Question

Anyone know if they can help me rq

Answer 1

Explanation:

`\begin{cases} 15x+31y=-3 \\\\ x=-y+3 \ \ | * 15\end{cases} \Leftrightarrow \ominus\begin{cases} 15x+31y=-3 \\\\ 15x=-15y+45 \ \ \end{cases} \Leftrightarrow \\\\\\ 15x-15x+31y=-3-(-15y)-45 \\\\ 31y=15y-48 \\\\ 31y-15y=-48 \\\\ 16y=-48 \ \ |/ 16 \\\\ y=-3 \ \ ; \ \ x=-y+3=3+3=6 \\\\`

`\huge \boldsymbol{\mathfrak {Unswer}}: x=6 \ \ ; \ \ y=-3`

Answer 2

We are given two equations, one of which has an isolated variable
`x`.

That screams to me that substitution would be a prefered strategy here, compared to elimination, although both work.

That means we'll be substituting our value of
`x`, which is given as
`-y+3`, into the first equation,
`15x+31y=-3`.

`15x+31y=-3`

`x=-y+3`

`15(-y+3)+31y=-3`

`-15y+45+31y=-3`

`16y=-48`

`y=-3`

With this value, we can plug it back into either of the two equations to solve for
`x`, I'll be substituting it back into the second equation, since it's easier.

`x=-y+3`

`y=-3`

`x=-(-3)+3`

`x=6`

So our solution is
`(6,-3)`, and to check we can plug it back into the first equation.

`15x+31y=-3`

`15(6)+31(-3)=-3`

`90-93=-3`

Which is true, so our solution is correct.

Hope this helps!