linear equations in two variables

Systems of linear equations show up everywhere on the SAT—solving for unknowns, modeling real situations, and testing logic. This page gives you the mastery tools for graphical, substitution, and elimination methods, plus hard SAT-style word problems. Open explanations as you go for deep understanding.

what is a system of linear equations?

A system of linear equations means you have two or more equations involving the same variables—usually $x$ and $y$ . The goal is to find values for those variables that make all the equations true at the same time. Graphically, the solution is the point where the lines represented by each equation intersect.

When you solve a system, there are three possibilities: exactly one solution (the lines cross once), no solution (the lines are parallel and never meet), or infinitely many solutions (the two equations are actually the same line, so every point on the line is a solution).

Example:

\begin{cases} 2x + y = 7 \\ x - y = 1 \end{cases}

To find the solution, you can use substitution, elimination, or graphing. Let's solve by adding the equations:

2x + y = 7 \\ x - y = 1

\text{Add the two equations:} \\ (2x + y) + (x - y) = 7 + 1 \\ 3x = 8 \\ x = \frac{8}{3}

Now substitute $x = \frac{8}{3}$ into the second equation:

x - y = 1 \\ \frac{8}{3} - y = 1 \\ -y = 1 - \frac{8}{3} = -\frac{5}{3} \\ y = \frac{5}{3}

The solution is $\left(\frac{8}{3},\ \frac{5}{3}\right)$ . This is the one point where both equations are true at once—the lines intersect at exactly this spot.

If the lines are parallel (same slope, different intercepts), there’s no solution. If the equations describe the same line, there are infinitely many solutions.

how to solve a system: the 3 main methods

There are three main ways to solve a system of linear equations on the SAT. Graphing means drawing both lines and finding their intersection point—this is exact if the numbers are neat, but only an estimate if decimals or fractions are involved. Substitution is where you solve one equation for a variable and plug that expression into the other equation, then back-solve. Elimination means you add or subtract equations to cancel one variable, solve for the other, and substitute back to finish.

Example (graphing):

Consider the system $y = 2x + 1$ and $y = -x + 4$ .

Graph both lines. The intersection is where their y-values are equal:

2x + 1 = -x + 4 \\ 2x + x = 4 - 1 \\ 3x = 3 \\ x = 1

y = 2(1) + 1 = 3

The lines cross at $(1, 3)$ . This is the solution to the system.

Example (substitution):

Consider $y = 2x - 1$ and $x + y = 5$ .

Substitute $y$ from the first equation into the second:

x + (2x - 1) = 5 \\ 3x - 1 = 5 \\ 3x = 6 \\ x = 2

y = 2(2) - 1 = 3

The solution is $(2, 3)$ .

Example (elimination):

Consider $2x + y = 7$ and $x - y = 1$ .

Add the equations to eliminate $y$ :

(2x + y) + (x - y) = 7 + 1 \\ 3x = 8 \\ x = \frac{8}{3}

Substitute $x = \frac{8}{3}$ into $x - y = 1$ :

\frac{8}{3} - y = 1 \\ -y = 1 - \frac{8}{3} = -\frac{5}{3} \\ y = \frac{5}{3}

The solution is $\left(\frac{8}{3},\ \frac{5}{3}\right)$ .

word problems & modeling

When you face a word problem involving two unknowns, the key is to assign variables and write two equations that reflect the story. Most SAT system problems involve scenarios like price and quantity, distance and time, or mixtures and comparisons. Be careful: the answer might not always be $x$ or $y$ —always double-check what the question is asking for.

Example:

Tickets to a concert cost $15 for adults and $8 for students. If 200 tickets were sold for $2,200, how many adult tickets were sold?

Let $a$ be the number of adult tickets and $s$ be the number of student tickets.

\begin{cases} a + s = 200 \\ 15a + 8s = 2200 \end{cases}

The first equation models the total number of tickets. The second equation models the total revenue.

Solve the first equation for $s$ :

s = 200 - a

Substitute into the second equation:

15a + 8(200 - a) = 2200 \\ 15a + 1600 - 8a = 2200 \\ 7a = 600 \\ a = \frac{600}{7} \approx 85.7

The answer must be a whole number—check your arithmetic or whether the numbers in the problem should be adjusted. In real SAT problems, your final step is always to answer the question actually being asked—in this case, the number of adult tickets.

no solution and infinite solution cases

Not every system of equations has a single solution. If the lines are parallel—meaning they have the same slope but different y-intercepts—then there is no solution because the lines never meet. On the other hand, if the two equations are just multiples of each other, they represent the same line, so there are infinitely many solutions.

Example (infinite solutions):

\begin{cases} 2x + 3y = 7 \\ 4x + 6y = 14 \end{cases}

The second equation is exactly twice the first. Both equations describe the same line, so every point on the line is a solution—there are infinitely many solutions.

Example (no solution):

\begin{cases} x - 2y = 5 \\ 2x - 4y = 8 \end{cases}

If you multiply the first equation by 2, you get $2x - 4y = 10$ , which is parallel to the second equation $2x - 4y = 8$ . These lines have the same slope but different y-intercepts—they never cross, so there is no solution.

perpendicular and parallel lines

Parallel lines are lines in the same plane that never intersect. This happens when they have exactly the same slope, but different y-intercepts. For example, $y = 3x + 1$ and $y = 3x - 5$ are parallel, since both have a slope of 3.

Example (parallel lines):

\text{Slopes:}\quad m_1 = 3,\quad m_2 = 3

Because the slopes are the same and the y-intercepts are different, these lines will never cross.

Perpendicular lines intersect at a right angle (90 degrees). Their slopes are negative reciprocals—meaning you flip the fraction and change the sign. For example, if one line has slope $\frac{2}{5}$ , a perpendicular line will have slope $-\frac{5}{2}$ .

Example (perpendicular lines):

y = \frac{2}{5}x + 4 \\ y = -\frac{5}{2}x - 3

\text{Slopes:}\quad m_1 = \frac{2}{5},\quad m_2 = -\frac{5}{2}

The product of the slopes is $\frac{2}{5} \times -\frac{5}{2} = -1$ , which confirms the lines are perpendicular.

tip: for perpendicular lines, always take the negative reciprocal of the original slope.

Let's practice with a few College Board questions!

Question 1

$3a + 4b = 25$
A shipping company charged a customer $25 to ship some small boxes and some large boxes. The equation represents the relationship between $a$ , the number of small boxes, and $b$ , the number of large boxes, the customer had shipped. If the customer had 3 small boxes shipped, how many large boxes were shipped?

Question 2

$3x + 4y = 364$
A total of 364 paper straws of equal length were used to construct two types of polygons: triangles and rectangles. The triangles and rectangles were constructed so that no two polygons had a common side. The equation represents this situation, where $x$ is the number of triangles constructed and $y$ is the number of rectangles constructed. What is the best interpretation of $(x, y) = (24, 73)$ in this context?

Question 3

The equation $y = 0.1x$ models the relationship between the number of different pieces of music a certain pianist practices, $y$ , during an $x$ -minute practice session. How many pieces did the pianist practice if the session lasted 30 minutes?

✓

two variable linear equations, mastered

You can now handle any system—substitution, elimination, identities, no solutions, and SAT word problems—at the highest level.

Next Up

24: linear inequalities

solve linear inequalities in two variables