1.18: Solving Linear Equations

The kind of the equation will determine the method we use to solve it. We will first discuss linear equations. These are equations that only contain the first power of a variable and nothing higher.

Example 16.1

Examples of linear equations:

1. \(x-4=6\) is a linear equation.
2. \(5 x-6=4 x+2\) is a linear equation.
3. \(x^{2}-2 x+1=0\) is not a linear equation, since the variable \(x\) is to the second power. This is a quadratic equation which we will study in chapter 20.

Critical Observation: We can add or subtract anything from an equation as long as we do it to both sides at the same time. This is a very essential tool to solve linear equations. It will help us isolate the variable on one side of the equation and the numbers on the other side of the equation.

If \(\quad a=b \quad\) then \(\quad a+c=b+c\).

If \(\quad a=b \quad\) then \(\quad a-c=b-c\).

Example 16.2

Isolate the variable in the given equation:

a) \(x-4=6\)

Here we add 4 to both sides of the equation to get

\[x-4+4=6+4\nonumber\]

which has the effect of isolating the \(x\) on one side of the equation and the numbers on the other since upon simplifying we see that

\[x=10\nonumber\]

It can be helpful to write this in a vertical form:

\[x-4=6\nonumber\]

\[+4 \quad+4\nonumber]\]

\[\Longrightarrow x=10\]

b) \(x+7=-2 .\) Here we add -7 from both sides because it will have the effect of isolating the \(x:\) (vertically written)

\[x+7=-2\nonumber\]

\[+-7 \quad-7\nonumber\]

\[\Longrightarrow x=-9\nonumber\]

c) \(5 x-6=4 x+2\)

Here \(x\) appears on both sides of the equation. If we subtract one of the terms from both sides, it will have the effect of isolating the \(x\) on one side.

We have a choice. We will subtract \(4 x\) from both sides so that an \(x\) is on the LHS. The alternative would have been to subtract \(5 x\) which would have left us with \(-x\) on the \(\mathrm{RHS}\) ( this would be somewhat inconvenient). We have (vertically written)

\[5 x-6=4 x+2\nonumber\]

\[-4 x-4 x\nonumber\]

\[\Longrightarrow x-6=2\nonumber\]

\[\quad+6 \quad+6\nonumber\]

\[\Longrightarrow x=8\nonumber\]

Note that each solution can be checked by plugging the number found into the original equation.

Example 16.3

Solve:

a) \(17-(4-2 x)=3(x+4)\)

To solve this equation, we first need to remove all parentheses and combine any like terms.

\[17-(4-2 x)=3(x+4)\nonumber\]

\[\Longrightarrow 17-4+2 x=3 x+12\nonumber\]

\[\Longrightarrow 13+2 x=3 x+12\nonumber\]

Following the example above, the solution is found (by subtracting \(2 x\) from both sides and subtracting 12 from both sides) to be \(x=1 .\) Now, we can check if our work is correct by substituting \(x=1\) in the original equation and seeing whether the \(\mathrm{RHS}\) and \(\mathrm{LHS}\) yield the same value:

RHS: \(17-(4-2 x)=17+(-1)(4+(-2 x))=17+(-1)(4+(-2 \cdot 1))=17+(-1)(4+(-2))=17+(-1)(2)=17+(-2)=15\)

LHS: \(3(x+4)=3(1+4)=3(5)=15\)

Since both values are equal, our solution of \(x=1\) is correct.

Critical Observation: We can multiply or divide an equation by any nonzero number as long as we do it to both sides at the same time. This is a very essential tool to solve linear equations where the coefficient of the variable is not 1.

If \(a=b\) then \(a \times c=b \times c\).

If \(a=b\) then \(\frac{a}{c}=\frac{b}{c} \quad\) when \(c \neq 0\).

Example 16.4

a) \(6 x=42\)

\[6 x=42\nonumber\]

\[\Longrightarrow \frac{6 x}{6}=\frac{42}{6}\nonumber\]

\[\Longrightarrow x=7\nonumber\]

b) \(-4 x-30=0\)

In this example we will first isolate the ' \(x\) -term' which is \(4 x\) before isolating \(x\).

\[-4 x-30=0\nonumber\]

\[\quad+30 \quad +30\nonumber\]

\[\Longrightarrow-4 x=30\nonumber\]

\[\Longrightarrow \frac{-4 x}{-4}=\frac{30}{-4}\nonumber\]

\[\Longrightarrow x=-\frac{15}{2}\nonumber\]

c) \(2 x=\frac{1}{4}\)

\[2 x=\frac{1}{4}\nonumber\]

\[\Longrightarrow \frac{2 x}{2}=\frac{1}{4} \div 2\nonumber\]

\[\Longrightarrow x=\frac{1}{4} \cdot \frac{1}{2}\nonumber\]

\[\Longrightarrow x=\frac{1 \cdot 1}{4 \cdot 2}\nonumber\]

\[\Longrightarrow x=\frac{1}{8}\nonumber\]

Note that dividing by 2 on both sides of the equation is the same as multiplying by \(\frac{1}{2}\). So, we can rewrite the solution like this:

\[2 x=\frac{1}{4}\nonumber\]

\[\Longrightarrow \frac{1}{2} \cdot 2 x=\frac{1}{2} \cdot \frac{1}{4}\nonumber\]

\[\Longrightarrow \frac{1 \cdot 2 x}{2}=\frac{1 \cdot 1}{4 \cdot 2}\nonumber\]

\[\Longrightarrow x=\frac{1}{8}\nonumber\]

Generally, multiplying a number by its reciprocal results in 1:

\[\frac{a}{b} \cdot \frac{b}{a}=\frac{a b}{b a}=1\nonumber\]

Let's use this fact in the next example.

d) \(\frac{2 x}{3}=\frac{5}{6}\)

\[\frac{2 x}{3}=\frac{5}{6}\nonumber\]

\[\Longrightarrow \frac{2}{3} \cdot x=\frac{5}{6}\nonumber\]

\[\Longrightarrow \frac{2}{3} \cdot x=\frac{5}{6}\nonumber\]

\[\Longrightarrow \frac{3}{2} \cdot \frac{2}{3} \cdot x=\frac{3}{2} \cdot \frac{5}{6}\nonumber\]

\[\Longrightarrow x=\frac{3 \cdot 5}{2 \cdot 6}\nonumber\]

\[\Longrightarrow x=\frac{5}{2 \cdot 2}\nonumber\]

\[\Longrightarrow x=\frac{5}{4}\nonumber\]

e) \(\frac{x}{5}+3=6\)

\[\frac{x}{5}+3=6\nonumber\]

\[\quad -3 \quad-3\nonumber\]

\[\Longrightarrow \frac{x}{5}=3\nonumber\]

\[\Longrightarrow 5 \cdot \frac{x}{5}=5 \cdot 3\nonumber\]

\[\Longrightarrow x=15\nonumber\]

f) \(5 x-6=2 x+3\)

\[5 x-6=2 x+3\nonumber\]

\[\quad+6 \quad +6 \quad\nonumber\]

\[\Longrightarrow 5 x=2 x+9\nonumber\]

\[\Longrightarrow-2 x \quad -2 x\nonumber\]

\[\Longrightarrow 3 x=9\nonumber\]

\[\Longrightarrow \frac{3 x}{3}=\frac{9}{3}\nonumber\]

\[\Longrightarrow x=3\nonumber\]

g) \(-3(x-1)=4(x+2)+2\)

We first remove the parentheses and collect like terms.

\[-3(x-1)=4(x+2)+2\nonumber\]

\[\Longrightarrow-3 x+3=4 x+8+2\nonumber\]

\[\Longrightarrow-3 x+3=4 x+10\nonumber\]

Now we proceed to solve the linear equation by isolating the variable:

\[-3 x+3=4 x+10\nonumber\]

\[\quad -3 \quad-3 \quad \nonumber\]

\[\Longrightarrow-3 x=4 x+7\nonumber\]

\[\quad -4 x \quad -4 x \quad \nonumber\]

\[\Longrightarrow-7 x=7\nonumber\]

\[\Longrightarrow \frac{-7 x}{-7}=\frac{7}{-7}\nonumber\]

\[\Longrightarrow x=-1\nonumber\]

h) \(10-3 x=-2(x-1)\)

\[10-3 x=-2(x-1)\nonumber\]

\[\Longrightarrow 10-3 x=-2 x+2\nonumber\]

\[\quad -10 \quad \quad-10\nonumber\]

\[\Longrightarrow-3 x=-2 x-8\nonumber\]

\[\quad +2x \quad + 2 x \quad \nonumber\]

\[\Longrightarrow-x=-8\nonumber\]

\[\Longrightarrow x=8\nonumber\]

Exit Problem

Solve: \(5 y-(7-2 y)=2(y+4)\)