# What are two consecutive integers whose product is 5 less than 5 times their sum. Domain for the smallest: {0, 6, 9}

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It looks like you've been given three possible numbers to try? I'll show you how to do this using algebra, and then I'll check my answers. That way, if you're just expected to test these three numbers to see which work, my check will be all you need and you can skip right to that.

We don't know what the integers are, so call the smaller one `x`. Since these are *consecutive *integers, the other one has to be `x+1`, one more than `x`. Product means multiply, and the product of these two integers is just `x(x+1)`, or `x^2+x` if you distribute.

Their sum is just `x+(x+1)`, or `2x+1` if you combine like terms. Five times their sum is `5(2x+1)=10x+5`. Five less than five times their sum is then `(10x+5)-5=10x`.

So we want to find some `x` such that `x^2+x=10x.` If we collect the variables on one side, we get `x^2-9x=0.` Now factor to get

`x(x-9)=0.` The only way to get a product to equal zero is if at least one factor is zero, so the solutions to this equation are `x=0` and `x-9=0`, in other words `x=9.`

This doesn't mean that 0 and 9 are the answers to the original problem though. We still need to check them.

**CHECK: **If `x=0` , then the next consecutive integer is 1. The product of 0 and 1 is zero. Five times the sum of 0 and 1 is 5, and 5 less than 5 is zero. The answers match, so the pair (0,1) is a solution.

If x=9, then the next integer is 10. The product of 9 and 10 is 90. The sum of 9 and 10 is 19, five times this sum is 95, and 5 less than this is 90. Again the answers match up and the pair (9,10) is another solution.

Just for completeness, (6)(7)=42 and 5(6+7)-5=60, so (6,7) isn't a solution.