SAT Math – Advanced Math
Linear and Quadratic Systems
Solving systems combining lines and parabolas
Systems of equations involve finding values that simultaneously satisfy two or more equations. On the SAT, you'll solve systems of linear equations, systems involving quadratics, and mixed linear-quadratic systems—determining where lines intersect lines, where parabolas intersect lines, and analyzing the number of solutions graphically and algebraically.
Success requires mastering substitution and elimination methods, interpreting solutions as intersection points, understanding that linear-quadratic systems can have 0, 1, or 2 solutions, and connecting algebraic solutions to graphical representations. These techniques aren't just equation-solving—they model supply-demand equilibrium, collision points in physics, break-even analysis in business, and any scenario where multiple constraints must be satisfied simultaneously.
Understanding Systems of Equations
What is a System of Equations?
A system consists of multiple equations that must all be true at the same time.
Graphically: Points where graphs intersect
Example: \(y = 2x + 1\) and \(y = x + 3\) intersect at (2, 5)
Types: Linear-linear, linear-quadratic, quadratic-quadratic
Substitution Method
Solve one equation for a variable, then substitute into the other equation.
2) Substitute expression into other equation
3) Solve resulting equation
4) Back-substitute to find other variable
Best for: When one equation is already solved or easily solved
Elimination Method
Add or subtract equations to eliminate one variable.
2) Multiply equations to match coefficients
3) Add/subtract to eliminate variable
4) Solve and back-substitute
Best for: Linear systems with convenient coefficients
Number of Solutions
Linear-quadratic systems can have different numbers of intersection points.
One solution: Line is tangent (touches at one point)
Two solutions: Line crosses parabola twice
Determine: Solve and check discriminant or count real solutions
Essential Methods and Strategies
Substitution Method Steps
For system: \(y = mx + b\) and \(y = ax^2 + bx + c\)
1. Both solved for y, so set them equal
2. \(mx + b = ax^2 + bx + c\)
3. Rearrange to standard form: \(ax^2 + (b-m)x + (c-b) = 0\)
4. Solve quadratic, then find y-values
Linear System (2×2)
Elimination example:
\(2x + 3y = 12\)
\(x - 3y = -3\)
Add equations → \(3x = 9\) → \(x = 3\)
Back-substitute to find y
Graphical Interpretation
Linear-linear: Lines intersect at one point (unless parallel)
Linear-quadratic: 0, 1, or 2 intersection points
Solution = coordinates of intersection point(s)
Special Cases
No solution: Equations lead to contradiction (like 0 = 5)
Infinite solutions: Equations are equivalent (same line)
Complex solutions: Algebraically valid but not real intersection points
Common Pitfalls & Expert Tips
❌ Forgetting to find both variables
After finding x, must substitute back to find y! Solution is an ordered pair (x, y), not just one value.
❌ Substitution errors with negatives
If \(y = x - 3\), substituting into \(x + 2y = 7\) gives \(x + 2(x - 3) = 7\). Watch those parentheses!
❌ Missing second solution in quadratic systems
Linear-quadratic systems often have TWO solutions. Don't stop after finding one x-value—use both solutions from the quadratic!
❌ Not checking solutions
Always verify by substituting back into BOTH original equations. Catches arithmetic errors!
✓ Expert Tip: Choose the easier method
If one equation is already solved for y, use substitution. If coefficients align nicely, try elimination. Don't force one method!
✓ Expert Tip: Set expressions equal when both solved for same variable
If \(y = 2x + 1\) and \(y = x^2\), immediately write \(2x + 1 = x^2\). Fastest approach for substitution!
✓ Expert Tip: Visualize the intersection
Think about what's happening graphically. Line crossing parabola = 2 points usually. Helps predict number of solutions!
Fully Worked SAT-Style Examples
Solve the system:
\(3x + 2y = 16\)
\(x - 2y = 0\)
Solution:
Step 1: Notice y-coefficients are opposites (2 and -2)
Add equations to eliminate y
Step 2: Add equations
\(3x + 2y = 16\)
\(+ (x - 2y = 0)\)
_________________
\(4x = 16\)
\(x = 4\)
Step 3: Substitute to find y
Use \(x - 2y = 0\):
\(4 - 2y = 0\)
\(-2y = -4\)
\(y = 2\)
Answer: \((4, 2)\)
Solve the system:
\(y = 3x - 5\)
\(2x + y = 10\)
Solution:
Step 1: First equation already solved for y
Substitute \(y = 3x - 5\) into second equation
Step 2: Substitute and solve for x
\(2x + (3x - 5) = 10\)
\(5x - 5 = 10\)
\(5x = 15\)
\(x = 3\)
Step 3: Find y
\(y = 3(3) - 5 = 9 - 5 = 4\)
Answer: \((3, 4)\)
Solve the system:
\(y = x + 2\)
\(y = x^2\)
Solution:
Step 1: Both equations solved for y, set them equal
\(x + 2 = x^2\)
Step 2: Rearrange to standard form
\(x^2 - x - 2 = 0\)
Step 3: Factor and solve
\((x - 2)(x + 1) = 0\)
\(x = 2\) or \(x = -1\)
Step 4: Find corresponding y-values
When \(x = 2\): \(y = 2 + 2 = 4\)
When \(x = -1\): \(y = -1 + 2 = 1\)
Answer: \((2, 4)\) and \((-1, 1)\)
How many solutions does this system have?
\(y = x^2 + 2x + 5\)
\(y = 2\)
Solution:
Set equations equal:
\(x^2 + 2x + 5 = 2\)
\(x^2 + 2x + 3 = 0\)
Check discriminant:
\(a = 1\), \(b = 2\), \(c = 3\)
\(b^2 - 4ac = 4 - 4(1)(3) = 4 - 12 = -8\)
Interpretation:
Negative discriminant means no real solutions
Graphically: horizontal line y = 2 doesn't intersect parabola
Answer: Zero real solutions (no intersection points)
Solve the system:
\(x + y = 5\)
\(x^2 + y^2 = 17\)
Solution:
Step 1: Solve first equation for y
\(y = 5 - x\)
Step 2: Substitute into second equation
\(x^2 + (5 - x)^2 = 17\)
\(x^2 + 25 - 10x + x^2 = 17\)
\(2x^2 - 10x + 8 = 0\)
\(x^2 - 5x + 4 = 0\)
Step 3: Factor
\((x - 4)(x - 1) = 0\)
\(x = 4\) or \(x = 1\)
Step 4: Find y-values
When \(x = 4\): \(y = 5 - 4 = 1\)
When \(x = 1\): \(y = 5 - 1 = 4\)
Answer: \((4, 1)\) and \((1, 4)\)
Find the solution(s):
\(y = -2x + 5\)
\(y = x^2 - 4x + 6\)
Solution:
Set equal:
\(-2x + 5 = x^2 - 4x + 6\)
\(0 = x^2 - 2x + 1\)
\(0 = (x - 1)^2\)
Solve:
\(x = 1\) (repeated root)
\(y = -2(1) + 5 = 3\)
Graphical meaning:
One solution (discriminant = 0) means line is tangent to parabola
They touch at exactly one point
Answer: \((1, 3)\)
The sum of two numbers is 10 and their product is 24. What are the numbers?
Solution:
Set up system:
Let x and y be the two numbers
\(x + y = 10\) (sum is 10)
\(xy = 24\) (product is 24)
Solve first for y:
\(y = 10 - x\)
Substitute:
\(x(10 - x) = 24\)
\(10x - x^2 = 24\)
\(x^2 - 10x + 24 = 0\)
\((x - 6)(x - 4) = 0\)
\(x = 6\) or \(x = 4\)
Find corresponding y-values:
If \(x = 6\), then \(y = 4\)
If \(x = 4\), then \(y = 6\)
Answer: The numbers are 4 and 6
Solution Count Summary
Linear-Linear System
• One solution (lines intersect)
• No solution (parallel lines)
• Infinite solutions (same line)
Linear-Quadratic System
• Two solutions (line crosses)
• One solution (line tangent)
• No solution (doesn't intersect)
Systems of Equations: Finding Simultaneous Solutions
Systems of equations represent the mathematical foundation for solving problems with multiple constraints—finding values that satisfy several conditions simultaneously. The SAT tests these skills because they're essential across quantitative disciplines: economists solve systems to find market equilibrium where supply equals demand, engineers use them to balance forces and voltages in circuits, chemists apply them to balance equations and calculate concentrations, and operations researchers employ them to optimize production under constraints. Linear systems with two equations and two unknowns provide the simplest case, solved efficiently by elimination or substitution depending on coefficient structure. Linear-quadratic systems introduce complexity—a line intersecting a parabola can produce zero, one, or two solutions, reflecting whether they miss, touch tangentially, or cross twice. Master substitution for systems where one equation is already solved, elimination for systems with aligned coefficients, and graphical interpretation to understand solution counts. Each solution represents an intersection point where all equations are simultaneously true—coordinates satisfying every constraint. The discriminant reveals solution count for quadratic systems without full calculation, enabling strategic problem-solving. These techniques transcend academic exercises: every time businesses find break-even points (where revenue equals cost), physicists calculate collision trajectories, or architects balance structural forces, they're solving systems. The fluency you develop—choosing methods strategically, handling both solutions in quadratic systems, verifying answers in original equations—distinguishes mechanical calculation from genuine mathematical understanding that empowers you to model and solve real-world problems involving multiple simultaneous relationships.