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

Dec 31, 2017

$a = 7$

#### Explanation:

$a \left(a + 3\right) + a \left(a - 6\right) + 35 = a \left(a - 5\right) + a \left(a + 7\right)$

${a}^{2} + 3 a + {a}^{2} - 6 a + 35 = {a}^{2} - 5 a + {a}^{2} + 7 a$ (distribute)

$2 {a}^{2} - 3 a + 35 = 2 {a}^{2} + 2 a$ (combine like terms)

$35 = 5 a$

$a = 7$

Dec 31, 2017

$a = 7$

#### Explanation:

$a \left(a + 3\right) + a \left(a - 6\right) + 35 = a \left(a - 5\right) + a \left(a + 7\right)$

Let's start by using the distributive property

$\left(a\right) \left(a\right) + \left(a\right) \left(3\right) + \left(a\right) \left(a\right) + \left(a\right) \left(- 6\right) + 35 = \left(a\right) \left(a\right) + \left(a\right) \left(- 5\right) + \left(a\right) \left(a\right) + \left(a\right) \left(7\right)$

$2 {a}^{2} - 3 a + 35 = 2 {a}^{2} + 2 a$

Then subtract $\textcolor{red}{2 {a}^{2}}$ from both sides

$\cancel{2 {a}^{2}} - 3 a + 35 - \cancel{\textcolor{red}{2 {a}^{2}}} = \cancel{2 {a}^{2}} + 2 a - \cancel{\textcolor{red}{2 {a}^{2}}}$

$- 3 a + 35 = 2 a$

Subtract $2 a$ from both sides

$- 3 a + 35 - 2 a = 2 a - 2 a$

$- 5 a + 35 = 0$

Subtract $35$ from bot sides

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

$- 5 a = - 35$

Divide both sides by $- 5$

$\frac{- 5 a}{-} 5 = \frac{- 35}{-} 5$

$a = 7$