Variables on Both Sides of the Equation: ax+b=cx+d

In this lesson, we will look at solving equations when there are variables on both sides of the equation, using both algebra tiles and pencil and paper. The second example is a word problem involving decimals and is solved using only pencil and paper.

 

Remember:

The key to solving multi-step equations is to follow the golden rule of Algebra

“Do unto the right-hand side of the equation
as you do unto the left-hand side of the equation.”

 

Try this problem:

Jeffrey spend the same amount of money each morning. On Sunday, he bought a newspaper for $\$2.50$ and also bought two doughnuts. On Monday, he bought a newspaper for $\$1.00$ and bought four doughnuts. What is the cost of one doughnut?

 

Step by Step Solution:

First step is to pick a variable to represent the doughnuts and write an equation.
   Lets use $d$ to represent the doughnuts.
   On Sunday Jeffrey $\$2.50$ for a newspaper and had 2 doughnuts: $2.50+2d$
   On Monday Jeffrey $\$1.00$ for a newspaper and had 4 doughnuts: $1.00+4d$
   He spends the same amount on Sunday and Monday,
   so our equation would be:

\[ 2.50+2d=1.00+4d \]

Now that we have an equation, we need to get the variables on one side of the equation and everything else on the other side.

\[ \begin{align*}
2.50+2d\color{blue}{-2d}&=1.00+4d\color{blue}{-2d}\\
2.50&=1.00+\color{blue}{2d} \\[1em]
2.50\color{blue}{-1.00}&=1.00 \color{blue}{-100}+2d\\
\color{blue}{1.50}&=2d \\[1em]
\frac{1.50}{\color{blue}{2}}&=\frac{2d}{\color{blue}{2}} \\
0.75&=d
\end{align*}\]

So one doughnut costs $75$ cents.