# How do you solve the system of equations 8x + 4y = 84 and 9x - 9y = 54?

$x = 9 , \setminus \setminus y = 3$

#### Explanation:

Given equations

$8 x + 4 y = 84$

$2 x + y = 21 \setminus \ldots \ldots \ldots . \left(1\right)$ &

$9 x - 9 y = 54$

$x - y = 6 \setminus \ldots \ldots \ldots . \left(2\right)$

Adding (1) & (2), we get

$2 x + y + x - y = 21 + 6$

$3 x = 27$

$x = \frac{27}{3}$

$x = 9$

setting $x = 9$ in (1), we get

$2 \left(9\right) + y = 21$

$y = 21 - 18$

$y = 3$

hence the solution of given equations is

$x = 9 , \setminus \setminus y = 3$

Jul 24, 2018

Let's solve the first equation first

$8 x + 4 y = 84$

All you have to do is take out the value of a variable

Let's take out the value of $y$ both $x$ and $y$ will be irritating in the next equation but $y$ will be less irritating (maybe)

Factorize out 4 and divide

$4 \left(2 x + y\right) = 84$

$2 x + y = 21$

color(red)(y=21-2x

Now to the second equation

$9 x - 9 y = 54$

Factorize out $9$

$9 \left(x - y\right) = 54$

$x - y = 6$

Put value of $y$

$x - \left(21 - 2 x\right) = 6$

Notice that after opening the brackets the values will go from plus to minus and from minus to plus

$x - 21 + 2 x = 6$

$3 x - 21 = 6$

$3 x = 6 + 21$

$3 x = 27$

$x = \frac{27}{3}$

color(darkorange)(x=9

$y = 21 - 2 x$

$y = 21 - 2 \times 9$

$y = 21 - 18$

color(darkorange)(y=3

Jul 24, 2018

$x = 9$ and $y = 3$

#### Explanation:

$9 \cdot \left(8 x + 4 y\right) + 4 \cdot \left(9 x - 9 y\right) = 9 \cdot 84 + 4 \cdot 54$

$72 x + 36 y + 36 x - 36 y = 756 + 216$

$108 x = 972$, so $x = \frac{972}{108} = 9$

Thus,

$9 \cdot 9 - 9 y = 54$

$81 - 9 y = 54$

$- 9 y = - 27$

$y = \frac{- 27}{- 9} = 3$