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Find two consecutive even integers such that 5 times their sum is 10 less than their product. (Enter your answers as a comma-separated list.)

 Apr 14, 2022

Best Answer 

 #1
avatar+1384 
+1

Let the smaller number be \(x\).

 

We have: \(5(x + x+2) = x(x+2) - 10\)

 

Simplifying, we have: \(10x+10 = x^2+2x-10\)

 

Converting the equation to quadratic form: \(x^2-8x-20=0\)

 

Factoring the equation: \((x+2)(x-10)=0\)

 

There are 2 values of \(x\) that work, 10 or -2, so there can be two sets: \(\color{brown}\boxed{10,12}\) or  \(\color{brown}\boxed{-2,0}\)

 Apr 14, 2022
 #1
avatar+1384 
+1
Best Answer

Let the smaller number be \(x\).

 

We have: \(5(x + x+2) = x(x+2) - 10\)

 

Simplifying, we have: \(10x+10 = x^2+2x-10\)

 

Converting the equation to quadratic form: \(x^2-8x-20=0\)

 

Factoring the equation: \((x+2)(x-10)=0\)

 

There are 2 values of \(x\) that work, 10 or -2, so there can be two sets: \(\color{brown}\boxed{10,12}\) or  \(\color{brown}\boxed{-2,0}\)

BuilderBoi Apr 14, 2022
 #2
avatar+122 
0

cerrectoooo

😎😎😎

 Apr 14, 2022

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