# Mathematics I (Math 1132)

Durham College, Mathematics
Free
• 42 lessons
• 0 quizzes
• 10 week duration
• ##### Measurements

An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

• ##### Fractions, Percentage, Ratios and Proportion

Emphasis here is placed on understanding fractions, percent, and using ratios to compare quantities and set up proportions to solve problems.

• ##### Geometry

This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.

## Mathematics I (Math 1132)

### Solve Equations Containing Fractions

Depending on the number of terms in the equation, these questions can go from being simple to extremely difficult. Easiest of these types is when you have a single term on the left side and a single term on the right. This of course was discussed earlier in this unit when we looked at equations having fractions, and the technique we use is called cross-multiplication.

However, this method has its limitations when when you have 3 terms, let say:

$\frac{a}{3}–8=6a\phantom{\rule{0ex}{0ex}}$

To solve for a, you can either combine the first two terms together, then cross-multiply (method 1), or find a common denominator – it’s 3 – and multiply each term by 3 to reduce it into an equation without fractions (method 2).

Let’s focus on method 1. Combing the first two terms we get:

$\frac{a}{3}–8\phantom{\rule{0ex}{0ex}}=\frac{a–24}{3}\phantom{\rule{0ex}{0ex}}$

Now our equation is suitable for cross-multiplication:

$\frac{a–24}{3}=\frac{6a}{1}\phantom{\rule{0ex}{0ex}}3\left(6a\right)=a–24\phantom{\rule{0ex}{0ex}}18a=a–24\phantom{\rule{0ex}{0ex}}18a–a=–24\phantom{\rule{0ex}{0ex}}17a=–24\phantom{\rule{0ex}{0ex}}a=\frac{–24}{17}\phantom{\rule{0ex}{0ex}}$

Had you gone with method 2, the solution would be the same:

Let’s try another example where method 1 or 2 need to be applied first.

$\frac{x–2}{4}=\frac{2x–4}{2}–6\phantom{\rule{0ex}{0ex}}$

Starting with method 1, notice that the right side has two terms that can be combined.

$\cdots =\frac{2x–4}{2}–6\phantom{\rule{0ex}{0ex}}=\frac{2x–4–2\left(6\right)}{2}\phantom{\rule{0ex}{0ex}}=\frac{2x–4–12}{2}\phantom{\rule{0ex}{0ex}}=\frac{2x–16}{2}\phantom{\rule{0ex}{0ex}}$

Now the equation becomes:

To see it done using method 2, watch the video below:

Try solving this last example on your own before continuing to another related concept involving algebraic fractions.

$\frac{3x–5}{2}=\frac{2\left(x–1\right)}{3}+4\phantom{\rule{0ex}{0ex}}$