Find three consecutive integers such that the product of the first two is 10 less than the product of the last two.

4

5

6

Step-by-step explanation:

Since we don't know the three consecutive numbers, let's call them x,x+1,and x+2.

Product of first two means x(x+1)=x^2+x

Product of last two means

(x+1)(x+2)=x^2+2x+1x+2=x^2+3x+2

"product of the first two is 10 less than the product of the last two"

So we have the equation x^2+x=x^2+3x+2-10

Simplify on right side: x^2+x=x^2+3x-8

Subtract x^2 on both sides: x=3x-8

Subtract 3x on both sides: -2x=-8

Divide both sides by -2: x=4

x=4

x+1=4+1=5

x+2=4+2=6

Check:

The product of first two is 10 less than the product of last two.

4(5)=20

5(6)=30

It's true that 30-10=20.

the solution is 12x-6y

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