Question

Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8

Answer 1

If $x$ is the first integer, then the next two are $x+2$ and $x+4$.

If the product of the first and second are greater than the third integer by 8, that means you have the following equation:

$x(x+2)=(x+4)+8\implies x^2+x-12=(x-3)(x+4)=0\implies x=3\text{ or }x=-4$

Ignore $x=-4$ since it's an even integer.

Check that $x=3$ is correct. The next two odd integers would be $5$ and $7$. You have

$3\times5=15=7+8$

so this is correct.

Answer 2

They consists of several equations with unknown. The three numbers will be 3, 5 and 7

System of equations

They consists of several equations with unknown

Let the first three consecutive odd integers be x, x + 2 and x+ 4

If the product of the first and the second exceeds the third by 8, hence;

x(x+2) = x + 4 + 8

x² +2x = x + 12

x² + x - 12 =0

Factorize

x² + 4x -3x - 12 =0

x(x+4) - 3(x+4) = 0

(x-3)(x+4) = 0
x = 3 and -4

The three numbers will be 3, 5 and 7

Learn more on equations here: https://brainly.com/question/13763238

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