Systems of Linear Equations
A system of linear equations is two or more equations considered together. The solution is the point that satisfies ALL the equations simultaneously. The SAT tests three methods (substitution, elimination, graphing) but also tests something subtler: recognizing when a system has no solution or infinite solutions without solving it. The hardest SAT problems give you a system with parameters and ask you to find conditions on those parameters — these can be solved in seconds if you know what to look for.
The three solution cases
Two linear equations form one of three patterns geometrically. Each pattern corresponds to a different number of solutions:
Recognizing which case a system falls into — without solving — is the key to the hardest SAT systems problems.
Method 1: Substitution
Solve one equation for one variable, then plug that expression into the other equation. Best when one equation is already solved for x or y.
Solve the system:
- The first equation already gives y in terms of x. Substitute (2x + 1) for y in the second equation:
3x + 2(2x + 1) = 16
- Distribute and combine like terms:
3x + 4x + 2 = 16 7x + 2 = 16 7x = 14 x = 2
- Substitute x = 2 back into the first equation to find y:
y = 2(2) + 1 = 5
Method 2: Elimination
Add or subtract the two equations (after multiplying by an integer if necessary) to make one variable cancel out. Best when both equations are in standard form (Ax + By = C).
Solve the system:
- The y coefficients are +3 and −3 — opposites. Add the two equations vertically; y will cancel:
2x + 3y = 13 + 4x − 3y = 5 ──────────────── 6x + 0 = 18 x = 3
- Substitute x = 3 into either original equation. Using the first:
2(3) + 3y = 13 6 + 3y = 13 3y = 7 y = 7/3
Solve the system:
- Neither variable's coefficients match yet. Multiply the FIRST equation by 2 so the y coefficients align:
3x + 2y = 11 × 2 ↓ 6x + 4y = 22
- Stack against the second (unchanged) equation. Subtract to eliminate y:
6x + 4y = 22 − 5x + 4y = 21 ──────────────── x + 0 = 1 x = 1
- Substitute x = 1 into either original. Using the first:
3(1) + 2y = 11 3 + 2y = 11 2y = 8 y = 4
Substitution when one equation is already in y = form (or x = form). One step setup, easy.
Elimination when both equations are in standard form (Ax + By = C). Look for matching or opposite coefficients first — those let you skip the multiplication step.
Graphing on Desmos when both methods feel slow. Plot both lines and read the intersection point directly — but only if the answer choices look like exact, clean values.
No-solution and infinite-solution systems
The SAT loves to ask "for what value of k does this system have no solution?" or "for what values of a and c does this system have infinitely many solutions?" The fastest, most reliable method: convert both equations to slope-intercept form and compare.
Solve each equation for y. Once both are in slope-intercept form, the relationship between slopes and y-intercepts tells you the case directly:
- Different slopes → one solution.
- Same slope, different y-intercepts → no solution (parallel lines).
- Same slope, same y-intercept → infinite solutions (same line).
For what value of k does the system below have no solution?
- Convert each equation to slope-intercept form (solve for y):
3x − 2y = 6 → y = (3/2)x − 3 6x + ky = 11 → y = (−6/k)x + 11/k
- For no solution, the slopes must be equal (parallel lines). Set them equal:
3/2 = −6/k Cross-multiply: 3k = 2 · (−6) 3k = −12 k = −4
- Verify that the y-intercepts differ (otherwise it would be infinite solutions, not no solution):
First line: y-intercept = −3 Second line: y-intercept = 11/(−4) = −2.75 −3 ≠ −2.75 ✓ Different intercepts confirmed.
For what values of a and c does this system have infinitely many solutions?
- Convert each equation to slope-intercept form:
2x + 5y = 10 → y = (−2/5)x + 2 ax + 15y = c → y = (−a/15)x + c/15
- For infinite solutions, slopes must match AND y-intercepts must match.
Match slopes:−2/5 = −a/15 Cross-multiply: −2 · 15 = −a · 5 −30 = −5a a = 6 - Match y-intercepts:
2 = c/15 c = 30
The linear combination shortcut
Some SAT problems ask for a value like 3x + 3y or x − y without asking for x and y individually. You can often find this without solving the full system — just by adding or subtracting the equations cleverly.
Look at what the question asks for. Check whether adding or subtracting the two equations (possibly after multiplying by a small integer) gives you exactly that expression.
If 5x + 7y = 20 and 3x + 2y = 8, what is the value of 8x + 9y?
- Look at the target: 8x + 9y. Check if simply adding the two equations yields this.
- Adding x coefficients: 5 + 3 = 8 ✓ Adding y coefficients: 7 + 2 = 9 ✓ Match!
5x + 7y = 20 + 3x + 2y = 8 ──────────────── 8x + 9y = 28
No need to find x and y individually.
If the question asks for something like "ax + by" rather than just x or y, check whether adding or subtracting the equations (possibly after multiplying by a small integer) gives you that combination directly. The SAT uses this pattern often — recognizing it saves serious time.
Word problems with systems
The hardest word problems set up two unknowns with two relationships. Pattern: identify the two quantities, write one equation for each constraint.
A movie theater sold 350 tickets for a total of $2,800. Adult tickets cost $10 and child tickets cost $6. How many adult tickets were sold?
- Define variables. Let a = number of adult tickets, c = number of child tickets.
- Write one equation per constraint:
a + c = 350 (total tickets) 10a + 6c = 2800 (total revenue)
- Use elimination. Multiply the first equation by 6 to align the c coefficients:
a + c = 350 × 6 ↓ 6a + 6c = 2100
- Subtract from the second equation to eliminate c:
10a + 6c = 2800 − 6a + 6c = 2100 ────────────────── 4a + 0 = 700 a = 175
Sample SAT-style problems
If 2x + 3y = 12 and x − y = 1, what is the value of x?
- From the second equation, solve for x: x = y + 1.
- Substitute into the first equation:
2(y + 1) + 3y = 12 2y + 2 + 3y = 12 5y = 10 y = 2
- Substitute back:
x = 2 + 1 = 3
The system below has no solution. What is the value of a?
- Convert to slope-intercept form:
ax + 4y = 8 → y = (−a/4)x + 2 6x − 8y = 3 → y = (3/4)x − 3/8
- For no solution, slopes must match:
−a/4 = 3/4 Cross-multiply: −a = 3 a = −3
- Check intercepts differ: 2 ≠ −3/8 ✓
If 4x − 3y = 11 and 2x + 5y = 1, what is the value of 6x + 2y?
- Target is 6x + 2y. Check if adding the two equations yields this.
- x coefficients: 4 + 2 = 6 ✓ y coefficients: −3 + 5 = 2 ✓ Match!
4x − 3y = 11 + 2x + 5y = 1 ──────────────── 6x + 2y = 12
A coffee shop sells small drinks for $3 and large drinks for $5. On a particular day, the shop sold 80 drinks for a total of $320. How many large drinks were sold?
- Let s = small drinks, L = large drinks. Write one equation per constraint:
s + L = 80 (total drinks) 3s + 5L = 320 (total revenue)
- Multiply the first equation by 3 to align s coefficients:
s + L = 80 × 3 ↓ 3s + 3L = 240
- Subtract from the second to eliminate s:
3s + 5L = 320 − 3s + 3L = 240 ────────────────── 0 + 2L = 80 L = 40
1. Solving for the wrong variable. Read what the question asks for. If it asks for x + y but you solve for x, you'll likely pick a wrong-answer choice that matches your partial work.
2. Sign errors when subtracting equations. When using elimination by subtraction, distribute the negative sign to ALL terms. (3x + 2y) − (x − y) = 2x + 3y, not 2x + y.
3. Forgetting to check the y-intercept condition. For "no solution," parallel slopes AND different y-intercepts. If both slopes AND intercepts match, that's infinite solutions, not no solution.