Title: Default for 3.1 Solving Linear Equations
Course: HELM Workbooks 2023-2024

Linear equations

Key Point

A linear equation is an equation of the form.

\(ax+b=0 \qquad a\ne 0\) .

where \(a\) and \(b\) are known numbers and \(x\) represents an unknown quantity to be found.

In the equation \(ax+b=0\), the number \(a\) is called the coefficient of \(x\), and the number \(b\) is called the constant term.

The following are examples of linear equations

- \(3x+4=0\),

- \(-2x+3=0\),

- \(-\dfrac{1}{2}x-3=0\).

Note that the unknown, \(x\), appears only to the first power, that is as \(x\), and not as \(x^2\), \(\sqrt{x}\), \(x^\frac{1}{2}\) etc. Linear equations often appear in a non-standard form, and also different letters are sometimes used for the unknown quantity. For example

- \( 2x=x+1\),

- \(3t-7=17\),

- \(13=3z+1\),

- \(1-\dfrac{1}{2}y=3\),

- \(2\alpha-1.5=0\),

are all examples of linear equations. Where necessary the equations can be rearranged and written in the form \(ax+b=0\). We will explain how to do this later in this Section.

Exercise

1. Write down the general form of a linear equation.
2. What is meant by the root or solution of a linear equation?
A root is a value for the unknown which satisfies the equation.

Solving a linear equation

To solve a linear equation we make the unknown quantity the subject of the equation. We obtain the unknown quantity on its own on the left-hand side. To do this we may apply the same rules used for transposing formulae. These are given again here.

Key Point

Operations which can be used in the process of solving a linear equation
- add the same quantity to both sides
- subtract the same quantity from both sides
- multiply both sides by the same quantity
- divide both sides by the same quantity
- take the reciprocal of both sides (invert)
- take functions of both sides; for example cube both sides.

A useful summary of these rules is ‘whatever we do to one side of an equation we must also do to the other’.

Example

Solve the equation \(x+14=5\).

Solution

Note that by subtracting 14 from both sides, we leave \(x\) on its own on the left. Thus

\(x+14-14 = 5-14\)
\(x = -9\)

Hence the solution of the equation is \(x=-9\). It is easy to check that this solution is correct by substituting \(x=-9\) into the original equation and checking that both sides are indeed the same. You should get into the habit of doing this.

Example

Solve the equation \(19y=38\).

Solution

In order to make \(y\) the subject we can divide both sides by \(19\):

\(19y = 38\)

\(\dfrac{19y}{19} = \dfrac{38}{19}\)

Cancelling the 19's gives \(y = \dfrac{38}{19}\)

So \(y = 2\)

Hence the solution of the equation is \(y=2\).

Example

Solve the equation \(4x+12=0\).

Solution

Starting from \(4x+12=0\) we can subtract \(12\) from both sides to obtain:

\(4x+12-12 = 0-12\)

So that \(4x =-12\)

If we now divide both sides by \(4\) we find \(\dfrac{4x}{4} = \dfrac{-12}{4}\)

Cancelling the \(4\)'s gives \(x =-3\)

Hence the solution of the equation is \(x=-3\).

Exercise

Solve the following equations.

1. \([value]\), therefore \(x=\)
2. \([value]\), therefore \(x=\)
3. \(\displaystyle[value]\), therefore \(x=\)
4. \(\displaystyle[value]\), therefore \(x=\)
Feedback: 1. \([value]\), therefore \(x=[value].\) 2. \([value]\), therefore \(x=[value].\) 3. \([value]\), therefore \(x=[value].\) 4. \([value]\), therefore \(x=[value].\)

Example

Solve the following equations:

1. \(x+3=\sqrt{7}\),
2. \(x+3=-\sqrt{7}\).

Solution

1. Subtracting \(3\) from both sides gives \(x=\sqrt{7}-3.\)
2. Subtracting \(3\) from both sides gives \(x=-\sqrt{7}-3.\)

Note that when asked to solve \(x+3=\pm\sqrt{7}\) we can write the two solutions as \(x=-3\pm\sqrt{7}\). It is usually acceptable to leave the solutions in this form (i.e. with the \(\sqrt{7}\) term) rather than calculate decimal approximations. This form is known as the surd form.

Example

Solve the equation \(\dfrac{2}{3}(t+7)=5\).

Solution

There are a number of ways in which the solution can be obtained. The idea is to gradually remove unwanted terms on the left-hand side to leave \(t\) on its own. By multiplying both sides by \(\dfrac{3}{2}\) we find:

\(\dfrac{3}{2} × \dfrac{2}{3}(t+7)=\dfrac{3}{2} × 5=\dfrac{3}{2} × \dfrac{5}{1}\)

and after simplifying and cancelling, \( t+7=\dfrac{15}{2}\)

Finally, subtracting \(7\) from both sides gives \(t =\dfrac{15}{2}-7=\dfrac{15}{2}-\dfrac{14}{2}=\dfrac{1}{2}\)

So the solution is \(\dfrac{1}{2}\).

Example

Solve the equation \(3(p-2)+2(p+4)=5\).

Solution

At first sight this may not appear to be in the form of a linear equation. Some preliminary work is necessary. Removing the brackets and collecting like terms we find the left-hand side yields \(5p+2\) so the equation is \(5p+2=5\) so that \(p=\dfrac{3}{5}\).

Example

Solve the equation \(2(x-5)=3-(x+6)\).

Solution

1. First remove the brackets on both sides
\(2x-10=3-x-6\).

We may write this as \(2x-10=-x-3\).

2. Rearrange the equation found in (a) so that terms involving \(x\) appear only on the left-hand side, and constants on the right. Start by adding \(10\) to both sides:
\(2x=-x+7\)
3. Now add \(x\) to both sides:
\(3x=7\)
4. Finally solve this to find \(x\):
\(x=\dfrac{7}{3}\)

Example

Solve the equation \(\dfrac{6}{1-2x}=\dfrac{7}{x-2}\).

Solution

This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation. By multiplying both sides by \((1-2x)\) and \((x-2)\) we find

\((1-2x)(x-2) × \dfrac{6}{1-2x} = (1-2x)(x-2) × \dfrac{7}{x-2}\)

Considering each side in turn and cancelling common factors:

\(6(x-2) = 7(1-2x)\)

Removing the brackets and rearranging to find \(x\) we have

\(6x-12=7-14x\)

Further rearrangement gives:

\(20x = 19\)

The solution is therefore \(x=\dfrac{19}{20}\).

Example

The figure below shows three branches of an electrical circuit which meet together at \(x\). Point \(x\) is known as a node. As shown in the figure the current in each of the branches is denoted by \(I\), \(I_1\) and \(I_2\). Kirchhoff's current law states that the current entering any node must equal the current leaving that node. Thus we have the equation \(I=I_1+I_2\).

[Image: Circuit diagram]

1. Given \(I_2=10A\) and \(I=18A\) calculate \(I_1\).
2. Suppose \(I=36A\) and it is known that current \(I_2\) is five times as great as \(I_1\). Find the branch currents.

Solution

1. Substituting the given values into the equation we find \(18=I_1+10\).
Solving for \(I_1\) we find \(I_1=18-10=8\).
Thus \(I_1\) equals \(8A\).
2. From Kirchhoff's law, \(I=I_1+I_2\). We are told that \(I_2\) is five times as great as \(I_1\), and so we can write \(I_2=5I_1\).
Since \(I=36\) we have
\(36= I_1+5I_1\)
Solving this linear equation \(36=6I_1\) gives \(I_1=6A\).
Finally, since \(I_2\) is five times as great as \(I_1\), we have \(I_2=5I_1=30A\)

Exercise

Solve the following equations.

1. \( [value] = [value]\), therefore \(x=\)
2. \( [value] = [value]+[value]\), therefore \(x=\)
3. \(\dfrac{[value]}{[value]}t=[value]\), therefore \(t=\)
Feedback: 1. \( [value] \Rightarrow [value]= [value]\), therefore \(x=[value].\) 2. \( [value] \Rightarrow [value] = [value]\), therefore \(x=[value].\) 3. \(\dfrac{[value]}{[value]}t=[value] \Rightarrow [value] = [value]\), therefore \(t=[value].\)

Exercise

Solve the following equations.

1. \([value] = [value][value]\), therefore \(y=\)
2. \( [value] = [value]\), therefore \(x=\)
3. \( [value] = [value]\), therefore \(x=\)
4. \( [value] = [value]\), therefore \(x=\)
5. \( [value] = [value]\), therefore \(x=\)
Feedback: 1. \([value] = [value][value]\), therefore \(y=[value].\) 2. \( [value] = [value]\), therefore \(x=[value].\) 3. \( [value] = [value]\), therefore \(x=[value].\) 4. \( [value] = [value]\), therefore \(x=[value].\) 5. \( [value] = [value]\), therefore \(x=[value].\)

Exercise

Solve the following equations.

1. \( [value]-[value]([value]+[value]) = [value]\), therefore \(t=\)
2. \( [value]\), therefore \(m=\)
3. \( \displaystyle [value]([value] + [value]) = [value]\), therefore \(x=\)
Feedback: 1. \(\begin{align} [value]-[value]([value]+[value])= [value] & \Rightarrow \\ [value] = [value] & \Rightarrow \\ t= [value]. \end{align}\) 2. \(\begin{align} [value] & \Rightarrow \\ [value] = [value] & \Rightarrow \\ m= [value]. \end{align}\) 3. \(\begin{align} [value]([value] + [value]) = [value] & \Rightarrow \\ [value] + [value]= [value] & \Rightarrow \\ x=[value]. \end{align}\)

Exercise

Solve the following equations.

1. If \(y = 2\) and \([value]\), find \(x\).
2. If \(y = -2\) and \([value]\), find \(x\).
3. If \(y = 0\) and \([value]\), find \(x\).
4. If \(x = -3\) and \([value]\), find \(y\).
5. If \(y = 10\) and \([value]\), find \(x\).
6. If \(\gamma = 2\) and \([value]=[value]\gamma +[value]\beta\), find \(\beta\).
Feedback: 1. If \(y = [value]\) and \([value]\) then, \([value]+[value]\cdot [value] = [value] \Rightarrow [value] \Rightarrow x= \displaystyle [value].\) 2. If \(y = [value]\) and \([value]\) then, \( [value] \Rightarrow [value] \Rightarrow x= \displaystyle [value].\) 3. If \(y = [value]\) and \([value]\) then, \([value] \Rightarrow x= \displaystyle [value].\) 4. If \(x = [value]\) and \([value]\) then, \([value] \Rightarrow x= \displaystyle [value].\) 5. If \(y = [value]\) and \([value]\) then, \([value] \Rightarrow x= \displaystyle [value].\) 6. If \(\gamma = [value]\) and \([value]=[value]\gamma +[value]\beta\) then, \([value]=[value] +[value]\beta \Rightarrow \beta= \displaystyle [value].\)

Exercise

Solve the following equations.

1. \( \displaystyle{[value]}\), therefore \(x=\)
2. \(\displaystyle{[value]}\), therefore \(x=\)
3. \( \displaystyle{[value]+[value] = [value]}\), therefore \(x=\)
4. \( \displaystyle{[value]}\), therefore \(x=\)
5. \( \displaystyle{[value]}\), therefore \(s=\)
Feedback: 1. \( \displaystyle{[value]}\) This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation by cross multiplying. Attention: the denominator should not be \(0\) so \( x \neq [value] .\) We find: \([value]=[value] \Rightarrow x = \displaystyle [value].\) 2. \(\displaystyle{[value]}\) Here we multiply all parts of the equation with the Lowest Common Denominator, which is \( [value] \times [value] = [value].\) We find: \( \displaystyle{[value]} \Rightarrow x=[value].\) 3. \( \displaystyle{[value]+[value] = [value]}\) Here we multiply all parts of the equation with the Lowest Common Denominator, which is \( [value] \times [value] = [value].\) We find: \( \displaystyle{[value]+[value] = [value]} \Rightarrow x = [value].\) 4. \( \displaystyle{[value]}\), therefore \(x=\) This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation by multiplying both sides by the Lowest Common Denominator, which is \((3x - [value])(x-1).\) Attention: the denominator should not be \(0\) so \( x \neq [value]\) and \( x \neq 1.\) We find: \([value] \Rightarrow x = [value].\) 5. \( \displaystyle{[value]}\) This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation by multiplying both sides by the Lowest Common Denominator, which is \(([value] + [value])(s+1).\) Attention: the denominator should not be \(0\) so \( s \neq [value]\) and \( s \neq -1.\) We find: \( \displaystyle{[value]} \Rightarrow s =[value].\)

Exercise

1. Solve the linear equation \(ax + b = 0\) for x.

2. Solve the linear equation \(\dfrac{1}{ax+b}= \dfrac{1}{cx+d}\) for x.
Feedback: 1. The linear equation \(ax + b = 0 \Rightarrow ax=-b \Rightarrow \displaystyle [value]\) for x. 2. The linear equation \(\dfrac{1}{ax+b}= \dfrac{1}{cx+d}\) can be rearranged into the standard form of a linear equation. By multiplying both sides by \( (ax+b)(cx+d) \) we find: \((ax+b)(cx+d)\dfrac{1}{ax+b}= (ax+b)(cx+d)\dfrac{1}{cx+d} \Rightarrow \) \( cx+d= ax+b \Rightarrow cx-ac = b-d\) \( \dfrac{b-d}{c-a}. \)

Exercise

Which of the following are linear equations?

1. [value]
2. [value]
3. [value]
4. [value]
Feedback: 1. [value] is a linaer equation in \(x.\) 2. [value] is a linear equation in \(t.\) 3. [value] is not a linear equation. There is a quadratic term [value]. 4. [value] is a linear equation in \(p\) where the constant term is zero.

To solve a linear equation means to find the value of \(x\) that can be substituted into the equation so that the left-hand side equals the right-hand side. Any such value obtained is known as a solution or root of the equation and the value of \(x\) is said to satisfy the equation.

Example

Consider the linear equation \(3x-2=10\).

1. Check that \(x=4\) is a solution.
2. Check that \(x=2\) is not a solution.

Solution

1. To check that \(x=4\) is a solution we substitute the value for \(x\) and see if both sides of the equation are equal. Evaluating the left-hand side we find \(3(4)-2\) which equals \(10\), the same as the right-hand side. So, \(x=4\) is a solution. We say that \(x=4\) satisfies the equation,
2. Substituting \(x=2\) into the left-hand side we find \(3(2)-2\) which equals \(4\). Clearly the left-hand side is not equal to \(10\) and so \(x=2\) is not a solution. The number \(x=2\) does not satisfy the equation.

Example

Test which of the given values are solutions of the equation \(18-4x=26\)

1. \(x=2\),
2. \(x=-2\),
3. \(x=8\).

Solution

1. Substituting \(x=2\), the left-hand side equals \(18-4\times 2=10\). But \(10 ≠ 26\) so \(x=2\) is not a solution.
2. Substituting \(x=-2\), the left-hand side equals \(18-4\times -2=26\). This is the same as the right hand side so \(x=-2\) is a solution.
3. Substituting \(x=8\), the left-hand side equals \(18-4\times 8=-14\). But \(-14 ≠ 26\) and so \(x=8\) is not a solution.

Exercise

Solve the following equations.

1. \(\displaystyle [value]\), therefore \(x=\)
2. \([value]\), therefore \(x=\)
3. \([value]\), therefore \(x=\)
Feedback: 1. \( \displaystyle [value]\), therefore \(x=\displaystyle [value].\) 2. \([value]\), therefore \(x=\displaystyle [value].\) 3. \([value]\), therefore \(x=[value].\)

Exercise

Solve the following equations.

1. \([value]\), therefore \(x=\)
2. \([value]\), therefore \(x=\)
3. \([value]\), therefore \(x=\)
Feedback: 1. \([value] \Rightarrow [value] = [value]\), therefore \(x= [value].\) 2. \([value] \Rightarrow [value]=[value]\), therefore \(x=[value].\) 3. \([value] \Rightarrow -x - [value]=[value] \Rightarrow [value] = [value]\), therefore \(x=[value].\)

Exercise

Solve the following equations.

1. \( \displaystyle [value] = [value] \), therefore \(m=\)
2. \( \displaystyle{[value] = [value]} \), therefore \(m=\)
Feedback: 1. \( \displaystyle [value] = [value] \Rightarrow [value] = [value]\), therefore \(m= \displaystyle [value].\) 2. \( \displaystyle{[value] = [value]} \Rightarrow [value] = [value](m+1) \), therefore \(m=[value].\)

Exercise

Solve the following equations.

1. \( [value]\), therefore \(x=\)
2. \( [value]\), therefore \(x=\)
Feedback: 1. \( [value]\), therefore \(x=[value].\) 2. \( [value]\), therefore \(x=[value].\)