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

Sagot :

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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