# How do you solve the system of linear equations 5x - 4y = 6 and 2x + 3y = 2?

Dec 6, 2017

$x = \frac{26}{23} \mathmr{and} y = - \frac{2}{23}$

#### Explanation:

$5 x - 4 y = 6$ --------------- Let this be equation (1).

$2 x + 3 y = 2$ --------------Let this be equation (2).

To eliminate one variable, we must make the coefficient of that variable same in both the equations. Say, if we want to eliminate $y$ in order to find $x$,

Multiply equation(1) by 3 and multiply equation(2) by 4, we get the following set:

(1) x 3 $\implies 15 x - 12 y = 18$ -----------(1')and

(2) x 4 $\implies 8 x + 12 y = 8$--------------(2'),

We see that the coefficients of $y$ are equal and opposite in sign, so we add these two equations(1') and(2'), in order to eliminate $y$:

$\implies 15 x + 8 x - \cancel{12 y} + \cancel{12 y} = 18 + 8$

$\implies 23 x = 26$

$\implies x = \frac{26}{23} = 1 \frac{3}{23}$

Substituting value of $x$ in any one equation, say in (1),

$\implies 5 x - 4 y = 6$

$\implies 5 \times \frac{26}{23} - 4 y = 6$

$\implies \frac{130}{23} - 4 y = 6$

$\implies - 4 y = 6 - \frac{130}{23}$

$\implies - 4 y = 6 \left(\frac{23}{23}\right) - \frac{130}{23} = \frac{138 - 130}{23}$

$\implies y = \frac{1}{-} 4 \times \frac{8}{23}$

$\implies y = \frac{1}{-} {\cancel{4}}^{1} \times {\cancel{8}}^{2} / 23$

$\implies y = - \frac{2}{23}$

So we have, $x = \frac{26}{23} \mathmr{and} y = - \frac{2}{23}$

Let us cross check by substituting values of $x$ and $y$ in equation (2),

$2 x + 3 y = 2$

$\implies$ Left hand side =$2 \left(\frac{26}{23}\right) + 3 \left(- \frac{2}{23}\right)$

$= \frac{52}{23} - \frac{6}{23} = \frac{46}{23} = 2 =$Right hand side =$2$
Hence values obtained are correct.