# How do you solve the simultaneous equations y = x^2 + 3x and y = 6 - 2x?

Aug 3, 2015

$\textcolor{red}{x = - 6 , y = 18}$ and $\textcolor{red}{x = 1 , y = 4}$

#### Explanation:

One way is to use the method of elimination.

Step 1. Enter the equations.

[1] $y = {x}^{2} + 3 x$
[2] $y = 6 - 2 x$

Step 2. Solve for one of the variables in terms of the other.

[2] $y = 6 - 2 x$

Since this is already done for us, we can go on to the next step.

Step 3. Substitute Equation 2 in Equation 1 and solve for $x$.

$6 - 2 x = {x}^{2} + 3 x$

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

$\left(x + 6\right) \left(x - 1\right) = 0$

$x + 6 = 0$ and $x - 1 = 0$

$x = - 6$ and $x - 1$

Step4. Substitute each value of $x$ in Equation 2

If $x = - 6$,
$y = 6 - 2 \left(- 6\right) = 6 + 12$
$y = 18$

If $x = 1$,
$y = 6 - 2 \left(1\right) = 6 - 2$
$y = 4$

Solutions: $x = - 6 , y = 18$ and $x = 1 , y = 4$

Check: Substitute the values of $x$ and $y$ in Equations 1 and 2.

Check **1**: (-6,18)

$18 = {\left(- 6\right)}^{2} + 3 \left(- 6\right)$
$18 = 36 - 18$
$18 = 18$

$18 = 6 - 2 \left(- 6\right)$
$18 = 6 + 12$
$18 = 18$

It checks!

Check **2**: (1,4)

$4 = {1}^{2} + 3 \left(1\right)$
$4 = 1 + 3$
$4 = 4$

$4 = 6 - 2 \left(1\right)$
$4 = 6 - 2$
$4 = 4$

It checks!

Our solutions are correct.