SOLUTION: PLEASE HELP! THANK YOU. Solve my substitution method. 6x+5y = 5 and x = 30 - 5y. I know that in the second equation x is already solved for so I would substitute this into equ

Algebra ->  Expressions-with-variables -> SOLUTION: PLEASE HELP! THANK YOU. Solve my substitution method. 6x+5y = 5 and x = 30 - 5y. I know that in the second equation x is already solved for so I would substitute this into equ      Log On


   



Question 387676: PLEASE HELP! THANK YOU.
Solve my substitution method.
6x+5y = 5 and
x = 30 - 5y. I know that in the second equation x is already solved for so I would substitute this into equation one right?
6(30 - 5y) + 5 = 5; 180 - 30y +5 = 5;
NOT SURE AFTER THIS.

Found 2 solutions by rapaljer, rfer:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct in your substitution, but you made an error. The first step should look like this:

6(30-5y) +5y = 5
180 -30y +5y=5
180-25y=5
180-180-25y = 5 - 180
-25y = -175

Divide both sides by -25:
y=7

Now, find x by substituting y-7 into the x equation:
x=30-5y
x=30-5(7)
x=30-35
x=-5

Check by substituting both numbers into the OTHER equation:

6x + 5y = 5
6(-5) + 5(7)= 5
-30 + 35 = 5
5 = 5

It checks!!

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Answer by rfer(16322) About Me  (Show Source):
You can put this solution on YOUR website!
you have done the hard part except you lost the y with the first 5
180-30y+5y=5
-25y=-175
y=7
now plug in the 7
solve for x
6x+5*7=5
6x+35=5
6x=-30
x=-5
-------------------
-5=30-5*7