# 5.1 Solving linear equations

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## Equations and inequalities: solving linear equations

The simplest equation to solve is a linear equation. A linear equation is an equation where the power of the variable(letter, e.g. $x$ ) is 1(one). The following are examples of linear equations.

$\begin{array}{ccc}\hfill 2x+2& =& 1\hfill \\ \hfill \frac{2-x}{3x+1}& =& 2\hfill \\ \hfill \frac{4}{3}x-6& =& 7x+2\hfill \end{array}$

In this section, we will learn how to find the value of the variable that makes both sides of the linear equation true. For example, what value of $x$ makes both sides of the very simple equation, $x+1=1$ true.

Since the definition of a linear equation is that if the variable has a highest power of one (1), there is at most one solution or root for the equation.

This section relies on all the methods we have already discussed: multiplying out expressions, grouping terms and factorisation. Make sure that you arecomfortable with these methods, before trying out the work in the rest of this chapter.

$\begin{array}{cccc}\hfill 2x+2& =& 1\hfill & \\ \hfill 2x& =& 1-2\hfill & \left(\mathrm{like terms together}\right)\hfill & \\ \hfill 2x& =& -1\hfill & \left(\mathrm{simplified as much as possible}\right)\hfill & \end{array}$

Now we see that $2x=-1$ . This means if we divide both sides by 2, we will get:

$x=-\frac{1}{2}$

If we substitute $x=-\frac{1}{2}$ , back into the original equation, we get:

$\begin{array}{ccc}\hfill \mathrm{LHS}& =& 2x+2\hfill \\ & =& 2\left(-\frac{1}{2}\right)+2\hfill \\ & =& -1+2\hfill \\ & =& 1\hfill \\ \hfill \mathrm{and}\\ \hfill \mathrm{RHS}& =& 1\hfill \end{array}$

That is all that there is to solving linear equations.

## Solving equations

When you have found the solution to an equation, substitute the solution into the original equation, to check your answer.

## Method: solving linear equations

The general steps to solve linear equations are:

1. Expand (Remove) all brackets that are in the equation.
2. "Move" all terms with the variable to the left hand side of the equation, and all constant terms (the numbers) to the right hand side of the equals sign.Bearing in mind that the sign of the terms will change from ( $+$ ) to ( $-$ ) or vice versa, as they "cross over" the equals sign.
3. Group all like terms together and simplify as much as possible.
4. If necessary factorise.
5. Find the solution and write down the answer(s).
6. Substitute solution into original equation to check answer.

Solve for $x$ : $4-x=4$

1. We are given $4-x=4$ and are required to solve for $x$ .

2. Since there are no brackets, we can start with rearranging and then grouping like terms.

3. $\begin{array}{cccc}\hfill 4-x& =& 4\hfill & \\ \hfill -x& =& 4-4\hfill & \left(\mathrm{Rearrange}\right)\hfill \\ \hfill -x& =& 0\hfill & \left(\mathrm{group like terms}\right)\hfill \\ \hfill âˆ´\phantom{\rule{1.em}{0ex}}x& =& 0\hfill & \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill 4-0& =& 4\hfill \\ \hfill 4& =& 4\hfill \end{array}$

Since both sides are equal, the answer is correct.

5. The solution of $4-x=4$ is $x=0$ .

Solve for $x$ : $4\left(2x-9\right)-4x=4-6x$

1. We are given $4\left(2x-9\right)-4x=4-6x$ and are required to solve for $x$ .

2. We start with expanding the brackets, then rearranging, then grouping like terms and then simplifying.

3. $\begin{array}{cccc}\hfill 4\left(2x-9\right)-4x& =& 4-6x\hfill & \\ \hfill 8x-36-4x& =& 4-6x\hfill & \left(\mathrm{expand the brackets}\right)\hfill \\ \hfill 8x-4x+6x& =& 4+36\hfill & \left(\mathrm{Rearrange}\right)\hfill \\ \hfill \left(8x-4x+6x\right)& =& \left(4+36\right)\hfill & \left(\mathrm{group like terms}\right)\hfill \\ \hfill 10x& =& 40\hfill & \left(\mathrm{simplify grouped terms}\right)\hfill \\ \hfill \frac{10}{10}x& =& \frac{40}{10}\hfill & \left(\mathrm{divide both sides by}\phantom{\rule{2pt}{0ex}}10\right)\hfill \\ \hfill x& =& 4\hfill & \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill 4\left(2\left(4\right)-9\right)-4\left(4\right)& =& 4-6\left(4\right)\hfill \\ \hfill 4\left(8-9\right)-16& =& 4-24\hfill \\ \hfill 4\left(-1\right)-16& =& -20\hfill \\ \hfill -4-16& =& -20\hfill \\ \hfill -20& =& -20\hfill \end{array}$

Since both sides are equal to $-20$ , the answer is correct.

5. The solution of $4\left(2x-9\right)-4x=4-6x$ is $x=4$ .

Solve for $x$ : $\frac{2-x}{3x+1}=2$

1. We are given $\frac{2-x}{3x+1}=2$ and are required to solve for $x$ .

2. Since there is a denominator of ( $3x+1$ ), we can start by multiplying both sides of the equation by ( $3x+1$ ). But because division by 0 is not permissible, there is a restriction on a value for x. ( )

3. $\begin{array}{cccc}\hfill \frac{2-x}{3x+1}& =& 2\hfill & \\ \hfill \left(2-x\right)& =& 2\left(3x+1\right)\hfill & \\ \hfill 2-x& =& 6x+2\hfill & \left(\mathrm{expand brackets}\right)\hfill \\ \hfill -x-6x& =& 2-2\hfill & \left(\mathrm{rearrange}\right)\hfill \\ \hfill -7x& =& 0\hfill & \left(\mathrm{simplify grouped terms}\right)\hfill \\ \hfill x& =& 0Ã·\left(-7\right)\hfill & \\ \hfill âˆ´\phantom{\rule{2.em}{0ex}}x& =& 0\hfill & \left(\mathrm{zero divided by any number is}\phantom{\rule{3pt}{0ex}}0\right)\hfill \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill \frac{2-\left(0\right)}{3\left(0\right)+1}& =& 2\hfill \\ \hfill \frac{2}{1}& =& 2\hfill \end{array}$

Since both sides are equal to 2, the answer is correct.

5. The solution of $\frac{2-x}{3x+1}=2$ is $x=0$ .

Solve for $x$ : $\frac{4}{3}x-6=7x+2$

1. We are given $\frac{4}{3}x-6=7x+2$ and are required to solve for $x$ .

2. We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying.

3. $\begin{array}{cccc}\hfill \frac{4}{3}x-6& =& 7x+2\hfill & \\ \hfill 4x-18& =& 21x+6\hfill & \left(\mathrm{each term is multiplied by}\phantom{\rule{3pt}{0ex}}3\right)\hfill \\ \hfill 4x-21x& =& 6+18\hfill & \left(\mathrm{rearrange}\right)\hfill \\ \hfill -17x& =& 24\hfill & \left(\mathrm{simplify grouped terms}\right)\hfill \\ \hfill \frac{-17}{-17}x& =& \frac{24}{-17}\hfill & \left(\mathrm{divide both sides by}\phantom{\rule{2pt}{0ex}}-17\right)\hfill \\ \hfill x& =& \frac{-24}{17}\hfill & \end{array}$
4. Substitute solution into original equation:

$\begin{array}{ccc}\hfill \frac{4}{3}Ã—\frac{-24}{17}-6& =& 7Ã—\frac{-24}{17}+2\hfill \\ \hfill \frac{4Ã—\left(-8\right)}{\left(17\right)}-6& =& \frac{7Ã—\left(-24\right)}{17}+2\hfill \\ \hfill \frac{\left(-32\right)}{17}-6& =& \frac{-168}{17}+2\hfill \\ \hfill \frac{-32-102}{17}& =& \frac{\left(-168\right)+34}{17}\hfill \\ \hfill \frac{-134}{17}& =& \frac{-134}{17}\hfill \end{array}$

Since both sides are equal to $\frac{-134}{17}$ , the answer is correct.

5. The solution of $\frac{4}{3}x-6=7x+2$ is,Â Â Â  $x=\frac{-24}{17}$ .

## Solving linear equations

1. Solve for $y$ : $2y-3=7$
2. Solve for $y$ : $-3y=0$
3. Solve for $y$ : $4y=16$
4. Solve for $y$ : $12y+0=144$
5. Solve for $y$ : $7+5y=62$
6. Solve for $x$ : $55=5x+\frac{3}{4}$
7. Solve for $x$ : $5x=3x+45$
8. Solve for $x$ : $23x-12=6+2x$
9. Solve for $x$ : $12-6x+34x=2x-24-64$
10. Solve for $x$ : $6x+3x=4-5\left(2x-3\right)$
11. Solve for $p$ : $18-2p=p+9$
12. Solve for $p$ : $\frac{4}{p}=\frac{16}{24}$
13. Solve for $p$ : $\frac{4}{1}=\frac{p}{2}$
14. Solve for $p$ : $-\left(-16-p\right)=13p-1$
15. Solve for $p$ : $6p-2+2p=-2+4p+8$
16. Solve for $f$ : $3f-10=10$
17. Solve for $f$ : $3f+16=4f-10$
18. Solve for $f$ : $10f+5+0=-2f+-3f+80$
19. Solve for $f$ : $8\left(f-4\right)=5\left(f-4\right)$
20. Solve for $f$ : $6=6\left(f+7\right)+5f$

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