Question 1000000

My question is based around 'Solving Systems of Equations by the Substitution Method'. 

I understand the concept of plugging in the x or y value into the one or the other equation but for some (very frustrating) reason whenever I follow these steps I don't come out with the right answer. I will list two below. 


1. 4x-y=-1 
2x+4y=13

I chose to take the top equation (4x-y=-1) and make that into -y=-4-1 and change that further by dividing by -y to get y=-4x+1. MY FIRST QUESTION is why did I have to change -1 to +1?

I then tried to plug y=-4x+1 into the 2x+4y=13 but I never get the right answer! Help?



2. x-3y=-11
6x-y=2 

please help?
<pre>4x - y = - 1 -------- eq (i) 
2x + 4y = 13 -------- eq (ii)
For the 1st equation, all you have to do is add y to both sides to get:
4x = - 1 + y
Then add 1 to get: 4x + 1 = y 
Now we have: y = 4x + 1 ------- eq (i)
2x + 4y = 13 --------- eq (ii)
2x + 4(4x + 1) = 13 ------- Substituting 4x + 1 for y in eq (ii)
2x + 16x + 4 = 13
18x = 9
x = {{{9/18}}}, or {{{highlight_green(x = 1/2)}}}

{{{y = 4(1/2) + 1}}} -------- Substituting {{{1/2}}} for x in eq (i)
y = 2 + 1, or {{{highlight_green(y = 3)}}}

Follow the same steps and you should be able to do no. 2