How do you solve the system of equations y = .5x + 20 and y = .15x?

Jul 1, 2016

Use substitution to get a solution of $\left(- \frac{400}{7} , - \frac{60}{7}\right)$.

Explanation:

We know that $y = y$, so that must also mean, if $y = .5 x + 20$ and $y = .15 x$:
$.5 x + 20 = .15 x$

This is now a linear equation:
$.5 x + 20 = .15 x$
$\cancel{.5 x} + 20 - \cancel{.5 x} = .15 x - .5 x$
$20 = - .35 x$
$x = \frac{20}{-} .35 = \frac{20}{-} \left(\frac{35}{100}\right)$
$= {\cancel{20}}^{4} \cdot - \frac{100}{\cancel{35}} ^ 7$
$= - \frac{400}{7} \approx - 57.14$

Since $y = .15 x$:
$y = .15 \cdot - \frac{400}{7}$
$= \frac{15}{\cancel{100}} ^ 1 \cdot - {\cancel{400}}^{4} / 7$
$= - \frac{60}{7} \approx - 8.57$

Thus, the ordered pair that is the solution to this equation is $\left(- \frac{400}{7} , - \frac{60}{7}\right)$.