Question

X + 4y = 13

x-y=3
You have to solve with the elimination method but i don’t know how to do that

Answer 1

`\bf \begin{array}{llrrlllll} x+4y&=13&&x&+4y&=13\\ x-y&=3\implies \stackrel{\textit{multiplying this one by }}{-1}&&-x&+y&=-3\\ \cline{4-6}\\ &&&0x&+5y&=10 \end{array} \\\\\\ 5y=10\implies y = \cfrac{10}{2}\implies \boxed{y = 2} \\\\\\ \stackrel{\textit{since we know that}}{x-y=3}\implies x-(2)=3\implies \boxed{x = 5} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (\stackrel{x}{5}~~,~~\stackrel{y}{2})~\hfill`

notice, is called the elimination method, because we eliminate one of the variables, either one, in this case the "x", by simply multiplying one of the equations by a value that gives us the negative equivalent of the other, and of course same - same = 0.

Answer 2

so when i think of the elimination method, it’s literally just cancelling numbers out then lining up the numbers and adding. so in this case, i would cancel out the x’s, but that’s just me, you could cancel out the y’s if you’d like. in order to cancel out the x’s you would multiply the second equation by -1, it would look like this

-1(x-y=3) i’ll just do the work and you can see what i mean. see picture for reference

the answer would be (5,2)