# How do you solve this system of equations: -\frac { 1} { 2} x + \frac { 1} { 5} y = 9 and 7x - \frac { 1} { 3} y = - \frac { 2} { 3}?

Feb 26, 2018

$x = \frac{86}{37} \mathmr{and} y = \frac{1880}{37}$

#### Explanation:

$- \setminus \frac{1}{2} x + \setminus \frac{1}{5} y = 9$----------(1)

and

$7 x - \setminus \frac{1}{3} y = - \setminus \frac{2}{3}$------------(2)

Multiplying equation (1) by 10 and equation (2) by 3 to eliminate the fractional part:

$- 5 x + 2 y = 90$--------------(1') and

$21 x - 1 y = - 2$ -----------(2')

Eliminate any one variable (say $y$):

(1') + (2') x 2 $\implies$

$- 5 x + \left(2 \times 21 x\right) + 2 y + \left(2 \times - 1 y\right) = 90 + \left(2 \times - 2\right)$

$\implies - 5 x + 42 x + 2 y - 2 y = 90 - 4$

$\implies 37 x = 86$

$\implies x = \frac{86}{37} = 2 \frac{12}{37}$

Substitute this value of $x$ in (1') or (2') to get the value of $y$

(2')$\implies 21 \times \left(\frac{86}{37}\right) - 1 y = - 2$

$\implies \frac{1806}{37} - y = - 2$

$\implies - y = - 2 - \frac{1806}{37}$

$- y = \frac{- 74 - 1806}{37}$

$y = \frac{1880}{37}$

Cross check the validity of $x \mathmr{and} y$ by substituting in left hand side (LHS)of any of (1) and (2):

LHS of (1)$\implies$

$- \setminus \frac{1}{2} x + \setminus \frac{1}{5} y = - \frac{1}{2} \left(\frac{86}{37}\right) + \frac{1}{5} \times \left(\frac{1880}{37}\right)$

$= - \frac{43}{37} + \frac{376}{37} = \frac{333}{37}$

$= 9$ = RHS (right hand side ) of equation (1)

$\therefore x = \frac{86}{37} \mathmr{and} y = \frac{1880}{37}$ are the correct values.