Linear equations and inequalities

This chapter applies to all AMC 8/10/12 test takers.

A linear equation describes a straight line on a graph. It usually involves one or two variables, and the variables are only to the first power (no exponents like $x^2$).

A system of linear equations is a set of two or more linear equations. Solving the system means finding values of the variables that make every equation true. Geometrically, that solution is the point where the lines intersect.

A linear inequality looks like a linear equation, but it uses symbols like $<$, $>$, $\le$, or $\ge$. On a graph, the boundary line separates the plane into two regions:

• One region where the inequality is true
• One region where it’s false

If the inequality is $\le$ or $\ge$, then points on the boundary line also satisfy the inequality.

The geometry connection will come up later. For now, we’ll focus on solving these problems algebraically.

Solving linear equations

Most linear equations are solved by isolating the variable. You do the same operation to both sides until the variable is alone.

That direct approach is most useful when the question really does ask for a single unknown. In the example below, it looks like there are two unknown test scores, but to make one score as low as possible, you can make the other score as high as possible (set it to $100$). Then you only have one unknown left.

Example: The question below is from 2019 AMC 8

Shauna takes five tests, each worth a maximum of $100$ points. Her scores on the first three tests are $76$ , $94$ , and $87$ . In order to average $81$ for all five tests, what is the lowest score she could earn on one of the other two tests?
A. $48$
B. $52$
C. $66$
D. $70$
E. $74$

Some problems don’t ask for $x$ or $y$ directly. Instead, they ask for an expression like $\frac{x+3y}{3x-y}$. In those cases, you often don’t need to find $x$ and $y$ separately - you just need enough information to rewrite the expression you’re asked for.

Example: The question below is from 2017 AMC 10B

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?
A. $-3$
B. $-1$
C. $1$
D. $2$
E. $3$

Solving linear inequalities

You solve inequalities the same way you solve equations, with one key rule:

• If you multiply or divide both sides by a negative number, you must flip the inequality sign.

Translating words into inequality symbols is also important:

• “no more than” means $\le$
• “at least” means $\ge$
• “between” two numbers often becomes a compound inequality like $0<x<10$

That last example means $x$ is greater than $0$ and less than $10$.

In the next problem, you can set up one inequality for each half of the game by comparing Candace’s shooting percentage to Steph’s.

Example: The question below is from 2022 AMC 8

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?

	First half	Second half
Steph	$\frac{15}{20}$	$\frac{10}{10}$
Candace	$\frac{\square }{12}$	$\frac{\square }{18}$

A. $7$
B. $8$
C. $9$
D. $10$
E. $11$

Word problems

Many AMC algebra word problems need to be translated into an equation (or a system of equations) before you can solve them. Here are some common translations to keep in mind.

Word	Algebraic meaning
is	$=$
more than	$+$
less than	$-$
twice $a$	$2a$
half of $a$	$a/2$
$x$ percent of $a$	$(x/100)a$

Try translating the next problem into a single equation with two variables, then isolate the ratio the question asks for.

Example: The question below is from 2019 AMC 10B

Alicia had two containers. The first was$\frac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\frac{3}{4}$full of water. What is the ratio of the volume of the first container to the volume of the second container?
A. $\frac{5}{8}$
B. $\frac{4}{5}$
C. $\frac{7}{8}$
D. $\frac{9}{10}$
E. $\frac{11}{12}$

You can also translate work-and-rate problems into equations using the following formula. You can rearrange it by dividing both sides by either rate or time to isolate the quantity you need.

$$\text{Work}=\text{Rate(Time)}$$

Solving systems of equations

How do you solve a system of linear equations?

There are two fundamental techniques used for systems of linear equations (SOLE). Sometimes you’ll use them to solve for a specific variable (like $x$). Other times, you’ll use them to find an expression the question asks for (like $2x+y$) without fully solving the system.

The two fundamental techniques used for SOLE equations are:

1. Isolate and substitute
2. Add or subtract the equations

SOLE technique #1: Isolate and substitute

The idea behind isolate and substitute is:

• Rewrite one equation so one variable is alone.
• Substitute that expression into the other equation.
• Solve the resulting one-variable equation.

Which of the following could be true?

Given:
$x+3y=12$
$y+3x=15$> Select all that apply.
A. $x>y$
B. $y>x$
C. $y=x$
D. $21y=33x$
E. $21x=33y$

The first step is to isolate one variable. Isolate $x$ in the first equation:

$$\begin{array}{rl}x+3y& =12\\ x& =12-3y\end{array}$$

Now substitute $x=12-3y$ into the second equation:

$$\begin{array}{rl}y+3x& =15\\ y+3(12-3y)& =15\end{array}$$

Solve for $y$:

$$\begin{array}{rl}y+3(12-3y)& =15\\ y+36-9y& =15\\ 36-8y& =15\\ 36& =15+8y\\ 21& =8y\\ 21/8& =y\end{array}$$

Now plug $y=21/8$ back into the first equation to find $x$:

$$\begin{array}{rl}x+3y& =12\\ x+3(21/8)& =12\\ x+63/8& =12\\ x& =12-63/8\\ x& =96/8-63/8\\ x& =33/8\end{array}$$

So the solution is:

$$\begin{array}{rl}x& =33/8\\ y& =21/8\end{array}$$

Now check each answer choice:

So we get:

Select all that apply.
A.$33/8>21/8$ TRUE
B. $21/8>33/8$ FALSE
C. $33/8=21/8$ FALSE
D. $21(21/8)=33(33/8)$ FALSE
E.$21(33/8)=33(21/8)$ TRUE

SOLE technique #2: Add or subtract the equations

Another common approach is to add or subtract the equations so that one variable cancels.

Solve for $y$:

$$\begin{array}{rl}4x-2y& =16\\ 4x+4y& =4\end{array}$$

Both equations contain $4x$, so subtracting one equation from the other will eliminate $x$.

$$\begin{array}{cccccc}& 4x& -2y& =& 16& \\ -(& 4x& +4y& =& 4& )\\ & & -6y& =& 12& \end{array}$$

Be sure you subtract the entire equation (equivalently, multiply the whole second equation by $-1$). Now solve:

$$\begin{array}{rl}-6y& =12\\ y& =-2\end{array}$$

The question only asks for $y$, so you can stop here. If you also wanted $x$, substitute back:

$$\begin{array}{rl}4x-2y& =16\\ 4x-2(-2)& =16\\ 4x+4& =16\\ 4x& =12\\ x& =3\end{array}$$

Sometimes the question asks for an expression instead of individual values. You can still use elimination and stop as soon as you reach the expression you need.

What is the value of $3x+y$?

Given:
$6x-7y=50$
$3x-8y=25$

Subtract the second equation from the first:

$$\begin{array}{cccccc}& 6x& -7y& =& 50& \\ -(& 3x& -8y& =& 25& )\\ & 3x& +y& =& 25& \end{array}$$

Since $3x+y$ is exactly what the question asks for, you’re done.

Sometimes you’ll need to multiply one equation first so the coefficients match.

Solve for $x$:

$$\begin{array}{rl}7x+10y& =2\\ -3x-5y& =3\end{array}$$

The $y$-coefficients are $10$ and $-5$. Multiply the second equation by $2$ so you get $-10y$:

$$\begin{array}{rl}-3x-5y& =3\\ 2(-3x-5y)& =2(3)\\ -6x-10y& =6\end{array}$$

Now add the equations:

$$\begin{array}{cccccc}& 7x& +10y& =& 2& \\ +(& -6x& -10y& =& 6& )\\ & x& & =& 8& \end{array}$$

So $x=8$.

SOE technique: Multiply or divide the equations

This next method can be used for systems of equations in general, not just linear equations. Instead of adding or subtracting, you can sometimes multiply or divide the equations to eliminate an expression.

Solve for $x$:

$$\begin{array}{rl}{y}^2/5& =x\\ 125/x& =y^2\end{array}$$

You could solve this by substitution, but multiplying the equations also works.

When you multiply (or divide) two equations, match left side with left side and right side with right side:

• Multiply the left side of the first equation ($y^2/5$) by the left side of the second equation ($125/x$)
• Multiply the right side of the first equation ($x$) by the right side of the second equation ($y^2$)

$$\begin{array}{ccccc}& y^2/5& =& x& \\ \times (& 125/x& =& y^2& )\\ & (y^2/5)(125/x)& =& (x)(y^2)& \end{array}$$

Now simplify:

$$\begin{array}{rl}(y^2/5)(125/x)& =(x)(y^2)\\ 125y^2/5x& =x{y}^2\\ 25y^2/x& =x{y}^2\\ 25y^2/x& =x{y}^2\\ 25y^2& =x^2{y}^2\\ 25& =x^2\end{array}$$

So there are two possibilities:

$$\begin{array}{rl}x& =5\\ x& =-5\end{array}$$

Now check which values actually work in the original system. From $y^2/5=x$, we know $x$ must be positive, so $x=-5$ is impossible.

$$\begin{array}{rl}{y}^2/5& =x\\ y^2/5& \ne -5\end{array}$$

So the only valid solution is $x=5$.

You can also solve the same system by substitution. For example, substitute $y^2=125/x$ into $y^2/5=x$:

$$\begin{array}{rl}{y}^2/5& =x\\ (125/x)/5& =x\\ 25/x& =x\\ 25& =x^2\end{array}$$

This again gives $x=5$ or $x=-5$, and the same check shows only $x=5$works.

This system can be solved through repeated substitution.

Example: The question below is from 2000 AMC 12

If$x,y,$ and $z$ are positive numbers satisfying

$$x+\frac{1}{y}=4,y+\frac{1}{z}=1,\text{and}z+\frac{1}{x}=\frac{7}{3}$$

Then what is the value of $xyz$?
A. $\frac{2}{3}$
B. $1$
C. $\frac{4}{3}$
D. $2$
E. $\frac{7}{3}$

Common themes

• Never divide or multiply an inequality by an unknown variable. You do not have enough information to know whether to flip the inequality.
• The exam is multiple choice. Do not be afraid to plug answer choices into the equations presented in the question instead of directly solving.
• Drawing out the graphs for inequalities or linear equations may help in estimating their solutions (which may lead to eliminating answer choices).
• When a question asks for the value of some combination of multiple variables, try to find an equivalency for said expression instead of directly solving for the values of each individual variable in the expression.
• The verb “to be” and all of its versions, like “is”, “was”, “would be”, and “equals”, all signify the equals sign. When translating a sentence into an equation, anything to the left of the verb should be on one side of the equation and anything to the right should be on the other. Also, periods often signify the end of an equation and the start of another.