Best Answer

Solve by the substitution method the given system of the linear equations of 3 variables.

2x + 7z = 64

8y + 6z = 40

3x - 2y + 4z = 46 (add the first equation with the second equation)

2x + 8y + 13z = 104

3x - 2y + 4z = 46 (multiply the second equation by 4)

2x + 8y + 13z = 104

12x - 8y + 16z = 184 (add both equations)

14x + 29z = 288

14x + 29z = 288

2x + 7z = 64 (multiply the second equation by -7)

14x + 29z = 288

-14x - 49z = -448 (add both equations)

-20z = -160 (divide both sides by -20)

z = 8

2x + 7z = 64 (substitute 8 for z)

2x + 7(8) = 64

2x + 56 = 64 (subtract 56 to both sides)

2x = 8(divide by 2 to both sides)

x = 4

8y + 6z = 40 (substitute 8 for z)

8y + 6(8) = 40

8y + 48 = 40 (subtract 48 to both sides)

8y = -8 (divide by 8 to both sides)

y= -1

Thus, the solutions are x = 4, y = -1, and z = 8.

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Q: 2x plus 7z equals 64 8y plus 6z equals 40 3x-2y plus 4z equals 46?

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Related questions

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well you first get the 43 and take away 3 to make it 40, then you divide it by 8 to give you 5 so thats how 2x becomes 5x. Then you add 2 x 5 which = 10 then you add the 3 that was taken away from the 40 so you come out with the solution of: 2x + 43 = 5x + 13.

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