# Solving Linear Equations

Loosely speaking, a linear equation is an equality statement containing only addition, subtraction, multiplication, and division. It does not need to include all of these operations, but it cannot include operations beyond them, such as exponentiation.

For example, these are linear equations:

$5x+9=5$
$14x-6=3x+2$
$-5x+2+7x=3x+8-x$

On the other hand, these are not linear equations:

$5x^2+9=5$
$14x-6=3 \sqrt{x} + 2$
$-5 \sin(x) + 2 + 7x = 3 |x| + 8$

The solution of a linear equation is the value that we can substitute for the variable to make the equation true.

Most linear equations have a single solution. We can find the solution by performing operations on both sides of the equation, to isolate the variable.

\begin{align*} \text{Given equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 5x+8=-2x+22 \\ \text{Add } 2x \text{ to both sides} \hspace{.5cm} &\Bigg| \hspace{.5cm} 7x+8=22 \\ \text{Subtract } 8 \text{ from both sides} \hspace{.5cm} &\Bigg| \hspace{.5cm} 7x=14 \\ \text{Divide both sides by } 7 \hspace{.5cm} &\Bigg| \hspace{.5cm} x=2 \end{align*}

To check our solution, we can substitute it in both sides of the equation and check that they evaluate to the same result:

\begin{align*} 5(2)+8 &= -2(2)+22 \\ 10+8 &= -4+22 \\ 18 &= 18 \end{align*}

However, some linear equations have no solutions. When we try to solve these equations, the variable vanishes and we are left with an untrue statement.

\begin{align*} \text{Given equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} 3x+1=2+3x \\ \text{Subtract } 2x \text{ from both sides} \hspace{.5cm} &\Bigg| \hspace{.5cm} 1=2 \end{align*}

This means that there is no number we can substitute for $x$ to make the given equation true. In fact, the right-hand side will always be $1$ more than the left-hand side: the left-hand side says to multiply the input by $3$ and add $1$, while the left-hand side says to multiply the input by $3$ and add $2$. Both sides multiply the input by $3$, but then add different amounts! We can never hope to get the results to be the same.

Even more interesting, some linear equations have infinitely many solutions. When we try to solve these equations, the variable still vanishes, but this time we are left with a true statement.

\begin{align*} \text{Given equation} \hspace{.5cm} &\Bigg| \hspace{.5cm} -2x+1=1-2x \\ \text{Add } 2x \text{ to both sides} \hspace{.5cm} &\Bigg| \hspace{.5cm} 1=1 \end{align*}

In other words, any number we substitute for $x$ will make the given equation true. The left-hand side and the right-hand side will always come out to the same result: the left-hand side tells us to multiply the input by $-2$ and add $1$, and the right-hand side tells us to multiply the input by $2$ and then subtract it from $1$. These are really just two ways of saying the same thing.

Practice Problems

Solve the following equations. (You can view the solution by clicking on the problem.)

$1) \hspace{.5cm} 2x+3=11$
Solution:
$x=4$

$2) \hspace{.5cm} -3x+7=6-2x$
Solution:
$x=1$

$3) \hspace{.5cm} -5+12=3x-2-x$
Solution:
$x=2$

$4) \hspace{.5cm} -5x+2=2x-1-7x$
Solution:
$\text{no solution}$

$5) \hspace{.5cm} -3x-17-12x=-5x+13$
Solution:
$x=-3$

$6) \hspace{.5cm} 18-x+1=10x+19-11x$
Solution:
$\text{infinitely many solutions}$

$7) \hspace{.5cm} 9(x+4) + 12(x-4) = 84$
Solution:
$x=5$

$8) \hspace{.5cm} 4(2x+3)-x=3(2x+4)+x$
Solution:
$\text{infinitely many solutions}$

$9) \hspace{.5cm} 5 \left[ 4(3x+1) - 3(3x+2) \right] + 25 = -75$
Solution:
$x=-6$

$10) \hspace{.5cm} 3 \left[ 2(x+1) - (x+2) \right] = 3x+1$
Solution:
$\text{no solution}$