# Mathematics for Technology I (Math 1131)

Durham College, Mathematics
Free
• 55 lessons
• 1 quizzes
• 10 week duration
• ##### Numerical Computation

Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

• ##### 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.

• ##### Trigonometry with Right Triangles

Here we focus on right angle triangles within quadrant I of an x-y plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

• ##### Trigonometry with Oblique Triangles

This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

• ##### Geometry

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

## Mathematics for Technology I (Math 1131)

### 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.

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