Solving multi-step equations are simply the extension of one-step equations and two-step equations. That's why if you know how to handle one-step and two-step equations, you should have no problems dealing with multi-step equations.

Make sure to always think about inverse operations. You use them as you navigate into the mess of untangling equations to get to the solution. For uniformity, keep your variables on the left side of the equation. This will help you be more focused and organized.

Example 1: Solve for \(x\).

$$7x - 3 = 5x + 15$$

Let's keep the variable on the left side. There is \(-3\) on the left, to get rid of it we must add \(3\) to both sides of the equation. Since we want the variable to be on the left side, we need to get rid of \(5x\) on the right. It can be accomplished by subtracting both sides of the equation by \(5\). Finally, the variable \(x\) is being multiplied by \(2\). The inverse of operation is to divide by \(2\). Let's divide both sides of the equation by \(2\) to solve the variable \(x\).

$$\eqalign{

7x - 3 &= 5x + 15 \cr

7x - 3 + 3 &= 5x + 15 + 3 \cr

7x &= 5x + 18 \cr

7x - 5x &= 5x - 5x + 18 \cr

2x &= 18 \cr

{{2x} \over 2} &= {{18} \over 2} \cr

x&= 9 \cr} $$

Example 2: Solve for \(y\).

$$3\left( {2y - 7} \right) = y + 9$$

This problem contains a grouping symbol (parenthesis) on the left side. Get rid of that first by distributing the outer number into the terms inside the parenthesis. After that, proceed to solve the equation as usual. Eliminate the \(-21\) on the left by adding \(+21\) to both sides of the equation. Follow by removing the variable on the right by subtracting both sides by \(y\). Finish it off by dividing both sides by \(5\).

$$\eqalign{

3\left( {2y - 7} \right) &= y + 9 \cr

6y - 21 &= y + 9 \cr

6y - 21 + 21 &= y + 9 + 21 \cr

6y &= y + 30 \cr

6y - y &= 30 \cr

5y &= 30 \cr

{{5y} \over 5} &= {{30} \over 5} \cr

y &= 6 \cr} $$

Example 3: Solve for \(h\).

$${{1 - 2h} \over 6} + {h \over 4} = 1$$

There are a few ways to approach this problem. But for me, the easiest is to get rid of the denominators right off the bat by multiplying both sides with the Least Common Multiple of the denominators which are \(6\) and \(4\).

So, let's find the LCM of \(6\) and \(4\) first. List the first few multiples of each number, then select the first common multiple.

The first few multiples of \(6\):

6, \(\boxed{12}\), 18, 24, 30, ...

The first few multiples of \(4\):

4, 8, \(\boxed{12}\), 16, 20, ...

Therefore, the LCM of \(6\) and \(4\) is \(12\).

Now, we multiply both sides of the equation by the LCM which is \(12\).

$$\eqalign{

{{1 - 2h} \over 6} + {h \over 4} &= 1 \cr

\color{red}{12}\left( {{{1 - 2h} \over 6} + {h \over 4}} \right) &= \color{red}{12}\left( 1 \right) \cr

2\left( {1 - 2h} \right) + 3h &= 12 \cr

2 - 4h + 3h &= 12 \cr

2 - h &= 12 \cr

2 - 2 - h &= 12 - 2 \cr

- h &= 10 \cr

{{ - h} \over { - 1}} &= {{10} \over { - 1}} \cr

h &= - 10 \cr} $$