Remember that when you are solving for \(x\) in a linear equation (that means that \(x\) does not have any exponents) you need to take all the numbers to one side and all the \(x^{\prime} s\) to the other side.

For example: Solve for \(\mathrm{x}\) :

\[

4 x+5=7-2(-2 x-5)

\]

First multiply out the bracket on the right:

\[

4 x+5=7+4 x+10

\] Remember that a negative times a negative is a positive.

when we take the \(x\) 's to the one side we notice that they cancel:

\[

4 x-4 x+5=17

\]

This means that there is no real solution for \(x\), or \(x\) does not exist. The two sides of the equation are not equal to each other.

If we changed the equation to say:

\[

4 x+5=7+2(-2 x-5)

\]

Then we could solve for \(x\) (Remember you cant change the equation in the exam - try to answer the question given to you. If the teacher realises that the question is wrong when she/he starts to mark it - you will get the marks anyway).

\[

4 x+5=7-4 x-10

\]

\(4 x+4 x=7-10-5 \quad\) Take the \(x\) 's to one side and the numbers (or constants) to the other side.

\(\begin{array}{ll}8 x=-8 & \text { Now divide both sides by the number attached to the } x \\ x=-1 & \text { Now you have your final answer }\end{array}\)