How do you solve the following system?: 23x +31y =1, -7x +5y = -2

Aug 15, 2017

$x = \frac{67}{332} \mathmr{and} y = - \frac{39}{332}$

Explanation:

The given equations are really uncomfortable and do not lend themselves to either substitution or elimination without using big numbers.

Let's equate one of the variables: solve for $y$ in each.

$23 x + 31 y = 1 \textcolor{w h i t e}{\times \times x} \mathmr{and} - 7 x + 5 y = - 2$

$31 y = 1 - 23 x \textcolor{w h i t e}{\times \times \times \times \times \times \times} 5 y = 7 x - 2$

$y = \frac{1 - 23 x}{31} \textcolor{w h i t e}{\times \times \times \times \times \times \times \times} y = \frac{7 x - 2}{5}$

Now as $y = y$, we can equate the right sides of each equation:

$\frac{1 - 23 x}{31} = \frac{7 x - 2}{5} \text{ } \leftarrow$ cross multiply

$31 \left(7 x - 2\right) = 5 \left(1 - 23 x\right)$

$217 x - 62 = 5 - 115 x$

$217 x + 115 x = 5 + 62$

$332 x = 67$

$x = \frac{67}{332}$

Now substitute this value for $x$ into either equation above.

$y = \frac{7 \left(\frac{67}{332}\right) - 2}{5}$

$y = \frac{\frac{469}{332} - 2}{5}$

$y = - \frac{195}{332} \div 5$

$y = - {\cancel{195}}^{39} / 332 \times \frac{1}{\cancel{5}}$

$y = - \frac{39}{332}$

Check by substituting the values for $x \mathmr{and} y$ into the the other equation.

$23 x + 31 y$

$23 \left(\frac{67}{332}\right) + 31 \left(- \frac{39}{332}\right)$

$= 1$

The equation checks out.