What are linear equations and inequalities?

Linear equations and inequalities are composed of constants and variables.

• Linear equations use the equal sign ($=$). (For example, $2x+1=5$ is a linear equation.)
• Linear inequalities use inequality signs ($>$, $<$, $\geq$, and $\leq$). (For example, $2x+1>5$ is a linear inequality)

Types of linear equations

The goal of solving a linear equation is to find the value of a variable; we isolate the variable step by step until only the variable is on one side of the equation and only a constant is on the other.

When solving linear equations, the most important thing to remember is that the equation will remain equivalent to the original equation only if we always treat both sides equally: whenever we do something to one side, we must do the exact same thing to the other side.

Linear Equations in One Variable

These examples walk through isolating $x$ in a one-variable linear equation, including cases that require combining like terms or distributing a coefficient.

1. If $2x+1=5$, what is the value of $x$?

Solution: Subtract $1$ from both sides to get $2x=4$, then divide both sides by $2$:

$$2x+1-1 = 5-1 \\ 2x = 4 \\ x = 2$$

2. If $2x-4=5-x$, what is the value of $x$?

Solution: Add $4$ to both sides, then add $x$ to both sides to collect the variable terms, and finally divide by the resulting coefficient:

$$2x-4+4 = 5-x+4 \\ 2x = 9-x \\ 2x + x = 9-x+x \\ 3x = 9 \\ x = 3$$

Recall that combining like terms follows the rule $ax \pm bx=(a\pm b)x$.

3. If $2(x+1)=5$, what is the value of $x$?

Solution: Distribute the $2$ to both terms inside the parentheses ($a(bx\pm c)=abx\pm ac$), then isolate $x$:

$$2(x)+2(1) = 5 \\ 2x+2 = 5 \\ 2x+2-2 = 5-2 \\ 2x = 3 \\ x = \frac{3}{2}$$

Equations with Fractions and Negative Numbers

Fractions and negative numbers can make linear equations trickier to solve. If an equation has only a fraction coefficient, it's often easiest to leave the fraction until the last step. If it has both fraction coefficients and fraction constants, clearing the fractions in the first step usually simplifies things.

1. If $\frac{1}{2}x+3=5$, what is the value of $x$?

Solution: Subtract $3$ from both sides, then multiply both sides by $2$ to clear the fraction last:

$$\frac{1}{2}x+3-3 = 5-3 \\ \frac{1}{2}x = 2 \\ \frac{1}{2}x \cdot 2 = 2 \cdot 2 \\ x = 4$$

2. What is the solution to the equation $\frac{1}{2}x+\frac{1}{3}=\frac{1}{5}$?

Solution: Multiply every term by $30$, the smallest common multiple of the denominators, to clear all the fractions immediately:

$$\left(\frac{1}{2}x+\frac{1}{3}\right)\cdot 30 = \frac{1}{5}\cdot 30 \\ 15x+10 = 6 \\ 15x+10-10 = 6-10 \\ 15x = -4 \\ x = -\frac{4}{15}$$

3. If $-2(x-5)=1$, what is the value of $x$?

Solution: Distribute $-2$ to both $x$ and $-5$, remembering that negative x negative positive, so $-2(-5)=10$:

$$-2(x)-2(-5) = 1 \\ -2x+10 = 1 \\ -2x+10-10 = 1-10 \\ -2x = -9 \\ x = \frac{9}{2}$$

Linear Equations in Two Variables

When an equation has two variables and the value of one of them is given, substitute the known value into the equation and solve for the other.

1. If $2x+5y=1$ and $x=3$, what is the value of $y$?

Solution: Substitute $x=3$ into the equation and solve for $y$:

$$2(3)+5y = 1 \\ 6+5y = 1 \\ 6+5y-6 = 1-6 \\ 5y = -5 \\ y = -1$$

Evaluating Expressions Using Linear Equations

Sometimes you're given a linear equation in one variable and asked to evaluate a different expression containing that variable. You can solve for the variable first and substitute it in, or — often faster — recognize the relationship between the equation and the expression and substitute directly.

1. If $2x+1=5$, what is the value of $8x+4$?

Solution: Solving $2x+1=5$ gives $x=2$, and substituting into $8x+4$ gives $8(2)+4=20$.

A quicker route: notice that $8x+4=4(2x+1)$. Since $2x+1=5$, substitute $5$ in directly:

$$8x+4 = 4(2x+1) \\ = 4(5) \\ = 20$$

Linear Inequalities

Solving inequalities works the same way as solving equations, with one key difference: dividing both sides by a negative number reverses the direction of the inequality sign.

1. What values of $x$ satisfy the inequality $2x+1>5$?

Solution: The coefficient of $x$ is positive, so the inequality sign keeps its direction:

$$2x+1-1 > 5-1 2x > 4 \\ x > 2$$

Checking $x=3$ (which satisfies $x>2$): $2(3)+1=7>5$ ✓. Checking $x=1$ (which does not): $2(1)+1=3$, and $3\ngtr 5$.

2. What values of $x$ satisfy the inequality $-2x+1>5$?

Solution: The coefficient of $x$ is negative, so the inequality sign reverses when dividing:

$$-2x+1-1 > 5-1 \\ -2x > 4 \\ x < -2$$

Checking $x=-3$ (which satisfies $x < -2$): $-2(-3)+1=7>5$ ✓. Checking $x=1$ (which does not): $-2(1)+1=-1$, and $-1\ngtr 5$.

How Many Solutions Can a Linear Equation Have?

Most linear equations have exactly one solution. But an equation can also have no solution or infinitely many solutions, depending on how its constants are set up:

• If the equation reduces to $x=a$ for some constant $a$, it has one solution.
• If the variable cancels out and what remains is $a=b$ for two different constants, it has no solution.
• If the equation reduces to $x=x$, it has infinitely many solutions.

1. In the equation $2x-1=ax-2$, if $a=2$, what value of $x$ satisfies the equation?

Solution: Substituting $a=2$ gives $2x-1=2x-2$. Subtracting $2x$ from both sides leaves $-1=-2$, which is never true — so no value of $x$ satisfies the equation.

2. In the equation $2x-4=a(x-2)$, if $a=2$, what value of $x$ satisfies the equation?

Solution: Substituting $a=2$ gives $2x-4=2(x-2)=2x-4$. Adding $4$ to both sides leaves $2x=2x$, which is always true — so all real values of $x$ satisfy the equation.

Things to Remember

• When solving linear equations, whatever you do to one side, you must do to the other, to keep the equation equivalent to the original.
• When solving linear inequalities:

  • If the coefficient of $x$ is positive, the inequality sign keeps its direction when dividing.
  • If the coefficient of $x$ is negative, the inequality sign reverses when dividing.
• When determining the number of solutions to a linear equation:

  • $x=a$ (constant) → one solution.
  • $a=b$ (different constants, variable eliminated) → no solution.
  • $x=x$ → infinitely many solutions.

Practice Problems