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How do you solve 3x-2y=-5 ; 3y-4x=8 for x and y. Explain please.

1 Answer

1 vote
3x-2y=-5
3y-4x=8

1) multiply the first equation by 3
3(3x-2y=-5) this will give you the new equation: 9x-6y=-15

2) multiply the second equation by 2
2(3y-4x=8) this should give you the equation: 6y-8x=16

3) combine both equations/ like terms
9x-6y=-15
6y-8x=16

4) -6y and 6y cancel out
9x=-15
-8x=16

5) 9x and -8x combine to make 1x or just x and -15 combined with 16 gives you just 1

6) we are now left with:
x=1

7) plug in the x to any of the two original equations ( i chose the first)
3x-2y=-5
3(1) - 2y = -5
3 - 2y = -5
-2y = -8
y = 4
When you plug in the x=1 you are given 3(1) - 2y = -5
Distribute the 3 and you should have 3 - 2y = -5
Subtract 3 from 3 (this cancels out) then from -5
This should leave you with -2y = -8 ( -3 and -5 add to -8)
Divide by -2 ( -2 divided by -2 cancels out)
-8 divided by -2 gives you 4 (two negatives make a positive)

So, y=4 and x=1
To check, plug in x=1 and y=4 into one equation. when you're done with that you can plug them into the other. when you plug them into the first equation you get -5=-5 which means they worked. when plugged into the second, the result is 8=8 which means x=1 and y=4 worked for both equations.



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