# How do you solve 5(4z+6)-2(z-4)=7z(z+4)-z(7z-2)-48?

Dec 18, 2016

$z = \frac{43}{6}$

#### Explanation:

First, expand the terms within parenthesis paying close attention to the terms inside and outside the parenthesis:

$\left(5 \cdot 4 z\right) + \left(5 \cdot 6\right) - \left(2 \cdot z\right) + \left(2 \cdot 4\right) = \left(7 z \cdot z\right) + \left(7 z \cdot 4\right) - \left(z \cdot 7 z\right) + \left(z \cdot 2\right) - 48$

$20 z + 30 - 2 z + 8 = 7 {z}^{2} + 28 z - 7 {z}^{2} + 2 z - 48$

Now we can group and combine like terms on each side of the equation:

$20 z - 2 z + 30 + 8 = 7 {z}^{2} - 7 {z}^{2} + 28 z + 2 z - 48$

$\left(20 - 2\right) z + 38 = \left(7 - 7\right) {z}^{2} + \left(28 + 2\right) z - 48$

$18 z + 38 = 0 {z}^{2} + 30 z - 48$

$18 z + 38 = 30 z - 48$

Next we can isolate the $z$ terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

$18 z + 38 \textcolor{red}{- 18 z + 48} = 30 z - 48 \textcolor{red}{- 18 z + 48}$

$18 z - 18 z + 38 + 48 = 30 z - 18 z - 48 + 48$

$0 + 38 + 48 = \left(30 - 18\right) z - 0$

$86 = 12 z$

Finally, we can solve for $z$ while keeping the equation balanced:

$\frac{86}{\textcolor{red}{12}} = \frac{12 z}{\textcolor{red}{12}}$

$\frac{2}{2} \cdot \frac{43}{6} = z$

$\frac{43}{6} = z$