Solving Systems of Linear Equations: Free SAT Math Lesson
A system of linear equations shows up constantly on the Digital SAT, often disguised as a word problem about two unknowns, and knowing how to solve one quickly, and recognize when it has no solution or infinitely many, is a high-value skill. This free lesson covers the three types of systems and how many solutions each one has, solving a system by substitution, solving a system by elimination, and finding the value of a constant that makes a system have no solution or infinitely many solutions. Each method is followed by a fully worked example, and then a set of original practice problems with instant feedback, so you can build fluency with every approach. Everything here is free, and you can keep practicing afterward with the full SAT Math Question Bank linked below.
1. Types of Systems of Equations
A set of linear equations with the same two variables is called a system of linear equations. Since each equation graphs as a line, a system of two linear equations can only ever relate to the other line in one of three ways, and each way tells you exactly how many solutions the system has.
Intersecting lines
Exactly one solution – consistent and independent
Coinciding lines
Infinitely many solutions – consistent and dependent
Parallel lines
No solution – inconsistent
| System type | Slopes and y-intercepts | Number of solutions |
|---|---|---|
| Intersecting lines (consistent, independent) | Different slopes | Exactly one |
| Coinciding lines (consistent, dependent) | Same slope, same y-intercept | Infinitely many |
| Parallel lines (inconsistent) | Same slope, different y-intercepts | None |
Worked Example: Without solving, determine the type of system for 4x − 2y = 8 and −6x + 3y = −12.
Rewrite both in slope-intercept form. First equation: −2y = 8 − 4x, so y = 2x − 4. Second equation: 3y = −12 + 6x, so y = 2x − 4.
Both equations simplify to the exact same line, y = 2x − 4, so this system has infinitely many solutions.
1. Without solving, determine the type of system for y = 4x − 1 and y = 4x + 6.
Explanation: Both equations have slope 4 but different y-intercepts (−1 and 6), so the lines are parallel and the system has no solution.
2. Without solving, determine the type of system for y = −2x + 5 and y = 3x − 1.
Explanation: The slopes, −2 and 3, are different, so the lines must intersect at exactly one point.
3. Without solving, determine the type of system for 4x − 2y = 8 and −6x + 3y = −12.
Explanation: This is the worked example above: both equations simplify to y = 2x − 4, so the lines coincide and the system has infinitely many solutions.
4. A system of two linear equations has the same slope and different y-intercepts. How many solutions does the system have?
Explanation: Same slope with different y-intercepts describes two distinct parallel lines, which never intersect, so the system has no solution.
2. Solving by Substitution
The substitution method works well when one equation is already solved for a variable, or can easily be solved for one. The steps are:
- Solve one of the equations for one of its variables, if it isn't already.
- Substitute that expression into the other equation and solve.
- Substitute the resulting value into either original equation to find the other variable.
Worked Example: Solve the system by substitution: y = 2x − 3 and 3x + y = 12.
3x + (2x − 3) = 12 (Substitute 2x − 3 for y)
5x − 3 = 12 (Combine like terms)
5x = 15, so x = 3
y = 2(3) − 3 = 3 (Substitute 3 for x in the first equation)
The solution is (3, 3).
1. Solve by substitution: y = x + 4 and 2x + y = 19. What is the value of x?
Explanation: Substitute x + 4 for y: 2x + (x + 4) = 19, so 3x + 4 = 19, 3x = 15, and x = 5.
2. Solve by substitution: x = 2y + 1 and x + 3y = 16. What is the value of y?
Explanation: Substitute 2y + 1 for x: (2y + 1) + 3y = 16, so 5y + 1 = 16, 5y = 15, and y = 3.
3. Solve by substitution: y = 5x − 7 and 4x − y = 5. What is the value of x?
Explanation: Substitute 5x − 7 for y: 4x − (5x − 7) = 5, so 4x − 5x + 7 = 5, −x + 7 = 5, −x = −2, and x = 2.
4. The sum of two numbers is 24, and one number is 3 times the other. Using substitution, find the larger number.
Explanation: Let the numbers be x and y, with x = 3y and x + y = 24. Substitute: 3y + y = 24, so 4y = 24, and y = 6. Then x = 3(6) = 18. The larger number is 18.
3. Solving by Elimination
The elimination method works well when adding or subtracting the two equations directly can make one variable cancel out. The steps are:
- Arrange both equations with like terms lined up in columns.
- If needed, multiply one or both equations so that the coefficients of one variable become opposites.
- Add the equations together, that variable will cancel, and solve for the remaining variable.
- Substitute that value into either original equation to find the other variable.
Worked Example: Solve the system by elimination: 2x − 3y = 3 and 5x + 2y = 17.
Multiply the first equation by 2 and the second equation by 3, so the y-coefficients become opposites:
4x − 6y = 6 (First equation × 2)
15x + 6y = 51 (Second equation × 3)
19x = 57, so x = 3 (Add the equations)
2(3) − 3y = 3 ⇒ 6 − 3y = 3 ⇒ y = 1 (Substitute into the first equation)
The solution is (3, 1).
1. Solve by elimination: x + y = 10 and x − y = 2. What is the value of x?
Explanation: Adding the equations directly eliminates y: (x + y) + (x − y) = 10 + 2, so 2x = 12, and x = 6.
2. Solve by elimination: 3x + 2y = 12 and 3x − y = 3. What is the value of y?
Explanation: Subtract the second equation from the first: (3x + 2y) − (3x − y) = 12 − 3, so 3y = 9, and y = 3.
3. Solve by elimination: 4x + 3y = 25 and 2x − 3y = −1. What is the value of x?
Explanation: Adding the equations directly eliminates y: (4x + 3y) + (2x − 3y) = 25 + (−1), so 6x = 24, and x = 4.
4. Solve by elimination: 5x − 2y = 16 and x + 3y = 10. What is the value of y?
Explanation: Multiply the second equation by 5: 5x + 15y = 50. Subtract the first equation: (5x + 15y) − (5x − 2y) = 50 − 16, so 17y = 34, and y = 2.
4. No Solution and Infinitely Many Solutions
Some SAT problems don't ask you to solve a system, they ask you to find the value of a constant that makes the system have no solution or infinitely many solutions. To answer these, rewrite both equations in slope-intercept form and compare. For no solution, the equations must have the same slope but different y-intercepts. For infinitely many solutions, the equations must have both the same slope and the same y-intercept.
Worked Example: For what value of c will the system below have no solution?
cx − 4y = 8
5x + 2y = −3
Rewrite both in slope-intercept form. First equation: −4y = 8 − cx, so y = c4x − 2. Second equation: 2y = −3 − 5x, so y = −52x − 32.
For no solution, the slopes must be equal: c4 = −52, so c = −10.
Checking the y-intercepts confirms they're different (−2 and −32), so the lines are parallel and the system has no solution when c = −10.
1. For what value of c will the system below have no solution?
cx − 4y = 8
5x + 2y = −3
Explanation: This is the worked example above. Setting the slopes equal gives c = −10, and the y-intercepts are confirmed to be different, so the system has no solution.
2. For what value of b will the system below have infinitely many solutions?
−3x + y = 6
6x − by = −12
Explanation: The first equation is y = 3x + 6. Rewriting the second: −by = −12 − 6x, so y = 6bx + 12b. For infinitely many solutions, both the slope and y-intercept must match: 6b = 3 gives b = 2, and checking the y-intercept, 122 = 6, which matches. So b = 2.
3. Does the system below have no solution, one solution, or infinitely many solutions?
3x − 2y = 7
y = 32x − 4
Explanation: Rewrite the first equation: −2y = 7 − 3x, so y = 32x − 72. Both equations have slope 32, but the y-intercepts, −72 and −4, are different. Same slope, different y-intercepts means no solution.
4. A system of two linear equations has the same slope and the same y-intercept for both equations. How many solutions does the system have?
Explanation: If both equations have the same slope and the same y-intercept, they describe the exact same line, so every point on that line is a solution, giving infinitely many solutions.
Common Mistakes to Avoid
- Only multiplying one side of an equation. When scaling an equation for elimination, every term, on both sides of the equal sign, must be multiplied by the same number.
- Adding when you should subtract, or the reverse. Check whether the coefficients you're eliminating are already opposites (add) or identical (subtract). Mixing this up is one of the most common elimination errors.
- Forgetting to solve for the second variable. After finding one variable's value, substitute it back into one of the original equations to find the other. A system's solution is an ordered pair, not a single number.
- Assuming a system with no obvious matching coefficients can't be solved by elimination. Any system can be set up for elimination by multiplying one or both equations, it just may take an extra step.
- Checking only the slopes when asked about no solution versus infinitely many. Equal slopes alone aren't enough to decide between the two cases, you also need to compare the y-intercepts: different y-intercepts mean no solution, and equal y-intercepts mean infinitely many solutions.
Frequently Asked Questions
How do I know whether to use substitution or elimination?
Use substitution when one equation is already solved for a variable, or can be solved for one in a single step, like y = 2x − 3. Use elimination when both equations are in standard form, Ax + By = C, and the coefficients are set up, or can easily be scaled, to cancel out.
Will substitution and elimination always give the same answer?
Yes. Both methods solve the exact same system of equations, so as long as the algebra is done correctly, they'll always arrive at the same solution. Picking a method is about efficiency, not about which one is "correct."
What does it mean for a system to be inconsistent?
An inconsistent system has no solution at all, its equations describe two parallel lines that never cross. If you solve an inconsistent system algebraically, you'll end up with a false statement, like 6 = 10, which signals that no value of the variables can satisfy both equations at once.
Where can I practice more problems like these?
The Systems of Equations quizzes in the SAT Math Question Bank include additional original problems on substitution, elimination, and no-solution systems, along with quizzes covering every other Heart of Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry skill on the Digital SAT.