SOLUTION: Find solution to each system of equations by substitution I am still struggling using substitution method given 2 equations. Any help (detailed instructions) you can give me so

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Question 121696This question is from textbook Elementary and Intermediate Algebra
: Find solution to each system of equations by substitution
I am still struggling using substitution method given 2 equations. Any help (detailed instructions) you can give me so i understand would be very much appreciated.
This is what I have gotten so far and its not working.
Problem: 6X+3Y=30 and 2X+3Y=18
2(30-3Y) over 6 = 3Y =18
60 + Y over 6 + 3Y =18
10/6 y =3y =18
5/3=3y =18
-3y=-5/3 +18
y = 5/3 +18
Then I lose it. Know its not right but don't know how to correct it.
It throws me when given a Y on one side and a X on the other side of the equation. How do i separate them? Then they throw a statement of multiply the equation by a number. Which number do I use? Then I run into another problem same as above only it has a -Y and a +Y number.
Problem: 1/2x + 2y = 5, 3X-1/2y =5 now I am dealing with fractions? Please help me.
Thank You.
This question is from textbook Elementary and Intermediate Algebra

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The simple rule is: whatever you do to
one side of an equation, you have to do
to the other side
6x+%2B+3y+=+30 and 2x+%2B+3y+=+18
---------------------
What I see right away is I can divide both sides of the
1st equation by 3 without getting any fractions
2x+%2B+y+=+10
Now, to get y alone on the left side,
I subtract 2x from both sides
y+=+10+-+2x
Now I can substitute this y for the y
in the 2nd equation
2x+%2B+3%2810+-+2x%29+=+18
2x+%2B+30+-+6x+=+18
30+-+4x+=+18
Now add 4x to both sides
30+=+4x+%2B+18
Now subtract 18 from both sides
30+-+18+=+4x
12+=+4x
divide both sides by 4
x+=+3 answer
Now you can use either equation to find y
2x+%2B+3y+=+18
2%2A3+%2B+3y+=+18
6+%2B+3y+=+18
subtract 6 from both sides
3y+=+12
y+=+4 answer
You can check these answers by substituting
back into original equations