Question

Solve the system of liner equations by substitution
6x-9=y
y= -3x

Answer 1

Hello there!


`\begin{bmatrix}6x-9=y\\ y=-3x\end{bmatrix}`


`\mathrm{Subsititute\:}y=-3x\ \textgreater \ \begin{bmatrix}6x-9=\left(-3x\right)\end{bmatrix}`


`Isolate\;x\;for\;6x-9=\left(-3x\right)`

Simply expand (-3x) by removing the parenthesis > -3x


`6x-9=-3x`


`\mathrm{Add\:}9\mathrm{\:to\:both\:sides} \ \textgreater \ 6x-9+9=-3x+9`

Simplify!

`6x=-3x+9`


`\mathrm{Add\:}3x\mathrm{\:to\:both\:sides} \ \textgreater \ 6x+3x=-3x+9+3x`

Simplify again!

`9x = 9`


`\mathrm{Divide\:both\:sides\:by\:}9 \ \textgreater \ (9x)/(9)=(9)/(9)`

Simplify again!

`x = 1`


`\mathrm{For\:}y=-3x,\mathrm{subsititute\:}x=1 \ \textgreater \ y=-3\cdot \:1\quad \Rightarrow \quad y=-3`

Therefore our solutions are...

`x=1,\:y=-3`

Hope this helps!
Done by substitution as requested.