Solve System of Equations

Solving Systems with Nonlinear Equations

This pattern gives you a system where at least one equation is nonlinear — usually a quadratic paired with a linear equation. The strategy is almost always substitution: set the two expressions for $y$ equal to each other, solve the resulting equation, and then find the other variable.

The Core Method

Given $y = \text{linear}$ and $y = \text{quadratic}$, set them equal:

Linear $=$ Quadratic → rearrange to standard form → solve for $x$ → plug back in for $y$

Worked Examples

Example 1. $a = -3$ and $b = a^2 + 5$. If $(a, b)$ is the solution, what is $b$?

A) $4$
B) $-4$
C) $14$
D) $2$

Substitute $a = -3$ into the second equation:
$b = (-3)^2 + 5 = 9 + 5 = 14$
Gotcha: Option B ($-4$) comes from computing $(-3)^2 = -9$ instead of $+9$. Remember: squaring any number (positive or negative) always gives a positive result.
The answer is C.

Example 2. $y = 54$ and $y = x^2 + 5$. What is a possible value of $x$?

A) $49$
B) $5$
C) $54$
D) $7$

Substitute $y = 54$: $54 = x^2 + 5$
$x^2 = 49$
$x = \pm 7$
Gotcha: Option A ($49$) is the value of $x^2$, not $x$. After isolating $x^2$, you still need to take the square root. Both $7$ and $-7$ are valid, but only $7$ appears in the choices.
The answer is D.

Example 3. $y = 2(x-5)(x+1)$ and $y = 12x - 60$. Which ordered pair is a solution?

A) $(0, 5)$
B) $(5, 0)$
C) $(-1, 0)$
D) $(5, -10)$

Set equal: $2(x-5)(x+1) = 12x - 60$
Factor the right side: $12(x - 5)$
$2(x-5)(x+1) = 12(x-5)$
If $x \neq 5$, divide both sides by $(x - 5)$: $2(x+1) = 12$, so $x + 1 = 6$, $x = 5$.
But we assumed $x \neq 5$... that means $x = 5$ is the only solution. Plug in: $y = 12(5) - 60 = 0$.
Solution: $(5, 0)$
Gotcha: Options A and C are x-intercepts or y-intercepts of individual equations, not intersection points of the system. The solution must satisfy both equations simultaneously.
The answer is B.

Example 4. $y = x + 2$ and $y = x^2 + 6x - 22$. What is the greatest possible value of $y$?

A) $-8$
B) $-6$
C) $3$
D) $5$

Set equal: $x + 2 = x^2 + 6x - 22$
$0 = x^2 + 5x - 24$
Factor: $(x + 8)(x - 3) = 0$, so $x = -8$ or $x = 3$
Find $y$ for each: $y = -8 + 2 = -6$ or $y = 3 + 2 = 5$
The greatest $y$ is $5$.
Gotcha: Don't confuse the $x$-values with the $y$-values. The question asks for the greatest $y$, not the greatest $x$. Always plug back into one of the original equations.
The answer is D.

Example 5. $6x + y = -10$ and $y = 2x^2 - 226$. What is one possible value of $x$?

A) $-6$
B) $12$
C) $9$
D) $-10$

From the first equation: $y = -6x - 10$
Substitute: $-6x - 10 = 2x^2 - 226$
$0 = 2x^2 + 6x - 216$
$0 = x^2 + 3x - 108$
Factor: $(x + 12)(x - 9) = 0$, so $x = -12$ or $x = 9$
The answer is C (either $x = 9$ works; $-12$ is not among the choices).

What to Do on Test Day

• Substitution is the go-to. If one equation is already solved for $y$ (or $x$), plug it into the other equation.
• After substitution, rearrange to standard form ($ax^2 + bx + c = 0$) and factor or use the quadratic formula.
• Squaring a negative gives a positive. $(-3)^2 = 9$, not $-9$.
• Don't stop at $x^2 = k$. Take the square root to get $x = \pm\sqrt{k}$, then check which values are among the answer choices.
• Two solutions are common. A linear-quadratic system typically has $0$, $1$, or $2$ intersection points. If asked for the "greatest" value, find both and pick the larger one.
• Plug back in to find $y$ after solving for $x$. Many wrong answers are $x$-values when the question asks for $y$, or vice versa.