Math Is Fun Forum

Monique

Member

Registered: 2007-02-17

Posts: 22

understanding this

a) how would I go about sloving for z

        -6z + 2 = 3z + 4z + 28   ???

b)  and how do I prove this solution is correct???

Kind Regards

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 52,200

Re: understanding this

Monique,
This an equation of degree 1 in 1 variable. z is a variable and since no squares or cubes or higher powers form part of the equation, the degree of the equation is 1.

-6z + 2 = 3z + 4z + 28
Move the -6z from the Left Hand side (LHS) to the Right Hand Side (RHS), as you move, the sign changes from '-' to '+'.
Hence,
2 = 3z + 4z + 28 + 6z
2 = 13z + 28

(When there are like terms like 3a, 4a etc. or 5z, 8 z etc, the aritmetic operations can be performed the same way as they are done with constants or numbers).

Move the 28 from the RHS to LHS, remember to change the sign.

2-28 = 13z

-26 = 13z

Hence, z = -2.

To prove the solution is correct, substitute the value of z you have obtained in the original equation.
The LHS is -6z + 2, that is -6(-2)+2= 12 + 2 = 14
The RHS is 3z + 4z + 28, that is 3(-2) + 4(-2) + 28,
that is -6 + (-8) + 28 = -14 + 28 = 14.
Since the LHS and RHS are equal, the solution is correct!

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