# How do you solve x - y = 4 and y = x² - 34 using substitution?

Mar 29, 2016

Since one of the variables is already isolated in one of the equations (y; equation II), we can immediately substitute.

#### Explanation:

$x - \left({x}^{2} - 34\right) = 4$

$x - {x}^{2} + 34 = 4$

$- {x}^{2} + x + 30 = 0$

You can solve this by factoring. With trinomials $y = a {x}^{2} + b x + c , a \ne 1$, you must find two numbers that multiply to $a \times c$ and that add to b. Two numbers that multiply to -30 and that add to 1 are +6 and -5.

$- {x}^{2} + 6 x - 5 x + 30 = 0$

$- x \left(x - 6\right) - 5 \left(x - 6\right) = 0$

$\left(- x - 5\right) \left(x - 6\right) = 0$

$x = - 5 \mathmr{and} 6$

Substituting both solutions for x in the original equation we get:

$y = {\left(- 5\right)}^{2} - 34 \mathmr{and} y = {\left(6\right)}^{2} - 34$

$y = 25 - 34 \mathmr{and} y = 36 - 34$

$y = - 9 \mathmr{and} y = 2$

Thus, the solution set is $\left\{- 5 , - 9\right\} \mathmr{and} \left\{6 , 2\right\}$. Always make to put the right x value with the right y value and to place them in the brackets in the form $\left\{x , y\right\}$.

Practice exercises:

1. Solve the following systems by substitution. Leave answers in exact form when necessary.

a) ${x}^{2} + 4 = y , 2 x + 3 y = - 8$

b) ${x}^{2} + 2 y - 5 = 0 , 3 y - 5 x = 6$

Good luck!