $- 3, - 6$ $color(blue)(-3,-6)$

Explanation:

Let the integers be represented by $a$ $a$ and $b$ $b$ .
We are told:
[1] $XXX a = 2 b + 9$ $color(white)("XXX")a=2b+9$ (One integer is nine more than two time the other integer)
and
[2] $XXX a \times b = 18$ $color(white)("XXX")a xx b = 18$ (The product of the integers is 18)

Based on [1], we know we can substitute $(2 b + 9)$ $(2b+9)$ for $a$ $a$ in [2];
giving
[3] $XXX (2 b + 9) \times b = 18$ $color(white)("XXX")(2b+9) xx b=18$

Simplifying with the target of writing this as a standard form quadratic:
[5] $XXX 2 b^{2} + 9 b = 18$ $color(white)("XXX")2b^2+9b=18$

[6] $XXX 2 b^{2} + 9 b - 18 = 0$ $color(white)("XXX")2b^2+9b-18=0$

You could use the quadratic formula to solve for $b$ $b$ or recognize the factoring:
[7] $XXX (2 b - 3) (b + 6) = 0$ $color(white)("XXX")(2b-3)(b+6)=0$
giving solutions:
$XXX b = \frac{3}{2}$ $color(white)("XXX")b=3/2$ which is not permitted since we are told the values are integers.
or
$XXX b = - 6$ $color(white)("XXX")b=-6$

If $b = - 6$ $b=-6$ then based on [1]
$XXX a = - 3$ $color(white)("XXX")a=-3$

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One integer is nine more than two times another integer. If the product of the integers is 18, how do you find the two integers?

1 Answer

Explanation:

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