Solve the system using the substitution method

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flora33

Junior Member
Joined
Mar 10, 2008
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All right, I thought I knew how to do this, but my answer seems wrong. Please take a look and let me see if you see something wrong:

-3x + y = 1
5x + 2y = -4

So I solve for y in the first equation because y has a coeffecient of 1:
-3x + 3x + y = 1 + 3x
y = 3x + 1

Then, substitute 3x + 1 for y in the second equation:
5x + 2(3x + 1)= -4
5x + 6x + 2 = -4
11x + 2 = -4
11x + 2 - 2 = -4 -2
11x = -6
11x/11 and -6/11
x = -6/11

Then, to find for y, substitute -6/11 for x in the substitution equation:
3(-6/11) + 1
=-7/11
So, the solution is (-6/11, -7/11)

I know that is wrong because I checked the original equations with these numbers and it did not match. What am I doing wrong? Please help!
 
Your answer is right. I plugged them in and I got -0.63636363636363 ... for both.
 
flora33 said:
All right, I thought I knew how to do this, but my answer seems wrong. Please take a look and let me see if you see something wrong:

-3x + y = 1
5x + 2y = -4

So I solve for y in the first equation because y has a coeffecient of 1:
-3x + 3x + y = 1 + 3x
y = 3x + 1

Then, substitute 3x + 1 for y in the second equation:
5x + 2(3x + 1)= -4
5x + 6x + 2 = -4
11x + 2 = -4
11x + 2 - 2 = -4 -2
11x = -6
11x/11 and -6/11
x = -6/11

Then, to find for y, substitute -6/11 for x in the substitution equation:
3(-6/11) + 1
=-7/11
So, the solution is (-6/11, -7/11)<<< These are correct - you must be making mistake/s during checking process.

I know that is wrong because I checked the original equations with these numbers and it did not match. What am I doing wrong? Please help!
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