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3 years ago

What is the solution to the system? [1] -3x+4y+2z=-3 [2] 2x-4y-z=0 [3] y=3x-13

Mathematics

1 answer:

Monica [59]3 years ago

8 0

Answer: x = 5

y = 2

z = 2

Step-by-step explanation:

-3x+4y+2z=-3 - ,- - - - - - 1

2x-4y-z=0 - - - - - - - - - - 2

y=3x-13 - - - - - - - - - - 3

We will use the method of substitution.

Substituting y = 3x -13 into equation 1 and equation 2, it becomes

-3x+4(3x-13)+2z=-3

-3x + 12x - 52 + 2z=-3+52

9x+2z = 49 - - - - - - -4

2x - 4(3x-13)-z= 0

2x - 12x + 52 - z = 0

-10x- z = -52 - - - - - - --5

Using elimination method to solve equation 4 and equation 5

Multiply equation 5 by 2 and equation 4 by 1

-20x - 2z = -104

9x+2z = 49

Adding both equations,

-11x = -55

x = -55/-11

x = 5

Substituting x = 5 into equation 3

y = 3x - 13

y = 3×5 -13 = 15-13

y = 2

Substituting x = 5 and y = 2 into equation 1,

-3x+4y+2z=-3

-3×5 + 4×2 +2z = -3

-15+8+2z = -3

-7+2z = -3

2z = -3+7

2z = 4

z = 4/2 = 2

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