7[z-(3z+29)+9]=2(z+10)

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Solution for 7[z-(3z+29)+9]=2(z+10) equation:


Simplifying
7[z + -1(3z + 29) + 9] = 2(z + 10)

Reorder the terms:
7[z + -1(29 + 3z) + 9] = 2(z + 10)
7[z + (29 * -1 + 3z * -1) + 9] = 2(z + 10)
7[z + (-29 + -3z) + 9] = 2(z + 10)

Reorder the terms:
7[-29 + 9 + z + -3z] = 2(z + 10)

Combine like terms: -29 + 9 = -20
7[-20 + z + -3z] = 2(z + 10)

Combine like terms: z + -3z = -2z
7[-20 + -2z] = 2(z + 10)
[-20 * 7 + -2z * 7] = 2(z + 10)
[-140 + -14z] = 2(z + 10)

Reorder the terms:
-140 + -14z = 2(10 + z)
-140 + -14z = (10 * 2 + z * 2)
-140 + -14z = (20 + 2z)

Solving
-140 + -14z = 20 + 2z

Solving for variable 'z'.

Move all terms containing z to the left, all other terms to the right.

Add '-2z' to each side of the equation.
-140 + -14z + -2z = 20 + 2z + -2z

Combine like terms: -14z + -2z = -16z
-140 + -16z = 20 + 2z + -2z

Combine like terms: 2z + -2z = 0
-140 + -16z = 20 + 0
-140 + -16z = 20

Add '140' to each side of the equation.
-140 + 140 + -16z = 20 + 140

Combine like terms: -140 + 140 = 0
0 + -16z = 20 + 140
-16z = 20 + 140

Combine like terms: 20 + 140 = 160
-16z = 160

Divide each side by '-16'.
z = -10

Simplifying
z = -10

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