Question

Solve each system by elimination by multiplying both equations. eliminate y first. show all work.

10x-3y=18
6x-10y=-22

Answer 1

First
`\mathrm{Multiply\:}10x-3y=18\mathrm{\:by\:}3:\quad 30x-9y=54`
Then
`\mathrm{Multiply\:}6x-10y=-22\mathrm{\:by\:}5:\quad 30x-50y=-110`
To get
`\begin{bmatrix}30x-9y=54\\ 30x-50y=-110\end{bmatrix}`
So...

`30x-50y=-110 \ \textgreater \ \begin{bmatrix}30x-9y=54\\ -41y=-164\end{bmatrix}`

`-41y=-164 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-41 \ \textgreater \ (-41y)/(-41)=(-164)/(-41)`
To get y = 4.

`\mathrm{For\:}30x-9y=54\mathrm{\:plug\:in\:} \:y=4 \ \textgreater \ 30x-9\cdot \:4=54`

`\mathrm{Multiply\:the\:numbers:}\:9\cdot \:4=36 \ \textgreater \ 30x-36=54 \ \textgreater \`
Next
`\mathrm{Add\:}36\mathrm{\:to\:both\:sides} \ \textgreater \ 30x-36+36=54+36 \ \textgreater \ \mathrm{Simplify} \ \textgreater \ 30x=90`
Finally
`\mathrm{Divide\:both\:sides\:by\:}30 \ \textgreater \ (30x)/(30)=(90)/(30) \ \textgreater \ x = 3`
Therefore our solutions are y = 4, x = 3