SOLUTION: Solve the system by Substitution: 3x + 3 = 5y 4x + 2y = -4

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Question 1155473: Solve the system by Substitution:
3x + 3 = 5y
4x + 2y = -4
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

3x+%2B+3+=+5y......eq.1
4x+%2B+2y+=+-4 ......eq.2
-----------------------
start with
3x+%2B+3+=+5y......eq.1, solve for y
3x%2F5+%2B+3%2F5+=+y......substitute in eq.2

4x+%2B+2%283x%2F5+%2B+3%2F5%29=+-4 ......eq.2,...solve for x
4x+%2B+6x%2F5+%2B+6%2F5=+-4........both sides multiply by 5
20x+%2B+6x+%2B+6=+-20
26x+=+-20-6
26x+=+-26
x+=+-1

go to
3x+%2B+3+=+5y......eq.1, substitute x
3%28-1%29+%2B+3+=+5y
-3+%2B+3+=+5y
0+=+5y
y=0

solution:
x+=+-1
y=0
intersection point is at: (-1,0)

+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+3x%2F5+%2B+3%2F5+%2C+-2x-2%29+


Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


I would never solve a system of equations like this using substitution; elimination is far easier.

However, if I did solve it using substitution, I would not solve for y and get an expression involving fractions and then substitute in the other equation.

Remember that, using substitution, you don't need to solve for y or x -- you can solve for 3y, or 7x, or whatever is convenient.

So here is how I would solve this using substitution.

3x%2B3+=+5y --> 6x%2B6+=+10y

4x%2B2y=-4 --> 20x%2B10y+=+-20

Now substitute "6x+6" for "10y" in the second equation:

20x%2B%286x%2B6%29+=+-20
26x%2B6+=+-20
26x+=+26
x+=+1

Then finish from there by any method you choose.