Systems of
Polynomial Equations
Learn to solve systems where curves meet lines and curves meet curves: the substitution and graphical strategies tested on the SAT.
SAT Math
What Is a System of Equations?
A system of equations is a set of two or more equations with the same variables. A solution is every ordered pair \((x,y)\) that satisfies all equations simultaneously.
Linear-Linear
Two straight lines: at most 1 intersection.
Linear-Quadratic
A line and a parabola: 0, 1, or 2 intersections.
Quadratic-Quadratic
Two parabolas (of the form \(y=ax^2+...\)): 0, 1, or 2 intersections.
Where a Line Meets a Parabola
Why Does the SAT Test This?
The SAT frequently pairs a linear equation with a quadratic equation and asks for the number of solutions, the solutions themselves, or conditions on a parameter. Mastering this skill earns you points in the Advanced Math domain.
Substitution
Solve one equation for \(y\), plug into the other, and reduce to a single-variable polynomial.
Graphical Reasoning
Use the discriminant to determine how many intersection points exist without solving.
The Substitution Strategy
1
Isolate \(y\) in the simpler equation (often the linear one).
2
Substitute that expression into the other equation.
3
Simplify to get a polynomial in \(x\) alone, then solve (factor or use the quadratic formula).
4
Back-substitute each \(x\)-value to find the corresponding \(y\).
The Discriminant Tells You How Many Solutions
After substitution you often land on a quadratic \(ax^2 + bx + c = 0\). The discriminant \(\Delta = b^2 - 4ac\) reveals everything:
\(\Delta > 0\)
Two distinct real solutions: the curves cross in two places.
\(\Delta = 0\)
One real solution (repeated root): the line is tangent to the curve.
\(\Delta < 0\)
No real solutions: the curves never intersect.
Visualising \(\Delta > 0\), \(\Delta = 0\), \(\Delta < 0\)
Substitution vs Elimination
| Substitution | Elimination | |
|---|---|---|
| Best when | One equation is already solved for \(y\) | Both equations are in standard form |
| Polynomial systems | Primary method: set \(f(x)=g(x)\) | Works when both have identical \(y\) terms |
| Risk | Algebra errors during expansion | Harder to apply with mixed-degree terms |
| SAT frequency | Very common | Less common for polynomial pairs |
Step-by-Step Approach
1
Identify the system type. Is it linear-quadratic, quadratic-quadratic, or does it involve a higher-degree polynomial?
2
Isolate \(y\) in the equation where it appears most simply (usually the linear one).
3
Substitute and simplify. Collect all terms on one side to form \(ax^2 + bx + c = 0\).
4
Solve the resulting equation. Factor if possible; otherwise use \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).
5
Back-substitute each \(x\) into either original equation to find \(y\). Write the solution(s) as ordered pairs.
6
Verify by plugging each pair into both original equations.
Example 1: Linear-Quadratic
1
Set equal: \(x^2 - 2x - 3 = x - 1\)
2
Rearrange: \(x^2 - 3x - 2 = 0\)
3
Factor: Hmm, discriminant: \(9 + 8 = 17\). Use the formula: \(x = \frac{3 \pm \sqrt{17}}{2}\)
4
Back-sub into \(y = x - 1\): \(y = \frac{3 \pm \sqrt{17}}{2} - 1 = \frac{1 \pm \sqrt{17}}{2}\)
Example 2: Clean Factoring
1
Set equal: \(x^2 + 2x = 3x + 6\)
2
Rearrange: \(x^2 - x - 6 = 0\)
3
Factor: \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\)
4
Back-sub: \(x=3 \Rightarrow y=15\). \(x=-2 \Rightarrow y=0\).
Example 3: Find \(k\) for Exactly One Solution
1
Set equal: \(x^2 = 4x + k \Rightarrow x^2 - 4x - k = 0\)
2
One solution means \(\Delta = 0\): \((-4)^2 - 4(1)(-k) = 0\)
3
Solve: \(16 + 4k = 0 \Rightarrow k = -4\)
Example 4: Quadratic-Quadratic
1
Set equal: \(x^2 + 4x + 1 = 2x^2 + x - 2\)
2
Rearrange: \(0 = x^2 - 3x - 3\)
3
Discriminant: \(\Delta = 9 + 12 = 21 > 0\), so two solutions.
4
Solve: \(x = \frac{3 \pm \sqrt{21}}{2}\). Back-substitute into either equation for \(y\).
Common Mistakes
✕ Forgetting to Rearrange
After setting the equations equal, students try to factor before moving all terms to one side. Always write \(ax^2+bx+c=0\) first.
✕ Dropping the \(y\)
Finding \(x\) but not back-substituting to get \(y\). The SAT asks for ordered pairs or for the \(y\)-coordinate specifically.
✕ Sign Errors in \(\Delta\)
Miscomputing \(b^2-4ac\) when \(a\) or \(c\) is negative. Wrap negative values in parentheses: \(-4(1)(-k) = +4k\).
Summary Cheat Sheet
| Scenario | Method | Key Check |
|---|---|---|
| Find the solutions | Substitution, then solve | Verify in both equations |
| How many solutions? | Compute \(\Delta = b^2 - 4ac\) | \(\Delta > 0\): 2, \(\Delta = 0\): 1, \(\Delta < 0\): 0 |
| Find \(k\) for tangency | Set \(\Delta = 0\) and solve for \(k\) | One repeated root |
| Quadratic-Quadratic | Subtract to reduce to linear or quadratic | Check if leading terms cancel |
| Sum/product of solutions | Vieta's: sum \(= -b/a\), product \(= c/a\) | No need to solve explicitly |
Question 1 of 5
How many solutions does the system \(y = x^2\) and \(y = -3\) have?
Setting the equations equal gives \(x^2 = -3\). This equation has no real solutions because the square of a real number cannot be negative. Graphically, the parabola \(y=x^2\) has its minimum at \((0,0)\) and never reaches the horizontal line \(y=-3\).
Question 2 of 5
What are the solutions to the system \(y = x^2 - 5x + 6\) and \(y = 0\)?
Set \(x^2 - 5x + 6 = 0\). Factor: \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\). Since \(y=0\) for both, the solutions are \((2,0)\) and \((3,0)\).
Question 3 of 5
For what value of \(k\) does the system \(y = x^2 + 1\) and \(y = 2x + k\) have exactly one solution?
Set \(x^2+1=2x+k\), giving \(x^2-2x+(1-k)=0\). For one solution, the discriminant \(\Delta\) must be 0. Here, \(a=1, b=-2, c=1-k\). So, \(\Delta = (-2)^2 - 4(1)(1-k) = 0\). This simplifies to \(4 - 4 + 4k = 0\), which means \(4k=0\), so \(k=0\).
Question 4 of 5
If \(y = x^2 + 3x\) and \(y = 2x^2 - x\), what is the sum of the \(x\)-coordinates of the solutions?
Set \(x^2+3x = 2x^2-x\), which simplifies to \(x^2-4x=0\), i.e., \(x(x-4)=0\). The solutions are \(x=0\) and \(x=4\). Their sum is \(0+4=\mathbf{4}\). Alternatively, by Vieta's formulas the sum is \(-(-4)/1 = 4\).
Question 5 of 5
The system \(y = x^2 - 4\) and \(y = mx\) has two solutions with \(x\)-values that are equal in magnitude but opposite in sign. What is \(m\)?
Set \(x^2-4=mx\), giving \(x^2-mx-4=0\). If the roots are \(r\) and \(-r\), their sum is \(0\). By Vieta's, the sum of roots is \(-b/a = -(-m)/1 = m\), so \(m=0\). Check: \(x^2-4=0 \Rightarrow x=\pm 2\), confirming equal magnitude and opposite sign.