Here's the question you clicked on:
brookester6
Use substitution to solve the problem... x=9y x+4=5(y+4) I already now that you can plug x in to the second equation and then distribute but what next?
\[\large \color{orangered}{x=9y}\]\[\large \color{orangered}{x}+4=5(y+4) \qquad \rightarrow \qquad \color{orangered}{9y}+4=5(y+4)\]Distributing gives us,\[\large 9y+4=5y+20\]So you've gotten this much done so far? Hmm ok. Subtract 5y from each side,\[9y-5y+4=\cancel{5y-5y}+20\]
\[\large 4y+4=20\]Subtract 4 from each side,\[4y+\cancel{4-4}=20-4\]
Understand the process of combining like terms? :D That's how the 9y and 5y combined.
You already have the first equation solved for x. So plug in what x is equal to, 9y, into the second equation and solve for y.
yea, now y = 4 and i plug it in to the top equation :) Thanks so much!!