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.)
+2668 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 \(\color{brown}\boxed{-2,0}\)
+2668 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 \(\color{brown}\boxed{-2,0}\)
