SOLUTION: Use Elimination to solve {{{x/3+4y=30}}} and {{{2x-y=5}}}

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Use Elimination to solve {{{x/3+4y=30}}} and {{{2x-y=5}}}      Log On


   

Question 59808This question is from textbook Alegbra Concepts and Applications
: Use Elimination to solve
x%2F3%2B4y=30
and
2x-y=5 This question is from textbook Alegbra Concepts and Applications

Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
(1) x%2F3%2B4y=30
(2) 2x-y=5
Add y to both sides of equation (2) to get 2x=5%2By
Then subtract 5 from both sides and you get y=2x-5
Replace y with 2x-5 in (1) and you get
(3) x%2F3%2B4%282x-5%29=30
Multiply 4 by 2x-5 to get 8x-20 so (3) becomes
x%2F3%2B8x-20=30
Add 20 to both sides to get
(4) x%2F3%2B8x=50
Add x%2F3 to 8x to get %288%2B1%2F3%29x
But, 8=24%2F3 so %288%2B1%2F3%29x=%2824%2F3%2B1%2F3%29x=%2825%2F3%29x
So, replacing x%2F3%2B8x with %2825%2F3%29x in (4) we get
%2825%2F3%29x=50
Now, multiply both sides by 3%2F25 and we get x=50%283%2F25%29 or x=6.
Since x=6, we replace x with 6 in (2) and get 2%2A6-y=5 or 12-y=5 so y=7.
To verify that x=6, y=7 is correct we use these values in (1) and (2):
From (1): x%2F3%2B4y=30 or 6%2F3%2B4%2A7=2%2B28 which is indeed 30, and
From (2): 2x-y=5 or 2%2A6-7=12-7 which is indeed 5.