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Article objectives:

  • The objective of this article is to explain how to solve linear equations with one variable.

Introduction

A linear equation is an equation where the exponents of variables are all 1. If the exponents are anything but 1 (0 doesn't count because then a constant results), the equation is not linear. Linear equations are easier to solve than most others, especially when only one variable is involved.

Example 1a: The equation \(5x + 3 = 5\) is linear since the terms with \(x\)s are of degree 1.

Example 1b: The equation \(x^2 + 5x = 3x - 3\) is not linear because \(x^2\) is of degree 2, not degree 1. The presence of linear terms is irrelevant.

Example 1c: The equation \(8 = 8\) is not linear because there are no variables written in the equation, let alone variables of degree 1.

To solve linear equations, typically one or more of the following operations are needed: addition, subtraction, multiplication, or division. Whatever operation(s) are chosen must be applied to both sides of the equation, so the sides remain equal. Otherwise all subsequent work will be meaningless.

Addition

We begin with an example where addition is required.

Example 2: Solve this equation:

$$x - 7 = 4$$

Solution: Solving for \(x\) consists of getting it on one side of the equation by itself. To do this, the "-7" must be removed, by adding 7 to both sides of the equation:

$$x - 7 + 7 = 4 + 7 \Rightarrow$$ $$x = 11$$

Subtraction

Solving linear equations with subtraction is very similar to using addition.

Example 3: Solve this equation for \(x\):

$$x + 8 = 13$$

Solution: \(x\) needs to be isolated. Removing the \(+8\) does this. Remove it by subtracting \(8\) from both sides of the equation:

$$x + 8 - 8 = 13 - 8 \Rightarrow$$ $$x = 5$$

Notice how \(8 - 8 = 0\), so the \(x\) remains the only thing on the left side of the equation.

Multiplication

Multiplying to solve an equation is only helpful when fractions are involved. So equations with no fractions such as \(x - 5 = 8\) and \(2x + 3 = 44\) are not going to be solved with multiplication!

Example 4: Solve this equation:

$$\frac{z}{4} = 2$$

Solution: First of all, don't panic since the variable is not \(x\). The procedure is the same, just solving for a different variable. Anyway, to see what to do, rewrite the left side as follows:

$$\frac{1}{4} \times z = 2$$

Since \(\frac{1}{4} \times 4 = 1\), multiplying both sides of the equation by \(4\) isolates \(z\):

$$\frac{1}{4} \times 4 \times z = 2 \times 4 \Rightarrow$$ $$z = 8$$

Example 5: The equation \(5x - 3 = 8\) cannot be solved with multiplication, because \(x\) is not multiplied by a fraction, but a constant, so other techniques are necessary.

Division

The equation in Example 5 would have required division to solve. However, it also requires addition since there are constants on both sides of the equation. Similar problems with multiple steps will be presented later.

In arithmetic, it is the backwards process of multiplication; the same principle goes for solving equations. Therefore the processes of solving equations with multiplication and division are similar. Using division can be treated like multiplying by a fraction, so that is the approach that will be used here.

Example 6: Solve this equation:

$$8x = -16$$

Solution: As usual the goal is to isolate \(x\). Multiply both sides of the equation by \(\frac{1}{8}\) since its coefficient will be \(1\):

$$8x \times \frac{1}{8} = -16 \times \frac{1}{8} \Rightarrow$$ $$x = -2$$

Solving in Multiple Steps

Often a combination of addition, subtraction, multiplication, and division are necessary to solve an equation. Sometimes an "x" will appear on both sides of the equation, and all of the "x"s need to be moved to the same side of the equation to get a definitive value for \(x\).

Example 7: Solve this equation:

$$x -5 = -x + 4$$

Solution: Begin by moving all of the \(x\)s to the same side of the equation, by adding \(x\) to both sides of the equation:

$$x + x - 5 = -x + x + 4 \rightarrow$$ $$2x - 5 = 4$$

Now, isolate the \(x\) by adding both sides of the equation by \(5\):

$$2x = 9$$

Finally divide both sides of the equation by \(2\) to get a definitive value for \(x\):

$$x = \frac{9}{2}$$

The final two examples involve fractions.

Example 8: Solve this equation for \(x\):

$$\frac{x}{4} + 7 = \frac{1}{4}$$

Solution A: There are multiple ways to begin, actually. The specific path is up to your preference. The first solution is to begin by isolating \(x\) directly, and then just get rid of everything else on the left side of the equation:

$$\frac{x}{4} = \frac{1}{4} - 7 \Rightarrow$$ $$\frac{x}{4} = \frac{1}{4} - \frac{28}{4} \Rightarrow$$ $$\frac{x}{4} = -\frac{27}{4} \Rightarrow$$ $$x = -27$$

Solution B: Some people prefer to eliminate fractions at the very beginning instead. To do this, multiply both sides of the equation by \(4\):

$$x + 28 = 1$$

Now just subtract \(28\) from both sides of the equation:

$$x = -27$$

The second solution is slightly easier, but both are perfectly viable, as long as you are careful to perform your algebraic manipulations correctly.

This final example is not initially a linear equation, but can be transformed into one and then be solved like one, just like the others.

Example 9: Solve this equation for \(x\):

$$7 = \frac{2}{x} - 4$$

Solution: Before messing with the fraction, add \(4\) to both sides of the equation to get all the constants on the same side:

$$11 = \frac{2}{x}$$

Here we multiply both sides of the equation by \(x\) to get \(x\) out of the denominator:

$$11x = 2$$

Finally isolate \(x\) by dividing both sides of the equation by \(11\):

$$x = \frac{2}{11}$$

There are more forms of linear equations than you may expect, but they are all rather simple when you consider a few straightforward algebraic manipulations are all you need. When the degrees of the terms get higher things get much more complicated quickly.