Equations With Rational Coefficients

Example 9

Solution

Begin by ensuring that all like terms have been collected in the numerator and denominator.  

The common denominator for \(5 \) and \(2\) is \(10\), so multiply both sides by \(10\), then divide.  At that point, it becomes straightforward:

• expand the brackets by multiplying each term inside by \(2\), 
• subtract \(2\) from both sides, and then
• divide by \(4\). 

\(\begin{align*} \frac{2x+1}{5}&=\frac{1}{2}\\ \frac{2x+1}{5}\times10&=\frac{1}{2}\times10\\ (2x+1)\times2&=1\times5\\ 4x+2&=5\\ 4x&=5-2\\ 4x&=3\\ x&=\frac{3}{4} \end{align*}\)

Let's take a closer look at the first two lines of this solution. Do you notice something special happening? Pay attention to the \(5\) in the denominator on the left side of the equation and the \(2\) in the denominator on the right side of the equation.

\(\begin{align*} \frac{2x+1}{\class{hl3}{5}}&=\frac{1}{\class{hl2}{2}}\\ \frac{2x+1}{\class{hl3}{5}}\times10&=\frac{1}{\class{hl2}{2}}\times10\\ (2x+1)\times \class{hl2}{2}&=1\times \class{hl3}{5} \end{align*}\)

When we multiply the numerator of each side by the denominator of the other side, this is called Cross-Multiplication.

Cross-multiplication is a useful technique, but it can be used only when there is a single fraction (rational expression) on the left side and the right side of an equal sign.

Example 10

Solution

There is a single rational expression on both sides of the equal sign, which means we can use cross-multiplication as the first step. Multiplying the left side by \(3\) (the denominator on the right) and the right side by \(5\) (the denominator on the left) gives us the following:

\(\begin{align*} \frac{3h+20}{\class{hl3}{5}}&=\frac{h}{\class{hl2}{3}}\\ (3h+20)(\class{hl2}{3})&=(h)(\class{hl3}{5})\\ 9h+60&=5h\\ 9h-5h&=-60\\ 4h&=-60\\ h&=\frac{-60}{4}\\ h&=-15 \end{align*}\)

This could also be solved by multiplying both sides through by the common denominator of \(15\).  Try this alternate method of solving if you'd like!

Example 11

Solution

The method of cross-multiplication is appealing but cannot always be used.  Can we use it as our first step in this example?

No!  We cannot cross-multiply immediately because there is not a single rational expression on the right side.  We could find a common denominator and write a single fraction on the right, then cross multiply OR we can multiply each term through by the common denominator of \(6\) (as shown below).

\(\begin{align*} \frac{m}{3}&=4-\frac{m}{6}\\ \frac{m}{3} \class{hl2}{\times6} &=4 \class{hl2}{\times 6} -\frac{m}{6} \class{hl2}{\times 6}\\ m\times \dfrac{6}{3} & = 4\times 6 - m\times \dfrac{6}{6}\\ 2m&=24-m\\ 2m+m&=24\\ 3m&=24\\ m&=\frac{24}{3}\\ m&=8 \end{align*}\)

Try This Revisited

Solution

To solve this equation, complete the following steps: