Multi-Step Equations

Solving multi-step equations is essentially an extension of solving one-step and two-step equations. If you’re comfortable with those, you should find multi-step equations manageable as well.

Always remember to consider inverse operations. They are essential as you work through the process of untangling equations to find the solution. To maintain consistency, keep your variables on the left side of the equation. This approach will help you stay focused and organized.

Example 1: Solve for \(x\).

$$7x – 3 = 5x + 15$$

Let’s keep the variable on the left side of the equation. To eliminate the \(-3\) on the left, we should add \(3\) to both sides. Since we want the variable to remain on the left side, we need to remove the \(5x\) on the right. We can do this by subtracting \(5x\) from both sides. Finally, the variable \(x\) is multiplied by \(2\), so to solve for \(x\), we need to do the inverse operation, which is to divide both sides by \(2\). Let’s divide by \(2\) to find the value of \(x\).

$$\eqalign{
 7x – 3 &= 5x + 15  \cr 
   7x – 3 + 3 &= 5x + 15 + 3  \cr 
   7x &= 5x + 18  \cr 
   7x – 5x &= 5x – 5x + 18  \cr 
   2x &= 18  \cr 
   {{2x} \over 2} &= {{18} \over 2}  \cr 
   x&= 9 \cr} $$

Example 2: Solve for \(y\).

$$3\left( {2y – 7} \right) = y + 9$$

First, address the grouping symbol (parenthesis) on the left side by distributing the outer number into the terms within the parenthesis. Then, solve the equation as you normally would. Start by eliminating the \(-21\) on the left side by adding \(+21\) to both sides of the equation. Next, remove the variable from the right side by subtracting \(y\) from both sides. Finally, divide both sides by \(5\) to complete the solution.

$$\eqalign{
   3\left( {2y – 7} \right) &= y + 9  \cr 
   6y – 21 &= y + 9  \cr 
   6y – 21 + 21 &= y + 9 + 21  \cr 
   6y &= y + 30  \cr 
  6y – y &= 30  \cr 
   5y &= 30  \cr 
   {{5y} \over 5} &= {{30} \over 5}  \cr 
   y &= 6 \cr} $$

Example 3: Solve for \(h\).

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

There are several ways to tackle this problem, but in my opinion, the simplest method is to eliminate the denominators immediately by multiplying both sides by 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 \(12\).

$$\eqalign{
   {{1 – 2h} \over 6} + {h \over 4} &= 1  \cr 
   \color{red}{12}\left( {{{1 – 2h} \over 6} + {h \over 4}} \right) &= \color{red}{12}\left( 1 \right)  \cr 
   2\left( {1 – 2h} \right) + 3h &= 12  \cr 
   2 – 4h + 3h &= 12  \cr 
   2 – h &= 12  \cr 
   2 – 2 – h &= 12 – 2  \cr 
    – h &= 10  \cr 
   {{ – h} \over { – 1}} &= {{10} \over { – 1}}  \cr 
   h &=  – 10 \cr} $$