Question

How do you solve `a(a+3)+a(a-6)+35=a(a-5)+a(a+7)`?

Answer 1

`a=7`

Explanation:

`a(a+3)+a(a-6)+35=a(a-5)+a(a+7)`

`a^2+3a+a^2-6a+35=a^2-5a+a^2+7a` (distribute)

`2a^2-3a+35=2a^2+2a` (combine like terms)

`35=5a`

`a=7`

Answer 2

`a=7`

Explanation:

`a(a+3)+a(a-6)+35=a(a-5)+a(a+7)`

Let's start by using the distributive property

`(a)(a) + (a)(3) + (a)(a)+(a)(-6)+35 = (a)(a)+(a)(-5)+(a)(a)+(a)(7)`

`2a^2 - 3a + 35 =2a^2+2a`

Then subtract `color(red)(2a^2)` from both sides

`cancel(2a^2) - 3a + 35 - cancelcolor(red)(2a^2)= cancel(2a^2) +2a - cancelcolor(red)(2a^2)`

`-3a + 35 = 2a`

Subtract `2a` from both sides

`-3a + 35 - 2a = 2a - 2a`

`-5a + 35 = 0`

Subtract `35` from bot sides

#-5a + 35 - 35 = 0 - 35

`-5a = -35`

Divide both sides by `-5`

`(-5a)/-5 = (-35)/-5`

`a = 7`