Introduction to algebra
Once you have an understanding of some key algebraic terms and techniques you can apply them to solve linear equations, as detailed in this page.
In brief, it covers the following:
- The definition of a linear equation
- How to rearrange a linear equation
- How to solve a linear equation in one variable
- How to write a linear equation in slope-intercept form
What is a linear equation?
Recall that an algebraic equation is similar to an algebraic expression, but it contains an equals (\(=\)) sign. For example, \(5 + d\) is an algebraic expression, while \(5 + d = 10\) is an algebraic equation.
A particular kind of algebraic equation is a linear equation, or first degree equation:
A linear equation is an equation where the variables are only raised to the first power. For example \(a\,–\,3 = 5\) is a linear equation, while \(a^2\,–\,3 = 5\) is not.
A linear equation with only one variable is called a linear equation in one variable. A linear equation with two variables is called a linear equation in two variables, and this pattern continues for more variables.
When you have a linear equation in one variable, you can find the value of that variable by solving the equation. To do this, you rearrange the equation so that the variable is by itself on one side of the equals sign. For example, to solve the equation \(c + 2 = 5\) you need to get \(c\) on its own on one side of the equals sign.
You can also solve linear equations that have two or more variables. However, to do this, you need the same number of equations as variables. This process is called solving simultaneous equations. This topic is not covered here, but if you would like to learn more,one example of a page you might find helpful is https://www.mathsisfun.com/algebra/systems-linear-equations.html.
Rearranging linear equations
To solve a linear equation in one variable, you need to rearrange the equation. This means you remove the constants that are attached to the variable. These constants may be added to the variable or multiplied by it.
To remove these constants, you must undo what has been done to the variable by using the opposite operation. For example you might:
| Remove a constant that was… | By… |
|---|---|
| Added | Subtracting it |
| Subtracted | Adding it |
| Multiplied by | Dividing by it |
| Divided by | Multiplying by it |
The important rule to remember is that any operation you perform on one side of the equation must also be performed on the other side. This keeps the equation balanced.
In other words, whatever you do to one side of an equation, you must also do to the other side in the same step. For example:
| On the left hand side (LHS) of \(=\), if you… | On the right hand side (RHS) of \(=\), you must… |
|---|---|
| Add \(2\) (\(+2\)) | Add \(2\) (\(+2\)) |
| Subtract \(5\) (\(-5\)) | Subtract \(5\) (\(-5\)) |
| Multiply by \(7\) (\(\times 7\)) | Multiply by \(7\) (\(\times 7\)) |
| Divide by \(3\) (\(\div 3\)) | Divide by \(3\) (\(\div 3\)) |
Returning to the equation \(c + 2 = 5\), to solve it you need to undo the addition of \(2\). You can do this by subtracting \(2\) from both sides of the equation, as shown below.
Note that for a simple equation like this, you may not need to use these steps. However, these methods are very useful when working with more complex equations. It is best to start with a simple example to understand the process.
\[\begin{aligned} && c + 2 &= 5 \\\ \therefore && c + 2 - 2 &= 5 - 2 && \textrm{ (undo addition by subtracting) } \\\ \therefore && c &= 3 && \textrm{ (simplify - equation is solved!) } \end{aligned}\]Note that the \(\therefore\) sign shown above means ‘therefore’.
At this stage, it is important to understand that as you become more confident in solving equations, you may develop different ways of showing your working that make sense to you. These methods may look a little different from the example above, and that is okay, as long as your working is correct.
For example, you might think of the ‘plus \(2\)’ as moving to the other side of the equation and changing to ‘subtract \(2\)’, as shown below:
\[\begin{aligned} && c + 2 &= 5 \\\ \therefore && c &= 5 - 2 && \textrm{ (addition of }2\textrm{ on left becomes subtraction of }2\textrm{ on right)} &\\\ \therefore && c &= 3 && \textrm{ (simplify - equation is solved!)} \end{aligned}\]Steps for solving linear equations
Often, you need to undo more than one operation to solve a linear equation in one variable. This can take some careful thinking for more complex equations. As mentioned earlier, there can be different ways to solve the same equation. However, the following steps are a useful starting point:
-
If your equation contains brackets, expand them first (if you need help wit this, refer to the Key terms and techniques page of this module).
-
If your equation contains like terms, group them together next. You may need to add or subtract terms from one side of the equation. Then simplify the equation (again, refer to the Key terms and techniques if needed).
- If you need to remove more than one constant from the variable, undo one operation at a time in this order:
- Remove anything that has been added to or subtracted from the variable
- Remove anything that the variable has been multiplied or divided by
- Check that your answer is correct. To do this, substitute the value you found back into the original equation and evaluate it. If the answer is not correct, go back, find where you made a mistake, and work through the steps again.
Example: Solve the linear equation \(10d - 2 = 7(d + 1)\)
Solution: Following the steps above, we have:
\[\begin{aligned} && 10d - 2 &= 7(d + 1) \\\ \therefore && 10d - 2 &= 7d + 7 && \textrm{ (expand brackets)} \\\ \therefore && 10d - 2 - 7d &= 7d + 7 - 7d && \textrm{ (remove variable from RHS of equation by subtracting)} \\\ \therefore && 3d - 2 &= 7 && \textrm{ (simplify)} \\\ \therefore && 3d - 2 + 2 &= 7 + 2 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3d &= 9 && \textrm{ (simplify)} \\\ \therefore && 3d/3 &= 9/3 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 3 && \textrm{ (simplify - equation is solved!)} \end{aligned}\]Finally, substituting \(3\) back into the original equation gives \((10)(3) - 2 = 30 - 2 = 28\) on the left hand side, and \(7(3 + 1) = 7(4) = 28\) on the right hand side. So the solution is correct.
Once you have worked through the example above, have a go at solving some or all of the following linear equations:
\(\begin{aligned} && c + 3 &= 9 \\\ \therefore && c + 3 - 3 &= 9 - 3 && \textrm{ (undo addition by subtracting)} \\\ \therefore && c &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && \frac{a}{3} &= 2 \\\ \therefore && 3(\frac{a}{3}) &= 3(2) && \textrm{ (undo division by multiplying)} \\\ \therefore && a &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 3p + 2 &= 17 \\\ \therefore && 3p + 2 - 2 &= 17 - 2 && \textrm{ (undo addition by subtracting)} \\\ \therefore && 3p &= 15 && \textrm{ (simplify)} \\\ \therefore && \frac{3p}{3} &= \frac{15}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && p &= 5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 3y - 5 &= 13 \\\ \therefore && 3y - 5 + 5 &= 13 + 5 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3y &= 18 && \textrm{ (simplify)}& \\\ \therefore && \frac{3y}{3} &= \frac{18}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && p &= 6 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 3a + 4 &= a + 6 \\\ \therefore && 3a + 4 - a &= a + 6 - a && \textrm{ (remove variable from RHS by subtracting)} \\\ \therefore && 2a + 4 &= 6 && \textrm{ (simplify)} \\\ \therefore && 2a + 4 - 4 &= 6 - 4 && \textrm{ (undo addition by subtracting)} \\\ \therefore && 2a &= 2 && \textrm{ (simplify)} \\\ \therefore && \frac{2a}{2} &= \frac{2}{2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 1 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 6n &= 20 + 2n \\\ \therefore && 6n - 2n &= 20 + 2n - 2n && \textrm{ (remove variable from RHS by subtracting)} \\\ \therefore && 4n &= 20 && \textrm{ (simplify)} \\\ \therefore && \frac{4n}{4} &= \frac{20}{4} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && n &= 5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 5a - 5 + 10a &= -5 \\\ \therefore && 15a - 5 &= -5 && \textrm{ (simplify like terms)} \\\ \therefore && 15a - 5 + 5 &= -5 + 5 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 15a &= 0 && \textrm{ (simplify)} \\\ \therefore && \frac{15a}{15} &= \frac{0}{15} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 0 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 5(d - 2) &= 10 \\\ \therefore && 5d - 10 &= 10 && \textrm{ (expand brackets)} \\\ \therefore && 5d - 10 + 10 &= 10 + 10 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 5d &= 20 && \textrm{ (simplify)} \\\ \therefore && \frac{5d}{5} &= \frac{20}{5} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 4 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && 3(d - 4) &= 12 \\\ \therefore && 3d - 12 &= 12 && \textrm{ (expand brackets)} \\\ \therefore && 3d - 12 + 12 &= 12 + 12 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 3d &= 24 && \textrm{ (simplify)} \\\ \therefore && \frac{3d}{3} &= \frac{24}{3} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && d &= 8 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && \frac{2a - 1}{6} &= 4 \\\ \therefore && 2a - 1 &= 24 && \textrm{ (simplify)} \\\ \therefore && 2a - 1 + 1 &= 24 + 1 && \textrm{ (undo subtraction by adding)} \\\ \therefore && 2a &= 25 && \textrm{ (simplify)} \\\ \therefore && \frac{2a}{2} &= \frac{25}{2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && a &= 12.5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
\(\begin{aligned} && \frac{-2t + 1}{3} + 2 &= 6 \\\ \therefore && \frac{-2t + 1}{3} + 2 - 2 &= 6 - 2 && \textrm{ (undo addition by subtracting)} \\\ \therefore && \frac{-2t + 1}{3} &= 4 && \textrm{ (simplify)} \\\ \therefore && 3(\frac{-2t + 1}{3}) &= 3(4) && \textrm{ (undo division by multiplying)} \\\ \therefore && -2t + 1 &= 12 && \textrm{ (simplify)} \\\ \therefore && -2t + 1 - 1 &= 12 - 1 && \textrm{ (undo addition by subtracting)} \\\ \therefore && -2t &= 11 && \textrm{ (simplify)} \\\ \therefore && \frac{-2t}{-2} &= \frac{11}{-2} && \textrm{ (undo multiplication by dividing)} \\\ \therefore && t &= -5.5 && \textrm{ (simplify - equation is solved!)} \end{aligned}\)
Slope-intercept form
When you graph a linear equation with two variables, it forms a straight line. The most common way to write this type of equation is in a form that clearly shows two important features of the line. This is called slope-intercept form. It is written as follows:
\[y = mx + c\]where:
- \(m\) is the gradient or slope of the line (for more information about gradient, see the Derivatives page in the Introduction to calculus module)
- \(c\) is the y-intercept, which is the point where the line crosses the y-axis (the vertical axis).
For example, the following linear equations are all in slope-intercept form:
\(y = 2x + 3\)
\(y = 4x - 5\)
\(y = -5x + 1\)
The graphs of these equations are shown below. Note that each graph has a different gradient and a different y-intercept.
If a linear equation in two variables is not written in slope-intercept form, you can rearrange it into this form by using the same techniques you use to solve linear equations.
Example: Rearrange the linear equation \(6x + 2y = 10\) to be in slope-intercept form.
Solution: Rearranging using the steps for solving linear equations gives:
\[\begin{aligned} && 6x + 2y &= 10 \\\ \therefore && 6x + 2y - 6x &= 10 - 6x && \textrm{ (remove } 6x \textrm{ from LHS of equation by subtracting)} \\\ \therefore && 2y &= 10 - 6x && \textrm{ (simplify)} \\\ \therefore && 2y/2 &= 10/2 - 6x/2 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= 5 - 3x && \textrm{ (simplify)} \\\ \therefore && y &= - 3x + 5&& \textrm{ (rearrange)} \end{aligned}\]Once you have worked through the example above, try some or all of the following problems that involve linear equations in slope-intercept form:
The gradient of this line is \(3\) and the y-intercept is \(-4\)
The gradient of this line is \(-0.4\) and the y-intercept is \(7\)
\(\begin{aligned} && -12x &= 6 - 3y \\\ \therefore && -12x + 3y &= 6 - 3y + 3y && \textrm{ (remove }-3y \textrm{ from RHS of equation by adding} \\\ \therefore && -12x + 3y &= 6 && \textrm{ (simplify)} \\\ \therefore && -12x + 3y + 12x &= 6 + 12x && \textrm{ (remove }-12x \textrm{ from LHS of equation by adding} \\\ \therefore && 3y &= 6 + 12x && \textrm{ (simplify)} \\\ \therefore && 3y/3 &= 6/3 + 12x/3 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= 4x + 2 && \textrm{ (simplify and rearrange)} \\\ \end{aligned}\)
\(\begin{aligned} && 4y + 20x + 8 &= 0 \\\ \therefore && 4y + 20x + 8 - 8 &= 0 - 8 && \textrm{ (remove } 8 \textrm{ from LHS of equation by subtracting} \\\ \therefore && 4y + 20x &= -8 && \textrm{ (simplify)} \\\ \therefore && 4y + 20x - 20x &= -8 - 20x && \textrm{ (remove } 20x \textrm{ from LHS of equation by subtracting} \\\ \therefore && 4y &= -8 - 20x && \textrm{ (simplify)} \\\ \therefore && 4y/4 &= -8/4 - 20x/4 && \textrm{ (undo multiplication by dividing)} \\\ \therefore && y &= -5x - 2 && \textrm{ (simplify and rearrange)} \\\ \end{aligned}\)