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Find 2 consecutive integers whose product if 5 less than 5 times their sum

Consecutive integers are two numbers, which have a common difference of 1.

So, the problem states that:

n(n+1)=5(n+(n+1))-10, where n is an integer.

Here's the answer.

n(n+1)=5(n+n+1)-10 --> Write the problem.

n(n+1)=5(2n+1)-10 --> Add n and n.

n(n+1)=10n+5-10 --> Distribute 5 into 2n+1.

n(n+1)=10n-5 --> Add 5 and -10.

n^2+n=10n-5 --> Distribute n into n+1.

n^2-9n+5=0 Subtract 10n-5 on both sides of the equation.

Ok. Now here n^2-9n+5=0 cannot factor into integers. Since it can not factor into integers, it is highly unlikely that they would be consecutive integers.

We can use the quadratic formula to solve the equation. The answers are n=(9±sqrt(61))/2.

Marley K's answer gave you a right idea, but a wrong quadratic.

If you're talking about even integers, then we can work out a similar process.

n(n+2)=5(n+n+2)-10 --> Write the problem.

n(n+2)=5(2n+2)-10 --> Add n and n.

n(n+2)=10n+10-10 --> Distribute 5 into 2n+2.

n(n+2)=10n-0 --> Add 10 and -10.

n(n+2)=10n --> Add 10n and 0.

n^2+2n=10n --> Distribute n into n+2.

n^2-8n=0 --> Subtract 10n on both sides of the equation.

n(n-8)=0 --> Factor out the n.

n=0 or n-8=0 --> Use the property if pq=0, then p=0 or q=0.

n=0 or n=8 Solve each equation for n.

Now we have a pair of consecutive even integers. (0,2) and (8,10).

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