# How do you solve the following system?: 8x +6y =9 , - 5x -7y = -2

May 3, 2017

With some arrangement. $x = \frac{51}{26}$ and $y = - \frac{29}{26}$

#### Explanation:

Expand the first equation with term 5
Expand the second equation with 8.

Now you have:

$40 x + 30 y = 45$
$- 40 x - 56 y = - 16$
Now sum these up:
$- 26 y = 29$

or $y = - \frac{29}{26}$

Now you can find x using the first or second equation:

$8 x - \frac{6 \cdot 29}{26} = 9$

$8 x = 9 + \frac{6 \cdot 29}{26}$

$8 x = \left(\frac{117}{13}\right) + \left(\frac{3 \cdot 29}{13}\right)$

$8 x = \frac{204}{13}$

$x = \frac{204}{104}$

or

$x = \frac{51}{26}$

$x = 1 \frac{25}{26}$, $y = - 1 \frac{21}{182}$

#### Explanation:

$8 x + 6 y = 9$ ....................(i)
$- 5 x - 7 y = - 2$ ....................(ii)

You can solve this system by using Elimination method.

You can eliminate either $x$ or $y$ here.
I will eliminate $x$.

So multiplying eq.(i) by $+ 5$ and eq(ii) by $+ 8$, you will get

$40 x + 30 y = 45$ ......................(iii) &
$- 40 x - 56 y = - 16$ ........................(iv)

Adding eq (iii) and (iv), you will get

$- 26 y = 29$

$\Rightarrow$ $y = - \frac{29}{26}$

Substituting $y = - \frac{29}{26}$ in eq(i), you will get

$8 x + \left(- \frac{29}{26}\right) \cdot 6 = 9$

$\Rightarrow$ $8 x = 9 + \frac{87}{13}$

$\Rightarrow$ $8 x = \frac{117 + 87}{13}$

$\Rightarrow$ $8 x = \frac{204}{13}$

$\Rightarrow$ $x = \frac{204}{104} = \frac{51}{26}$