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31 votes
31 votes
Anyone know if they can help me rq

Anyone know if they can help me rq-example-1
User Arsenbonbon
by
2.6k points

2 Answers

10 votes
10 votes

Answer:

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

User Lampslave
by
2.6k points
14 votes
14 votes

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!

User Arthur Klezovich
by
2.3k points