# SAT Linear Equations Worksheet PDF with Answers | 1600.now Source: https://1600.now/sat-linear-equations-one-variable-worksheet --- ## SAT Linear Equations Worksheet Solve a one-variable linear equation by performing the same operation on both sides. Distribute across every term, clear fractions carefully, and check the final value in the original equation. Written by [Luke Finigan](https://1600.now/about) This 12-question worksheet includes word problems, constants, and equations with one, no, or infinitely many solutions. Download a student PDF for an attempt without answers or choose the full explained version. 12 questions · about 25 minutes · Algebra › Linear equations in one variable [Download 4-page PDF](https://1600.now/downloads/worksheets/sat-linear-equations-one-variable-worksheet-student-letter.pdf) [Share to Google Classroom](https://classroom.google.com/share?url=https%3A%2F%2F1600.now%2Fsat-linear-equations-one-variable-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. ### Simplify before deciding how many solutions exist Distribute first, then collect the variable terms on one side and the constants on the other. For $3(2x-5)=21$, distribution gives $6x-15=21$, then $6x=36$ and $x=6$. Substitution gives $3(12-5)=21$. After simplifying, an equation of the form $ax=b$ has one solution when $a\ne0$. If $a=0$ and $b\ne0$, it has no solution. If both are zero, every allowed value works. A parameter question asks which coefficient produces one of these cases. In a word problem, define what $x$ measures and distinguish an amount remaining from an amount used. Reread the requested expression at the end: finding $x$ does not finish a question asking for $2x+1$. Try it yourself How many solutions does $5x+2=5x-3$ have? A One B None C Infinitely many ### Questions 1. **Question 1** If $3(2x - 5) = 21$, what is $x$? A. 9 B. 6 C. 3 D. 12 **B.** Distribute: $6x - 15 = 21$. Add 15 to get $6x = 36$, so $x = 6$. Checking gives $3(12 - 5) = 21$. 2. **Question 2** If $\dfrac{x}{4} + 3 = 8$, what is $x$? A. 11 B. 32 C. 5 D. 20 **D.** Subtract 3: $\dfrac{x}{4} = 5$. Multiply both sides by 4 to obtain $x = 20$. 3. **Question 3** If $5x+7=2x-11$, what is $x$? A. −6 B. 4 C. 6 D. −4 **A.** Subtract $2x$ and 7 from both sides: $3x=-18$. Divide by 3 to obtain $x=-6$. 4. **Question 4** A repair costs a fixed fee of $35 plus $18 per hour. A customer pays $107 before tax. How many hours of work are charged? A. 2 B. 3 C. 4 D. 5 **C.** Set $35+18h=107$. Then $18h=72$, so $h=4$ hours. The fixed fee is paid once. 5. **Question 5** For which value of $k$ does the equation $kx+4=7x-2$ have no solution? A. −2 B. 7 C. −7 D. 4 **B.** No solution occurs when the $x$ coefficients are equal but constants differ. With $k=7$, subtracting $7x$ gives the false statement $4=-2$. 6. **Question 6** The equation $2(3x+a)=6x+10$ is true for every real $x$. What is $a$? A. 5 B. 3 C. 10 D. 2 **A.** Expand to $6x+2a=6x+10$. For every $x$, the constants must agree: $2a=10$, so $a=5$. 7. **Question 7** If $4(x+2)-3x=19$, what is $x$? A. 3 B. 7 C. 11 D. 27 **C.** Distribute and combine like terms: $4x+8-3x=19$ becomes $x+8=19$. Subtract 8 to obtain $x=11$. 8. **Question 8** If $\dfrac{2x-3}{5}=7$, what is $x$? A. 16 B. 19 C. 20 D. 35 **B.** Multiply both sides by 5: $2x-3=35$. Adding 3 gives $2x=38$, so $x=19$. 9. **Question 9** A tank initially contains 120 liters of water and drains at 8 liters per minute. After how many minutes will 56 liters remain? A. 7 B. 8 C. 15 D. 22 **B.** Set $120-8t=56$. The tank loses 64 liters, so $t=\dfrac{64}{8}=8$ minutes. 10. **Question 10** If $\dfrac{x-2}{3}=\dfrac{x+4}{5}$, what is $x$? A. 4 B. 8 C. 11 D. 22 **C.** Multiply by 15 to get $5(x-2)=3(x+4)$. Thus $5x-10=3x+12$, giving $2x=22$ and $x=11$. 11. **Question 11** For which value of $a$ does $5x+2=5(x+a)$ have infinitely many solutions? A. $-\dfrac{2}{5}$ B. $\dfrac{2}{5}$ C. 2 D. $\dfrac{5}{2}$ **B.** The right side expands to $5x+5a$. The equations agree for every $x$ only if $2=5a$, so $a=\dfrac{2}{5}$. 12. **Question 12** If $7x-4=3x+20$, what is $2x+1$? A. 6 B. 11 C. 12 D. 13 **D.** First solve $4x=24$, giving $x=6$. The question asks for $2x+1$, which is $12+1=13$. 0/12 correct ### Check your linear 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 | | --- | --- | --- | | 32 for $\dfrac{x}{4} + 3 = 8$ (problem 2) | You multiplied 8 by 4 before dealing with the $+3$. That solves $\dfrac{x}{4} = 8$, a different equation. | Undo the addition first: $\dfrac{x}{4} = 5$, so $x = 20$. Check 32: $\dfrac{32}{4} + 3 = 11$, not 8. | | 6 for $5x + 7 = 2x - 11$ (problem 3) | A sign slipped while you moved the constants. Subtracting 7 from $-11$ gives $-18$, not 18. | From $3x = -18$, $x = -6$. Check 6 in the original: the left side is $5(6) + 7 = 37$, and the right side is $2(6) - 11 = 1$. | | 10 in problem 6 | The 2 outside the parentheses multiplies $a$ too, so the left side is $6x + 2a$, not $6x + a$. | Match the constants: $2a = 10$, so $a = 5$. Problem 11 has the same trap: $5(x + a) = 5x + 5a$, so $5a = 2$ and $a = \dfrac{2}{5}$, not 2. | | 16 for $\dfrac{2x - 3}{5} = 7$ (problem 8) | You subtracted 3 when you needed to add it. After multiplying by 5, the equation is $2x - 3 = 35$. | Add 3 to get $2x = 38$, so $x = 19$. Check: $\dfrac{2(19) - 3}{5} = \dfrac{35}{5} = 7$. | | 7 or 15 in problem 9 | $\dfrac{56}{8} = 7$ is how long 56 liters take to drain, and $\dfrac{120}{8} = 15$ is when the tank is empty. The question says 56 liters remain. | Find what drains: $120 - 56 = 64$ liters, and $\dfrac{64}{8} = 8$ minutes. Check: $120 - 8(8) = 56$. | | 6 or 12 in problem 12 | Those are $x$ and $2x$. The question asks for $2x + 1$. | Solve $4x = 24$ for $x = 6$, then finish the expression: $2(6) + 1 = 13$. | #### Worked example: the equation with no solution Problem 5 asks for the value of $k$ that gives $kx + 4 = 7x - 2$ no solution. Collect the $x$-terms on one side and the constants on the other: $kx - 7x = -2 - 4$, which is $(k - 7)x = -6$. If $k - 7$ isn't 0, you can divide by it and get exactly one solution, $x = \dfrac{-6}{k - 7}$. If $k = 7$, the left side is $0 \cdot x$, which is 0 for every $x$, and $0 = -6$ is never true. So the answer is 7. With $k = -7$, the equation becomes $-14x = -6$, which has the solution $x = \dfrac{3}{7}$. Problem 11 is the other case: when the $x$-terms and the constants both match, every $x$ works. ### Apply the check to fresh questions On another linear-equations question, finish by substituting into the original equation and checking the requested quantity. The practice link provides fresh problems on the same skill. [Practice linear equations questions](https://1600.now/bank/math/skill/Linear%20equations%20in%20one%20variable) [Review the linear equations in one variable method](https://1600.now/sat-skill/linear-equations-one-variable) ### More Math worksheets [All worksheets](https://1600.now/sat-worksheets) - [Linear Equations in Two Variables](https://1600.now/sat-linear-equations-two-variables-worksheet) - [Systems of Equations](https://1600.now/sat-systems-of-linear-equations-worksheet) - [Linear Functions](https://1600.now/sat-linear-functions-worksheet)