5(2z-5)=9(z+2)+z-43

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Solution for 5(2z-5)=9(z+2)+z-43 equation:


Simplifying
5(2z + -5) = 9(z + 2) + z + -43

Reorder the terms:
5(-5 + 2z) = 9(z + 2) + z + -43
(-5 * 5 + 2z * 5) = 9(z + 2) + z + -43
(-25 + 10z) = 9(z + 2) + z + -43

Reorder the terms:
-25 + 10z = 9(2 + z) + z + -43
-25 + 10z = (2 * 9 + z * 9) + z + -43
-25 + 10z = (18 + 9z) + z + -43

Reorder the terms:
-25 + 10z = 18 + -43 + 9z + z

Combine like terms: 18 + -43 = -25
-25 + 10z = -25 + 9z + z

Combine like terms: 9z + z = 10z
-25 + 10z = -25 + 10z

Add '25' to each side of the equation.
-25 + 25 + 10z = -25 + 25 + 10z

Combine like terms: -25 + 25 = 0
0 + 10z = -25 + 25 + 10z
10z = -25 + 25 + 10z

Combine like terms: -25 + 25 = 0
10z = 0 + 10z
10z = 10z

Add '-10z' to each side of the equation.
10z + -10z = 10z + -10z

Combine like terms: 10z + -10z = 0
0 = 10z + -10z

Combine like terms: 10z + -10z = 0
0 = 0

Solving
0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

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