# SAT Nonlinear Equations and Systems Worksheet PDF with Answers

Source: https://1600.now/sat-nonlinear-equations-and-systems-worksheet

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## SAT Nonlinear Equations and Systems Worksheet

Solve nonlinear equations by reducing them to a form you can factor, split into cases, or solve with the quadratic formula. Check every candidate in the original equation after squaring or clearing a denominator.

Written by [Luke Finigan](https://1600.now/about)

The 12 questions cover quadratics, absolute values, radicals, rational equations, and intersections of lines with parabolas. Choose a student PDF or the version with explained answers.

12 questions · about 25 minutes · Advanced Math › Nonlinear equations in one variable and systems of equations in two variables

[Download 4-page PDF](https://1600.now/downloads/worksheets/sat-nonlinear-equations-and-systems-worksheet-student-letter.pdf)

[Share to Google Classroom](https://classroom.google.com/share?url=https%3A%2F%2F1600.now%2Fsat-nonlinear-equations-and-systems-worksheet) · This worksheet is free to print and share under [CC BY-NC-ND 4.0](https://creativecommons.org/licenses/by-nc-nd/4.0/). Tutors may also use it in paid sessions. If you post it online, please link to this page instead of re-uploading the PDF. These are original questions, not College Board items.

### Keep restrictions with the equation

For $|x-a|=b$ with $b>0$, solve $x-a=b$ and $x-a=-b$. For a quadratic, move all terms to one side and factor before dividing by a variable; division by $x$ can discard the solution $x=0$.

A square-root expression is nonnegative. In $\sqrt{3x+10}=x$, that immediately requires $x\ge0$. Squaring may produce an algebraic candidate outside that condition. Substitution into the original equation is the final test.

Write excluded denominator values before clearing fractions. For a nonlinear system, substitute one equation into the other, solve for the possible $x$-values, and compute each corresponding $y$-value. Each intersection must satisfy both equations.

Try it yourself: What are the solutions to $x^2=3x$? · A $3$ only B $0$ and $3$ C $-3$ and $3$

### Questions

1. **Question 1**

   What are the solutions to $x^2 - 7x + 10 = 0$?

   A. $-5$ and $-2$ B. $-2$ and 5 C. 2 and 5 D. 1 and 10

   **C.** Factor: $(x - 2)(x - 5) = 0$, so $x = 2$ or $x = 5$. Choice A has the signs reversed; those are the solutions to $x^2 + 7x + 10 = 0$.
2. **Question 2**

   What are the solutions to $|x - 4| = 9$?

   A. 13 only B. 5 and 13 C. $-13$ and 5 D. $-5$ and 13

   **D.** The expression inside the absolute value bars equals 9 or $-9$. If $x - 4 = 9$, then $x = 13$. If $x - 4 = -9$, then $x = -5$.
3. **Question 3**

   What are all solutions to $2x^{2}=8x$?

   A. 0 and 4 B. 4 only C. −4 and 4 D. 0 and −4

   **A.** Move every term to one side: $2x^{2}-8x=0$, so $2x(x-4)=0$ and $x=0$ or $x=4$. Dividing both sides by $x$ loses the solution 0.
4. **Question 4**

   What value of $x$ satisfies $\sqrt{x+5}=4$?

   A. −1 B. 3 C. 11 D. 21

   **C.** Square both sides: $x+5=16$, so $x=11$. Check: $\sqrt{16}$ is 4. Choice B comes from doubling 4 instead of squaring it.
5. **Question 5**

   What is the solution to $\dfrac{6}{x-1}=\dfrac{3}{x-4}$?

   A. −2 B. 3 C. 5 D. 7

   **D.** Cross-multiply: $6(x-4)=3(x-1)$, so $6x-24=3x-3$ and $x=7$. Check: $\dfrac{6}{6}=1$ and $\dfrac{3}{3}=1$. Choice A multiplies each numerator by its own denominator: $6(x-1)=3(x-4)$.
6. **Question 6**

   How many distinct real solutions does $2x^{2}-4x+5=0$ have?

   A. Zero B. Exactly one C. Exactly two D. Infinitely many

   **A.** The discriminant is $b^{2}-4ac=(-4)^{2}-4(2)(5)=16-40=-24$. A negative discriminant means there are no real solutions.
7. **Question 7**

   Which point $(x,y)$ is a solution to the system of equations $x^{2}+y^{2}=25$ and $y=x+1$?

   A. (0, 5) B. (3, 4) C. (4, 3) D. (−3, −2)

   **B.** A solution must satisfy both equations. For (3, 4), $3^{2}+4^{2}=25$ and $4=3+1$. Choices A and C satisfy only the first equation, and choice D satisfies only the second.
8. **Question 8**

   What are all values of $x$ that satisfy $x=\sqrt{3x+10}$?

   A. −2 only B. $-2$ and 5 C. 5 only D. 2 and 5

   **C.** Squaring gives $x^{2}=3x+10$, so $(x-5)(x+2)=0$. Check both: $\sqrt{25}$ is 5, so 5 works. $\sqrt{4}$ is 2, not −2, so −2 does not.
9. **Question 9**

   What is the larger solution to $x^{2}-6x+4=0$?

   A. $-3+\sqrt{5}$ B. $3+\sqrt{5}$ C. $3+\sqrt{13}$ D. $6+2\sqrt{5}$

   **B.** Complete the square: $x^{2}-6x+9=5$, so $(x-3)^{2}=5$. The solutions are $3+\sqrt{5}$ and $3-\sqrt{5}$. Choice D skips dividing by 2 in the quadratic formula.
10. **Question 10**

    In the $xy$-plane, the graphs of $y=x^{2}+6x$ and $y=c$ intersect at exactly one point, where $c$ is a constant. What is the value of $c$?

    A. −9 B. −3 C. 0 D. 9

    **A.** A horizontal line meets this upward-opening parabola exactly once, at its vertex. The vertex has $x=-\dfrac{6}{2}=-3$ and $y=(-3)^{2}+6(-3)=-9$, so $c=-9$. Choice B is the $x$-coordinate of the vertex.
11. **Question 11**

    What value of $x$ satisfies $\dfrac{x}{x-3}=\dfrac{3}{x-3}+2$?

    A. 0 B. 3 C. 6 D. No value of $x$ satisfies the equation.

    **D.** Multiplying both sides by $x-3$ gives $x=3+2(x-3)$, so $x=3$. But $x=3$ makes both denominators 0, so it is not a solution, and the equation has no solution.
12. **Question 12**

    The graphs of $y=x^{2}-2x-3$ and $y=-x+k$ in the $xy$-plane intersect at a point with $x$-coordinate 4, where $k$ is a constant. What is the $x$-coordinate of the other intersection point?

    A. −4 B. −3 C. 3 D. 9

    **B.** At $x=4$, the parabola has $y=16-8-3=5$, so $5=-4+k$ and $k=9$. Setting $x^{2}-2x-3=-x+9$ gives $x^{2}-x-12=0$, or $(x-4)(x+3)=0$. The other intersection has $x=-3$. Choice D is the value of $k$.

0/12 correct

### Check your nonlinear equations answers

Each row is a wrong answer choice from one of the sheet's problems. Find the one you picked, then run the check in the last column.

| What your answer looked like | What actually happened | Fix it next time |
| --- | --- | --- |
| 13 only in problem 2 | An absolute value equation has two cases. $\|x - 4\| = 9$ means $x - 4$ is 9 or $-9$. | Solve both: $x = 13$ or $x = -5$. Check $-5$: $\|-5 - 4\| = \|-9\| = 9$. |
| 4 only in problem 3 | Dividing both sides of $2x^2 = 8x$ by $x$ throws away $x = 0$, which also works. | Move everything to one side and factor: $2x(x - 4) = 0$, so $x = 0$ or $x = 4$. Check 0: $2(0)^2 = 0$ and $8(0) = 0$. |
| 3 in problem 4 | You doubled 4 instead of squaring it, so you solved $x + 5 = 8$. | Square both sides: $x + 5 = 16$, so $x = 11$. Check: $\sqrt{16} = 4$. |
| $-2$ in problem 5 | You multiplied each numerator by its own denominator: $6(x - 1) = 3(x - 4)$. Cross-multiplying pairs each numerator with the other fraction's denominator. | $6(x - 4) = 3(x - 1)$ gives $3x = 21$, so $x = 7$. Check: $\dfrac{6}{7 - 1} = 1$ and $\dfrac{3}{7 - 4} = 1$. |
| $-2$ and 5 in problem 8, or 3 in problem 11 | Squaring in problem 8 and multiplying by $x - 3$ in problem 11 each produced a value that fails the original equation. $\sqrt{3(-2) + 10} = 2$, not $-2$, and $x = 3$ makes both denominators in problem 11 equal 0. | Check every candidate in the original equation. Problem 8 keeps only 5, and problem 11 has no solution. |
| $6 + 2\sqrt{5}$ in problem 9 | That's the numerator of the quadratic formula before you divide by $2a = 2$. | $x = \dfrac{6 \pm \sqrt{20}}{2} = \dfrac{6 \pm 2\sqrt{5}}{2} = 3 \pm \sqrt{5}$, so the larger solution is $3 + \sqrt{5}$. |

#### Worked example: a parabola and a line with an unknown constant

Problem 12 says the graphs of $y = x^2 - 2x - 3$ and $y = -x + k$ meet at a point with $x$-coordinate 4, and asks for the other intersection. Find $k$ first. On the parabola, $x = 4$ gives $y = 16 - 8 - 3 = 5$, so $(4, 5)$ is on the line too: $5 = -4 + k$, and $k = 9$.

Now set the two expressions equal: $x^2 - 2x - 3 = -x + 9$, which becomes $x^2 - x - 12 = 0$, or $(x - 4)(x + 3) = 0$. The root $x = 4$ is the point you were given, so the answer is $-3$. Check: at $x = -3$, the parabola gives $9 + 6 - 3 = 12$ and the line gives $3 + 9 = 12$. The choice 9 is $k$, and 3 has the wrong sign.

### Apply the check to fresh questions

For another nonlinear equation, carry restrictions alongside your work and test every candidate. Use the practice link for fresh radical, rational, and system questions.

[Practice nonlinear equations questions](https://1600.now/bank/math/skill/Nonlinear%20equations%20in%20one%20variable%20and%20systems%20of%20equations%20in%20two%20variables)

[Review the nonlinear equations and systems method](https://1600.now/sat-skill/nonlinear-equations-and-systems)

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