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Question 217891: The problem is:
3x+2y=4
2x-3y=6
I can only get so far in this problem and then I'm stuck. Here's how far I can get:
3x=-2y+4
3x/3=-2/3y+4/3
x=-2/3y+4/3
Now if I substitute this into the 2nd equation, I have:
2(-2/3y+4/3)-3y=6
This is where I get stuck, it doesn't make any sense. I have looked in the back of the student manual that works with my textbook ( Miller//O'Neil//Hyde: Beginning & Intermediate Algebra, 2nd edition) and it just doesn't click. Any help would be greatly appreciated!
Lisa
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! 3x=-2y+4
3x/3=-2y/3+4/3
x=-2y/3+4/3
Now if I substitute this into the 2nd equation 2x-3y=6, I have:
2(-2y/3+4/3)-3y=6
-4y/3+8/3-3Y=6
-4Y/3-3Y=6-8/3 combine the y terms by finding a common denominator (3).
(-4Y-3Y*3)/3=(6*3-8)/3
(-4Y-9Y)/3=(18-8)/3 cancel out the 3 in the denominators.
-13y=10
y=-10/13 ans.
Replace y by (-10/13) in the first equation.
3x=-2(-10/13)+4
3x=20/13+4 divide both sides by 3.
x=20/13*1/3+4/3
x=20/39+4/3 find the common denominator (39).
x=(20+4*13)/39
x=(20+52)/39
x=72/39 reduce the fraction.
x=24/13 ans.
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