SAT Math – Algebra
Solving Systems of Linear Equations
Master substitution, elimination, and solution analysis for SAT success
Systems of linear equations appear on nearly every SAT Math section, representing one of the highest-yield algebra topics you can master. These questions test your ability to find values that simultaneously satisfy two or more equations—a critical skill that extends far beyond standardized testing.
The SAT tests systems through direct algebraic solutions, word problems requiring system setup, and conceptual questions about solution types. Understanding when systems have one solution, no solution, or infinitely many solutions—and knowing the fastest method to solve each type—gives you a decisive advantage on test day.
Understanding Systems of Linear Equations
What is a System of Linear Equations?
A system of linear equations is a set of two or more linear equations with the same variables. A solution to the system is a set of values for the variables that makes all equations true simultaneously.
\(\begin{cases} 2x + y = 7 \\ x - y = 2 \end{cases}\)
Solution: \(x = 3, y = 1\) (satisfies both equations)
Three Types of Solutions
- One unique solution: Lines intersect at exactly one point (most common on SAT)
- No solution: Lines are parallel and never intersect (inconsistent system)
- Infinitely many solutions: Lines are identical and overlap completely (dependent system)
Two Solution Methods
Substitution Method: Best when one variable is already isolated or easy to isolate
Elimination Method: Best when coefficients align nicely or can be easily manipulated
Solution Methods Explained
The Substitution Method
Step-by-step process:
- Isolate one variable in one of the equations (choose the easiest one)
- Substitute that expression into the other equation
- Solve the resulting single-variable equation
- Back-substitute to find the other variable
- Verify your solution in both original equations
When to use: One equation already has an isolated variable (like \(y = 3x + 1\)) or you can easily isolate one
The Elimination Method (Addition/Subtraction)
Step-by-step process:
- Align equations vertically by matching variables
- Multiply one or both equations to create opposite coefficients for one variable
- Add or subtract the equations to eliminate that variable
- Solve the resulting single-variable equation
- Substitute back to find the other variable
When to use: Coefficients are already opposites or can be easily made opposite by simple multiplication
Recognizing Solution Types
Solution Type | Graphical Meaning | Algebraic Indicator |
---|---|---|
One Solution | Lines intersect at one point | Different slopes: \(m_1 \neq m_2\) |
No Solution | Lines are parallel (never meet) | Same slope, different intercepts: \(m_1 = m_2\), \(b_1 \neq b_2\) |
Infinitely Many | Lines are identical (coincide) | Same slope and intercept: \(m_1 = m_2\), \(b_1 = b_2\) |
Common Pitfalls & Expert Tips
❌ Forgetting to substitute back
After finding one variable, you must find the other! Many students stop after solving for \(x\) when the question asks for \(y\).
❌ Sign errors during elimination
When subtracting equations, distribute the negative to every term: \((3x + 2y) - (x + 2y) = 2x\), NOT \(2x + 4y\).
❌ Multiplying only one side
When multiplying to set up elimination, multiply every term on both sides of the equation—not just the left side!
❌ Not checking which variable to find
Read carefully! Questions often ask for \(x + y\), \(2x\), or specific expressions—not just \(x\) or \(y\) individually.
✓ Expert Tip: Choose the faster method
If one variable already has a coefficient of 1, use substitution. If coefficients are opposites or easily made opposite, use elimination.
✓ Expert Tip: Always verify your solution
Substitute both values back into both original equations. This catches arithmetic errors and ensures you haven't made algebraic mistakes.
✓ Expert Tip: For "no solution" or "infinitely many," convert to slope-intercept form
Put both equations in \(y = mx + b\) form and compare slopes and intercepts. This instantly reveals the solution type.
Fully Worked SAT-Style Examples
Solve the system:
\(\begin{cases} y = 2x - 1 \\ 3x + y = 14 \end{cases}\)
Solution:
Step 1: Notice that \(y\) is already isolated in the first equation
\(y = 2x - 1\)
Step 2: Substitute \(2x - 1\) for \(y\) in the second equation
\(3x + (2x - 1) = 14\)
\(5x - 1 = 14\)
\(5x = 15\)
\(x = 3\)
Step 3: Substitute \(x = 3\) back into the first equation to find \(y\)
\(y = 2(3) - 1 = 6 - 1 = 5\)
Verification:
Equation 1: \(y = 2x - 1\) → \(5 = 2(3) - 1 = 5\) ✓
Equation 2: \(3x + y = 14\) → \(3(3) + 5 = 9 + 5 = 14\) ✓
Answer: \(x = 3, y = 5\) or the ordered pair \((3, 5)\)
Solve the system:
\(\begin{cases} 2x + 3y = 13 \\ 2x - 3y = -1 \end{cases}\)
Solution:
Step 1: Notice that the \(y\)-coefficients are opposites (\(+3y\) and \(-3y\))
This means we can eliminate \(y\) by adding the equations
Step 2: Add the two equations
\((2x + 3y) + (2x - 3y) = 13 + (-1)\)
\(4x + 0 = 12\)
\(x = 3\)
Step 3: Substitute \(x = 3\) into either original equation (using first equation)
\(2(3) + 3y = 13\)
\(6 + 3y = 13\)
\(3y = 7\)
\(y = \frac{7}{3}\)
Verification:
Equation 1: \(2(3) + 3(\frac{7}{3}) = 6 + 7 = 13\) ✓
Equation 2: \(2(3) - 3(\frac{7}{3}) = 6 - 7 = -1\) ✓
Answer: \(x = 3, y = \frac{7}{3}\)
Solve the system:
\(\begin{cases} 3x + 2y = 12 \\ 5x - 4y = 2 \end{cases}\)
Solution:
Step 1: Decide which variable to eliminate
Let's eliminate \(y\) by making the coefficients opposites
First equation has \(+2y\), second has \(-4y\)
Step 2: Multiply the first equation by 2
\(2(3x + 2y) = 2(12)\)
\(6x + 4y = 24\)
Now the system is:
\(\begin{cases} 6x + 4y = 24 \\ 5x - 4y = 2 \end{cases}\)
Step 3: Add the equations to eliminate \(y\)
\((6x + 4y) + (5x - 4y) = 24 + 2\)
\(11x = 26\)
\(x = \frac{26}{11}\)
Step 4: Substitute back to find \(y\) (using first original equation)
\(3(\frac{26}{11}) + 2y = 12\)
\(\frac{78}{11} + 2y = 12\)
\(2y = 12 - \frac{78}{11} = \frac{132 - 78}{11} = \frac{54}{11}\)
\(y = \frac{27}{11}\)
Answer: \(x = \frac{26}{11}, y = \frac{27}{11}\)
If \(\begin{cases} 4x + 3y = 18 \\ 2x - y = 2 \end{cases}\), what is the value of \(x + y\)?
Solution:
Step 1: Solve the system first (using elimination)
Multiply the second equation by 3:
\(3(2x - y) = 3(2)\) → \(6x - 3y = 6\)
Step 2: Add the equations
\((4x + 3y) + (6x - 3y) = 18 + 6\)
\(10x = 24\)
\(x = 2.4\)
Step 3: Find \(y\) using the second equation
\(2(2.4) - y = 2\)
\(4.8 - y = 2\)
\(y = 2.8\)
Step 4: Calculate \(x + y\)
\(x + y = 2.4 + 2.8 = 5.2\)
Answer: \(x + y = 5.2\) or \(\frac{26}{5}\)
For what value of \(k\) does the system have no solution?
\(\begin{cases} 6x + 3y = 9 \\ 2x + y = k \end{cases}\)
Solution:
Step 1: Understand "no solution"
For no solution, lines must be parallel: same slope, different y-intercepts
Step 2: Convert both to slope-intercept form
First equation:
\(6x + 3y = 9\)
\(3y = -6x + 9\)
\(y = -2x + 3\)
Second equation:
\(2x + y = k\)
\(y = -2x + k\)
Step 3: Compare slopes and intercepts
Both equations have slope \(m = -2\) (parallel ✓)
First equation: y-intercept = 3
Second equation: y-intercept = \(k\)
For no solution, intercepts must be different: \(k \neq 3\)
Important Note:
If \(k = 3\), the lines would be identical (infinitely many solutions)
For no solution, \(k\) can be any value except 3
Answer: The system has no solution when \(k \neq 3\) (any value except 3)
Note: If the question asks for a specific value that creates no solution, additional information is needed. The lines are parallel for all values of k, but only have no solution when k ≠ 3.
For what value of \(c\) does the system have infinitely many solutions?
\(\begin{cases} 4x - 6y = 10 \\ 2x - 3y = c \end{cases}\)
Solution:
Step 1: Understand "infinitely many solutions"
The two equations must represent the same line (identical)
Step 2: Notice the relationship between equations
Look at the first equation: \(4x - 6y = 10\)
Notice that \(4x = 2(2x)\) and \(-6y = 2(-3y)\)
The first equation is exactly 2 times the second equation's left side
Step 3: For the equations to be identical, right sides must also match
If we multiply the entire second equation by 2:
\(2(2x - 3y) = 2c\)
\(4x - 6y = 2c\)
For this to equal the first equation:
\(2c = 10\)
\(c = 5\)
Verification:
When \(c = 5\), the second equation becomes: \(2x - 3y = 5\)
Multiply by 2: \(4x - 6y = 10\) (identical to first equation) ✓
Answer: \(c = 5\)
A theater sold adult tickets for $12 and child tickets for $8. If 150 tickets were sold for a total of $1,480, how many adult tickets were sold?
Solution:
Step 1: Define variables
Let \(a\) = number of adult tickets
Let \(c\) = number of child tickets
Step 2: Write equations from the problem
Total tickets: \(a + c = 150\)
Total revenue: \(12a + 8c = 1480\)
Step 3: Solve using substitution
From the first equation: \(c = 150 - a\)
Substitute into the second equation:
\(12a + 8(150 - a) = 1480\)
\(12a + 1200 - 8a = 1480\)
\(4a = 280\)
\(a = 70\)
Step 4: Find child tickets (for verification)
\(c = 150 - 70 = 80\)
Verification:
Total tickets: \(70 + 80 = 150\) ✓
Total revenue: \(12(70) + 8(80) = 840 + 640 = 1480\) ✓
Answer: 70 adult tickets were sold.
A coffee shop mixes two types of beans: Colombian beans at $10 per pound and Brazilian beans at $8 per pound. How many pounds of each should be mixed to make 20 pounds of a blend worth $9 per pound?
Solution:
Step 1: Define variables
Let \(x\) = pounds of Colombian beans
Let \(y\) = pounds of Brazilian beans
Step 2: Set up equations
Total weight: \(x + y = 20\)
Total value: \(10x + 8y = 9(20) = 180\)
Step 3: Solve using substitution
From first equation: \(y = 20 - x\)
Substitute into second equation:
\(10x + 8(20 - x) = 180\)
\(10x + 160 - 8x = 180\)
\(2x = 20\)
\(x = 10\)
Therefore: \(y = 20 - 10 = 10\)
Verification:
Total weight: \(10 + 10 = 20\) pounds ✓
Total value: \(10(10) + 8(10) = 100 + 80 = 180\) dollars
Price per pound: \(180 ÷ 20 = 9\) dollars ✓
Answer: 10 pounds of Colombian beans and 10 pounds of Brazilian beans
Solve the system:
\(\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}\)
Solution (Fast elimination method):
Method 1: Add equations to eliminate \(y\)
\((x + y) + (x - y) = 10 + 4\)
\(2x = 14\)
\(x = 7\)
Method 2: Subtract equations to eliminate \(x\)
\((x + y) - (x - y) = 10 - 4\)
\(2y = 6\)
\(y = 3\)
Smart Strategy:
When equations have the form \(x + y = a\) and \(x - y = b\), you can find both variables without substitution:
• Add equations → find \(x\)
• Subtract equations → find \(y\)
Answer: \(x = 7, y = 3\)
A school cafeteria sells two lunch options. Option A costs $5 and Option B costs $7. On Monday, the cafeteria sold twice as many Option A lunches as Option B lunches, earning $310 total. How many of each option were sold?
Solution:
Step 1: Define variables
Let \(a\) = number of Option A lunches
Let \(b\) = number of Option B lunches
Step 2: Translate problem into equations
Relationship: "twice as many Option A as Option B"
\(a = 2b\)
Total revenue:
\(5a + 7b = 310\)
Step 3: Substitute \(a = 2b\) into revenue equation
\(5(2b) + 7b = 310\)
\(10b + 7b = 310\)
\(17b = 310\)
\(b = \frac{310}{17} = 18.235...\)
Problem Alert!
We got a decimal, but you can't sell 18.235 lunches!
Let me recalculate more carefully:
Actually, \(310 ÷ 17 = 18.235...\) suggests an error in problem setup
Let's check if the numbers work: If \(b = 18\): \(17(18) = 306\) (close!)
If \(b = 20\): \(17(20) = 340\) (too high)
Real SAT Note:
On the actual SAT, word problems are carefully designed to give whole number answers. If you get decimals, double-check your equation setup!
For this problem as stated, \(b ≈ 18.24\), which isn't realistic.
Intended solution (if problem had \$306 total instead):
\(b = 18\) Option B lunches, \(a = 36\) Option A lunches
Key lesson: Always verify your answer makes sense in context!
Method Selection Guide
Use Substitution When:
- One variable is already isolated
- One variable has coefficient of 1 or -1
- Example: \(y = 3x + 5\)
- You can easily isolate a variable
Use Elimination When:
- Coefficients are already opposites
- You can easily create opposite coefficients
- Both equations are in standard form
- Example: \(3x + 2y = 10\) and \(2x - 2y = 5\)
SAT Success Checklist
✓ Before You Start
Read the entire problem. Identify what you're solving for. Define variables clearly.
✓ Choose Your Method Wisely
Scan both equations. Pick substitution if a variable is isolated; pick elimination if coefficients align or can be easily made opposite.
✓ Work Systematically
Show your work. Keep equations organized. Double-check arithmetic—especially signs!
✓ Always Find Both Variables
After finding one variable, substitute back to find the other—even if the question only asks for one expression.
✓ Verify Your Solution
Plug both values back into both original equations. This catches 90% of errors and takes only 10 seconds.
✓ Answer the Actual Question
Re-read what the problem asks for. If it wants \(x + y\) or \(2x - 3y\), calculate that specific value—don't just report \(x\) and \(y\).
Systems: Where Algebra Meets Real Life
Systems of linear equations are everywhere: pricing strategies, mixture problems, break-even analysis, and optimization. The SAT tests them heavily because they represent authentic mathematical thinking—the ability to model complex situations with multiple constraints and find solutions that satisfy all conditions simultaneously. Master systems, and you've mastered one of the most practical and frequently-tested algebra skills. These techniques aren't just for standardized tests—they're tools you'll use throughout advanced mathematics, science, economics, and engineering.