Systems of
Equations
Master the methods for solving two or more equations simultaneously: substitution, elimination, and graphing to find the point where lines meet.
SAT Math
What Is a System of Equations?
A system of equations is a set of two or more equations that share the same variables. A solution to the system is any ordered pair \((x, y)\) that satisfies every equation in the set at the same time.
The Graphical Interpretation
Each linear equation represents a line on the coordinate plane. Solving a system means finding where those lines intersect.
Three Possible Outcomes
One Solution
Lines intersect at exactly one point. The system is consistent and independent.
No Solution
Lines are parallel (same slope, different intercept). The system is inconsistent.
Infinitely Many
Lines are identical (same equation). The system is consistent and dependent.
Method 1: Substitution
Isolate one variable in one equation, then substitute that expression into the other equation. This reduces the system to a single equation in one variable.
1
Isolate a variable: pick the equation where a variable has coefficient 1 or \(-1\).
2
Substitute the expression into the other equation.
3
Solve the resulting single-variable equation.
4
Back-substitute to find the other variable, then verify in both equations.
Method 2: Elimination
Add or subtract the equations to cancel one variable. If coefficients don't already match, multiply one or both equations by a constant first.
1
Align both equations in standard form \(ax + by = c\).
2
Multiply one or both equations so a variable has equal (or opposite) coefficients.
3
Add or subtract the equations to eliminate that variable.
4
Solve for the remaining variable, then back-substitute.
Substitution vs Elimination
| Substitution | Elimination | |
|---|---|---|
| Best when | A variable already has coefficient 1 or \(-1\) | Both equations are in standard form with "nice" coefficients |
| Key move | Isolate and plug in | Add / subtract equations |
| Risk | Messy fractions if coefficients aren't 1 | Sign errors when multiplying entire equations |
| SAT tip | Great for "solve for \(y\) in terms of \(x\)" questions | Fastest for "find \(x + y\)" or "find \(3x - 2y\)" directly |
How to Approach Any System
1
Read the question carefully. Does it ask for \(x\), \(y\), or a combined expression like \(x + y\)?
2
Choose a method. Substitution if a variable is easy to isolate; elimination if coefficients line up.
3
Solve and simplify. Watch your arithmetic: distribute carefully and combine like terms.
4
Check your answer. Plug the solution back into both original equations.
5
Answer what's asked. If the question asks for \(2x + y\), compute it: don't just report \(x\) and \(y\).
Example 1: Substitution
1
Variable \(y\) is already isolated in the first equation: \(y = 3x - 1\).
2
Substitute into the second: \(2x + (3x - 1) = 9\).
3
Simplify: \(5x - 1 = 9 \implies 5x = 10 \implies\) \(x = 2\).
4
Back-substitute: \(y = 3(2) - 1 = 5\). Solution: \((2,\,5)\).
Example 2: Elimination
1
Notice the \(4y\) and \(-4y\) are already opposites.
2
Add the equations: \((3x + 5x) + (4y - 4y) = 18 + (-2)\).
3
Result: \(8x = 16 \implies\) \(x = 2\).
4
Back-substitute: \(3(2) + 4y = 18 \implies 4y = 12 \implies\) \(y = 3\). Solution: \((2,\,3)\).
Example 3: SAT-Style Problem
1
Multiply the second equation by 3: \(12x + 3y = 39\).
2
Add to the first: \((2x + 12x) + (-3y + 3y) = 7 + 39\).
3
\(14x = 46 \implies x = \frac{46}{14} = \frac{23}{7}\).
4
Substitute into Eq 2: \(4\!\cdot\!\frac{23}{7} + y = 13 \implies y = 13 - \frac{92}{7} = \frac{91-92}{7} = -\frac{1}{7}\).
5
\(x + y = \frac{23}{7} - \frac{1}{7} =\) \(\frac{22}{7}\).
Special Cases: No Solution & Infinite
No Solution
Eliminating both variables gives a false statement like \(0 = 5\). The lines are parallel.
Infinitely Many
Eliminating both variables gives a true identity like \(0 = 0\). The equations describe the same line.
Finding Constants on the SAT
1
No solution means the lines are parallel: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).
2
Compute: \(\frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3}\) and \(\frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2}\). Since \(\frac{1}{3} \neq \frac{1}{2}\), the constant-side ratios already differ.
3
Set \(\frac{b_1}{b_2} = \frac{a_1}{a_2}\): \(\frac{3}{k} = \frac{1}{3} \implies k = 9\).
4
Verify all three ratios: \(\frac{2}{6} = \frac{3}{9} = \frac{1}{3}\), but \(\frac{7}{14} = \frac{1}{2} \neq \frac{1}{3}\). The coefficient ratios match but the constant ratio differs, confirming no solution. So \(k = 9\).
Systems in Word Problems
The SAT frequently disguises systems as word problems. The key skill is translating words into equations.
1
Define variables: Let \(x\) and \(y\) represent the unknowns mentioned in the problem.
2
Write two equations from two different pieces of information in the problem.
3
Solve using substitution or elimination.
4
Answer in context: make sure your answer matches what the question asks (e.g., "number of adult tickets").
Common Mistakes
✕ Sign Errors
Forgetting to distribute the negative when subtracting an entire equation: \(-(2x + 3y)\) is \(-2x - 3y\), not \(-2x + 3y\).
✕ Solving for the Wrong Variable
You find \(x = 4\) but the question asks for \(y\) or \(x + y\). Always re-read the question before selecting an answer.
✕ Forgetting to Multiply All Terms
When multiplying an equation by a constant, every term (including the constant on the right side) must be multiplied.
Summary of Methods
| Feature | Substitution | Elimination | Graphing |
|---|---|---|---|
| Speed on SAT | Medium | Fast | Slow (calculator only) |
| Best for | One variable isolated | Standard form, matching coefficients | Checking work visually |
| Handles special cases? | Yes (contradiction or identity) | Yes (contradiction or identity) | Yes (parallel or overlapping lines) |
| SAT frequency | High | Very high | Rare (but useful on Desmos) |
Question 1 of 5
If \( x + 2y = 10 \) and \( 3x - 2y = 6 \), what is the value of \( x \)?
Add the two equations: \((x + 3x) + (2y - 2y) = 10 + 6\), giving \(4x = 16\), so \(x = 4\).
Question 2 of 5
The system \( 4x + 6y = 24 \) and \( 2x + 3y = 12 \) has:
Divide the first equation by 2: you get \(2x + 3y = 12\), which is identical to the second equation. Since the two equations describe the same line, there are infinitely many solutions.
Question 3 of 5
If \( y = 2x + 1 \) and \( y = -x + 7 \), what is the value of \( x + y \)?
Set them equal: \(2x + 1 = -x + 7 \implies 3x = 6 \implies x = 2\). Then \(y = 2(2) + 1 = 5\). So \(x + y = 2 + 5 = 7\).
Question 4 of 5
A store sells pens for $2 each and notebooks for $5 each. A student buys 8 items for a total of $25. How many notebooks did the student buy?
Let \(p\) = pens, \(n\) = notebooks. The system is \(p + n = 8\) and \(2p + 5n = 25\). From the first: \(p = 8 - n\). Substitute: \(2(8-n) + 5n = 25 \implies 16 - 2n + 5n = 25 \implies 3n = 9 \implies n = 3\).
Question 5 of 5
For what value of \( k \) does the system \( 6x - 9y = 12 \) and \( 2x - 3y = k \) have infinitely many solutions?
Divide the first equation by 3: \(2x - 3y = 4\). For infinitely many solutions, the two equations must be identical. The second equation is \(2x - 3y = k\), so we need \(k = 4\).