# Thread: Solve each of the systems by substitution.

1. ## Solve each of the systems by substitution.

1. 5x – 2y = -5 I think I did this wrong can someone help please?
y – 5x = 3

y = 5x + 3 y =5(-2.2) + 3
5x – 2(5x + 3) = -5 y = -11 + 3 = -8
5x – 10x + 6 = -5
-5x + 6 = -5
- 6 = -6
-5x = -11

x =-11/-5 = -2.2

so the solution would be (-2.2, -8)

2. 4x – 12y = 5
-x + 3y = -1

2. Originally Posted by Patience
1. 5x – 2y = -5 I think I did this wrong can someone help please?
y – 5x = 3

y = 5x + 3 y =5(-2.2) + 3
5x – 2(5x + 3) = -5 y = -11 + 3 = -8
5x – 10x + 6 = -5
-5x + 6 = -5
- 6 = -6
-5x = -11

x =-11/-5 = -2.2

so the solution would be (-2.2, -8)
where did you get the -2.2 from, how did it suddenly appear, was there more info in the question?

5x – 2y = -5 .......................(1)
y – 5x = 3 ..........................(2)

from (2), we see y = 5x + 3, substitute 5x + 3 for y in (1)

=> 5x - 2(5x + 3) = -5
=> 5x - 10x - 6 = -5
=> -5x = 1
=> x = -1/5

but y = 5x + 3
=> y = 5(-1/5) + 3
=> y = 2

3. Originally Posted by Patience
2. 4x – 12y = 5
-x + 3y = -1
Is there a typo here?
4x – 12y = 5..................(1)
-x + 3y = -1..................(2)

from (2) we see x = 3y + 1, substitute 3y + 1 for x in (1)

=> 4(3y + 1) - 12y = 5
=> 12y + 4 - 12y = 5
=> 4 = 5
see what i mean about the typo. One of your equations was a multiple of the other, which means we can't find a solution by this method, as there would be infinately many. I've seen a simialr thing happen with another simultaneous equations question you submitted (see number 1 here http://www.mathhelpforum.com/math-he...stitution.html ), either you made a typo in both questions, or your text book is just naughty to be giving you things like this