ACCUPLACER Advanced Algebra & Functions Quiz: Writing Linear Equations
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Writing Linear EquationsQuestion 1 of 20

A line is parallel to y=3x−7y = 3x - 7 and has the same y-intercept as the line 4x−2y=124x - 2y = 12. What is the equation of this line?

y=3x−6y = 3x - 6
y=2x−7y = 2x - 7
y=3x+12y = 3x + 12
y=−13x−6y = -\frac{1}{3}x - 6
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Writing Linear Equations

Practice Writing Linear Equations in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Writing Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A line is parallel to y=3x−7y = 3x - 7 and has the same y-intercept as the line 4x−2y=124x - 2y = 12. What is the equation of this line?

  1. y=3x−6y = 3x - 6 (correct answer)
  2. y=2x−7y = 2x - 7
  3. y=3x+12y = 3x + 12
  4. y=−13x−6y = -\frac{1}{3}x - 6
Explanation: A line parallel to y=3x−7y = 3x - 7 must have the same slope, so m=3m = 3. To find the y-intercept of 4x−2y=124x - 2y = 12, set x=0x = 0: 4(0)−2y=12  ⟹  −2y=12  ⟹  y=−64(0) - 2y = 12 \implies -2y = 12 \implies y = -6. So, the y-intercept is b=−6b = -6. Combining the slope m=3m = 3 and y-intercept b=−6b = -6 gives the equation y=3x−6y = 3x - 6.

Question 2

What is the equation of the line that passes through the x-intercept of 3x−5y=153x - 5y = 15 and the y-intercept of y=2x+8y = 2x + 8?

  1. y=−85x+8y = -\frac{8}{5}x + 8 (correct answer)
  2. y=−34x−3y = -\frac{3}{4}x - 3
  3. y=85x+8y = \frac{8}{5}x + 8
  4. y=−58x+5y = -\frac{5}{8}x + 5
Explanation: First, find the required intercepts. For 3x−5y=153x - 5y = 15, the x-intercept occurs when y=0y = 0, so 3x=15  ⟹  x=53x = 15 \implies x = 5. The point is (5,0)(5, 0). For y=2x+8y = 2x + 8, the y-intercept occurs when x=0x = 0, so y=8y = 8. The point is (0,8)(0, 8). Now, find the equation of the line passing through (5,0)(5, 0) and (0,8)(0, 8). The y-intercept is b=8b = 8. The slope is m=8−00−5=−85m = \frac{8 - 0}{0 - 5} = -\frac{8}{5}. Using the slope-intercept form, the equation is y=−85x+8y = -\frac{8}{5}x + 8.

Question 3

A line passes through the point (p,q)(p, q) and is parallel to the line given by ax+by=cax + by = c, where a,b≠0a, b \neq 0. What is the equation of this line?

  1. ax+by=ap+bqax + by = ap + bq (correct answer)
  2. bx−ay=bp−aqbx - ay = bp - aq
  3. ax+by=cax + by = c
  4. ax−by=ap−bqax - by = ap - bq
Explanation: Any line parallel to ax+by=cax + by = c must have the form ax+by=Dax + by = D for some constant DD, because parallel lines in standard form have the same coefficients for x and y. To find the specific value of DD for the line that passes through (p,q)(p, q), substitute these coordinates into the equation: a(p)+b(q)=Da(p) + b(q) = D. Therefore, the constant DD is equal to ap+bqap + bq. The equation of the line is ax+by=ap+bqax + by = ap + bq.

Question 4

A specialty printing company charges a one-time setup fee for a design, plus a fixed price for each t-shirt printed. A customer is quoted a price of $190 for 20 shirts and $400 for 50 shirts.

Which of the following equations represents the total cost, CC, as a function of the number of shirts, nn?

  1. C=3.5n+120C = 3.5n + 120
  2. C=7n+50C = 7n + 50 (correct answer)
  3. C=8n+30C = 8n + 30
  4. C=10.55nC = 10.55n
Explanation: The problem provides two points on a linear function: (20, 190) and (50, 400). First, calculate the slope (price per shirt): m=400−19050−20=21030=7m = \frac{400 - 190}{50 - 20} = \frac{210}{30} = 7. Then, use the point-slope form with one of the points, such as (20, 190): C−190=7(n−20)C - 190 = 7(n - 20). Distributing the 7 gives C−190=7n−140C - 190 = 7n - 140. Adding 190 to both sides yields C=7n+50C = 7n + 50. This means the setup fee is $50 and each shirt costs $7.

Question 5

An empty water tank is being filled. It has a capacity of 1,000 gallons. The water level rises at a rate of 2.5 inches per minute. For this particular tank, each inch of water height corresponds to 10 gallons of water.

Which equation models the volume of water VV, in gallons, in the tank after tt minutes?

  1. V(t)=10tV(t) = 10t
  2. V(t)=2.5tV(t) = 2.5t
  3. V(t)=12.5tV(t) = 12.5t
  4. V(t)=25tV(t) = 25t (correct answer)
Explanation: The problem requires finding the rate of change of volume (gallons per minute), which is the slope of the linear function. The rate of height change is 2.5 inches/minute. The conversion factor is 10 gallons/inch. To find the rate in gallons per minute, multiply these values: Rate = (2.5 inches/minute) × (10 gallons/inch) = 25 gallons/minute. Since the tank starts empty, the initial value (y-intercept) is 0. Therefore, the equation is V(t)=25tV(t) = 25t.

Question 6

A line passes through the point (−6,2)(-6, 2) and has a slope of 23\frac{2}{3}. Which of the following is the equation of the line in standard form Ax+By=CAx + By = C, where A,BA, B, and CC are integers with A>0A > 0?

  1. 2x−3y=−182x - 3y = -18 (correct answer)
  2. 2x+3y=−62x + 3y = -6
  3. 2x−3y=−122x - 3y = -12
  4. 3x+2y=−143x + 2y = -14
Explanation: Start with the point-slope form: y−y1=m(x−x1)y - y_1 = m(x - x_1). Substituting the given point and slope gives y−2=23(x−(−6))y - 2 = \frac{2}{3}(x - (-6)), which is y−2=23(x+6)y - 2 = \frac{2}{3}(x + 6). To eliminate the fraction, multiply the entire equation by 3: 3(y−2)=2(x+6)3(y - 2) = 2(x + 6). Distribute on both sides: 3y−6=2x+123y - 6 = 2x + 12. Rearrange into standard form Ax+By=CAx + By = C by moving the x-term to the left and the constant to the right: −2x+3y=18-2x + 3y = 18. Since the convention is for AA to be positive, multiply the entire equation by -1 to get 2x−3y=−182x - 3y = -18.

Question 7

A piece of machinery is purchased for $65,000. Its value depreciates linearly over 10 years to a scrap value of $5,000.

Which equation represents the value VV of the machine tt years after its purchase, for 0≤t≤100 \le t \le 10?

  1. V(t)=−6000t+5000V(t) = -6000t + 5000
  2. V(t)=6000t+5000V(t) = 6000t + 5000
  3. V(t)=−6500t+65000V(t) = -6500t + 65000
  4. V(t)=−6000t+65000V(t) = -6000t + 65000 (correct answer)
Explanation: This is a linear relationship where time tt is the independent variable and value VV is the dependent variable. We have two points: at t=0t=0, V=65000V=65000, and at t=10t=10, V=5000V=5000. The point (0,65000)(0, 65000) gives us the y-intercept, b=65000b = 65000. The slope mm is the rate of depreciation: m=5000−6500010−0=−6000010=−6000m = \frac{5000 - 65000}{10 - 0} = \frac{-60000}{10} = -6000. Using the slope-intercept form V(t)=mt+bV(t) = mt + b, the equation is V(t)=−6000t+65000V(t) = -6000t + 65000.

Question 8

Line L1L_1 passes through the point (4,1)(4, 1) and is perpendicular to the line y=−2x+7y = -2x + 7. Line L2L_2 is parallel to the line 3x−y=53x - y = 5 and has the same x-intercept as line L1L_1. What is the equation of line L2L_2?

  1. y=3x−2y = 3x - 2
  2. y=12x+2y = \frac{1}{2}x + 2
  3. y=3x−6y = 3x - 6 (correct answer)
  4. y=−2x+4y = -2x + 4
Explanation: This is a multi-step problem. First, find the equation of L1L_1. The slope of y=−2x+7y = -2x + 7 is -2. The perpendicular slope is 12\frac{1}{2}. Using point (4,1)(4, 1), L1L_1 is y−1=12(x−4)  ⟹  y=12x−1y - 1 = \frac{1}{2}(x - 4) \implies y = \frac{1}{2}x - 1. Second, find the x-intercept of L1L_1 by setting y=0y = 0: 0=12x−1  ⟹  1=12x  ⟹  x=20 = \frac{1}{2}x - 1 \implies 1 = \frac{1}{2}x \implies x = 2. So L2L_2 passes through (2,0)(2, 0). Third, find the slope of L2L_2. L2L_2 is parallel to 3x−y=53x - y = 5, which is y=3x−5y = 3x - 5. The slope is 3. Finally, write the equation for L2L_2 with slope 3 passing through (2,0)(2, 0): y−0=3(x−2)  ⟹  y=3x−6y - 0 = 3(x - 2) \implies y = 3x - 6.

Question 9

The monthly cost to run a factory is a linear function of the number of units produced. In May, 1,500 units were produced and the cost was $28,000. In June, 2,000 units were produced and the cost was $34,000. What is the equation for the cost CC as a function of the number of units uu?

  1. C(u)=10u+13000C(u) = 10u + 13000
  2. C(u)=17u+2500C(u) = 17u + 2500
  3. C(u)=12u+10000C(u) = 12u + 10000 (correct answer)
  4. C(u)=15u+4000C(u) = 15u + 4000
Explanation: The information gives two points: (1500, 28000) and (2000, 34000). First, calculate the slope (cost per unit): m=34000−280002000−1500=6000500=12m = \frac{34000 - 28000}{2000 - 1500} = \frac{6000}{500} = 12. This is the variable cost. Next, use the point-slope form with the point (1500, 28000): C−28000=12(u−1500)C - 28000 = 12(u - 1500). Distribute the 12: C−28000=12u−18000C - 28000 = 12u - 18000. Add 28000 to both sides to find the equation in slope-intercept form: C(u)=12u+10000C(u) = 12u + 10000. The fixed cost is $10,000.

Question 10

A community garden sells small plots, ss, for $45 and large plots, ll, for $75. The garden's total revenue from selling all available plots is $9,000. Which of the following equations, in simplified standard form, represents this relationship?

  1. 45s+75l=900045s + 75l = 9000
  2. 75s+45l=900075s + 45l = 9000
  3. 3s+5l=6003s + 5l = 600 (correct answer)
  4. 5s+3l=6005s + 3l = 600
Explanation: The total revenue is the sum of the revenue from small plots and large plots. Revenue from small plots is 45s45s and revenue from large plots is 75l75l. The total revenue equation is 45s+75l=900045s + 75l = 9000. To simplify this equation, find the greatest common divisor (GCD) of 45, 75, and 9000. The GCD of 45 and 75 is 15. Since 9000 is also divisible by 15 (9000 / 15 = 600), we can divide the entire equation by 15. This gives 45s15+75l15=900015\frac{45s}{15} + \frac{75l}{15} = \frac{9000}{15}, which simplifies to 3s+5l=6003s + 5l = 600.

Question 11

The value of a collectible item, VV, is modeled by a linear function of time in years, tt. The value decreases by $150 for every 5-year increase in time. The item's value was $2,500 in the year 2010.

Which equation gives the value V(t)V(t) of the item, where tt is the number of years after 2000?

  1. V(t)=−30t+2500V(t) = -30t + 2500
  2. V(t)=−30t+2800V(t) = -30t + 2800 (correct answer)
  3. V(t)=30t+2200V(t) = 30t + 2200
  4. V(t)=−150t+4000V(t) = -150t + 4000
Explanation: First, determine the slope (rate of change). The value decreases by $150 over 5 years, so the slope is m=ΔVΔt=−1505=−30m = \frac{\Delta V}{\Delta t} = \frac{-150}{5} = -30 dollars per year. Next, identify a point on the line. The year 2010 is t=10t = 10 years after 2000, and the value was $2,500. So, we have the point (10,2500)(10, 2500). Using the point-slope form V−V1=m(t−t1)V - V_1 = m(t - t_1), we get V−2500=−30(t−10)V - 2500 = -30(t - 10). Distributing gives V−2500=−30t+300V - 2500 = -30t + 300. Adding 2500 to both sides yields V(t)=−30t+2800V(t) = -30t + 2800.

Question 12

A linear function g(x)g(x) satisfies the conditions g(−2)=5g(-2) = 5 and g(4)=−7g(4) = -7. Which of the following defines g(x)g(x)?

  1. g(x)=−2x+1g(x) = -2x + 1 (correct answer)
  2. g(x)=2x+9g(x) = 2x + 9
  3. g(x)=−12x+4g(x) = -\frac{1}{2}x + 4
  4. g(x)=−2x+9g(x) = -2x + 9
Explanation: The given conditions correspond to the points (−2,5)(-2, 5) and (4,−7)(4, -7). First, find the slope of the line passing through these points: m=−7−54−(−2)=−126=−2m = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2. Next, use the point-slope form with the point (−2,5)(-2, 5): y−5=−2(x−(−2))y - 5 = -2(x - (-2)). This simplifies to y−5=−2(x+2)  ⟹  y−5=−2x−4y - 5 = -2(x + 2) \implies y - 5 = -2x - 4. Adding 5 to both sides gives y=−2x+1y = -2x + 1. Thus, g(x)=−2x+1g(x) = -2x + 1.

Question 13

Which of the following is an equation of the line passing through the points (12,−1)(\frac{1}{2}, -1) and (−13,23)(-\frac{1}{3}, \frac{2}{3})?

  1. y=−2xy = -2x (correct answer)
  2. y=−2x−2y = -2x - 2
  3. y=2x−2y = 2x - 2
  4. y=25x−65y = \frac{2}{5}x - \frac{6}{5}
Explanation: First, calculate the slope mm using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. m=23−(−1)−13−12=23+1−26−36=53−56m = \frac{\frac{2}{3} - (-1)}{-\frac{1}{3} - \frac{1}{2}} = \frac{\frac{2}{3} + 1}{-\frac{2}{6} - \frac{3}{6}} = \frac{\frac{5}{3}}{-\frac{5}{6}}. To divide by a fraction, multiply by its reciprocal: m=53⋅(−65)=−3015=−2m = \frac{5}{3} \cdot (-\frac{6}{5}) = -\frac{30}{15} = -2. Now use the point-slope form with the point (12,−1)(\frac{1}{2}, -1): y−(−1)=−2(x−12)y - (-1) = -2(x - \frac{1}{2}). This simplifies to y+1=−2x+1y + 1 = -2x + 1. Subtracting 1 from both sides gives y=−2xy = -2x.

Question 14

A company's profit follows a linear relationship with the number of units sold. When 50 units are sold, the profit is $2,000. When 80 units are sold, the profit is $3,200. What will be the profit when 100 units are sold?

  1. $4,000 (correct answer)
  2. $4,200
  3. $4,400
  4. $4,600
Explanation: Using points (50, 2000) and (80, 3200), the slope is (3200 - 2000)/(80 - 50) = 1200/30 = 40 dollars per unit. Using point-slope form with (50, 2000): P - 2000 = 40(u - 50), which gives P = 40u - 2000 + 2000 = 40u. Verification: at u = 50, P = 40(50) = 2000 ✓; at u = 80, P = 40(80) = 3200 ✓. At u = 100: P = 40(100) = 4000. Choice B adds an incorrect base fee. Choice C uses wrong rate calculation. Choice D compounds multiple calculation errors.

Question 15

A cell phone plan charges a monthly fee plus a rate per minute of usage. In January, 120 minutes of usage cost $45 total. In February, 200 minutes of usage cost $65 total. What equation represents the total monthly cost $CC forfor mm $ minutes of usage?

  1. C=0.25m+15C = 0.25m + 15
  2. C=0.25m+25C = 0.25m + 25 (correct answer)
  3. C=0.30m+9C = 0.30m + 9
  4. C=0.20m+21C = 0.20m + 21
Explanation: Using points (120, 45) and (200, 65), the rate per minute is (65 - 45)/(200 - 120) = 20/80 = 0.25 dollars per minute. Using point-slope form with (120, 45): C - 45 = 0.25(m - 120), which gives C = 0.25m - 30 + 45 = 0.25m + 25. The monthly fee is $25 and the rate is $0.25 per minute. Choice A uses the correct rate but calculates the monthly fee as $15 instead of $25. Choice C uses an incorrect rate of $0.30. Choice D uses an incorrect rate of $0.20 and wrong monthly fee.

Question 16

A water tank initially contains 150 gallons and drains at a constant rate. After 4 hours, it contains 90 gallons, and after 7 hours, it contains 45 gallons. Which equation represents the amount of water ww (in gallons) in the tank after tt hours?

  1. w=150−15tw = 150 - 15t (correct answer)
  2. w=150−20tw = 150 - 20t
  3. w=180−15tw = 180 - 15t
  4. w=135−15tw = 135 - 15t
Explanation: First, find the rate of change using the two given points: (4, 90) and (7, 45). The slope is (45 - 90)/(7 - 4) = -45/3 = -15 gallons per hour. Using point-slope form with (4, 90): w - 90 = -15(t - 4), which simplifies to w = -15t + 60 + 90 = 150 - 15t. We can verify: at t = 0, w = 150 ✓. Choice B uses an incorrect rate of -20. Choice C incorrectly calculates the y-intercept as 180. Choice D uses the correct rate but wrong y-intercept of 135.

Question 17

The temperature of a cooling object decreases linearly over time. At t=3t = 3 minutes, the temperature is 85°C85°C, and the temperature decreases at a rate of 12°C12°C per minute. Which equation gives the temperature TT (in degrees Celsius) as a function of time tt (in minutes)?

  1. T=85−12(t−3)T = 85 - 12(t - 3)
  2. T=121−12tT = 121 - 12t (correct answer)
  3. T=49−12tT = 49 - 12t
  4. T=97−12tT = 97 - 12t
Explanation: Given that the temperature decreases at 12°C per minute, the slope is -12. Using point-slope form with (3, 85): T - 85 = -12(t - 3), which expands to T - 85 = -12t + 36, so T = -12t + 121 or T = 121 - 12t. Choice A is in point-slope form but wasn't simplified to slope-intercept form. Choice C incorrectly calculates the y-intercept as 49. Choice D incorrectly calculates the y-intercept as 97.

Question 18

A line passes through C(−3,5)C(-3, 5) and is perpendicular to the line that passes through A(7,−1)A(7, -1) and B(4,8)B(4, 8). What is the equation of this line?

  1. y=−3x−4y = -3x - 4
  2. y=13x+6y = \frac{1}{3}x + 6 (correct answer)
  3. y=−13x+4y = -\frac{1}{3}x + 4
  4. y=3x+14y = 3x + 14
Explanation: First, find the slope of the line through points A and B: mAB=8−(−1)4−7=9−3=−3m_{AB} = \frac{8 - (-1)}{4 - 7} = \frac{9}{-3} = -3. The line we want is perpendicular to this, so its slope is the negative reciprocal of -3, which is m⊥=13m_{\perp} = \frac{1}{3}. Now, use the point-slope form with the point C(−3,5)C(-3, 5) and the perpendicular slope: y−5=13(x−(−3))y - 5 = \frac{1}{3}(x - (-3)). This simplifies to y−5=13(x+3)  ⟹  y−5=13x+1y - 5 = \frac{1}{3}(x + 3) \implies y - 5 = \frac{1}{3}x + 1. Adding 5 to both sides gives y=13x+6y = \frac{1}{3}x + 6.

Question 19

What is the equation of the line that passes through the point (−7,4)(-7, 4) and is perpendicular to the y-axis?

  1. x=4x = 4
  2. x=−7x = -7
  3. y=−7y = -7
  4. y=4y = 4 (correct answer)
Explanation: The y-axis is a vertical line. A line perpendicular to a vertical line must be a horizontal line. The equation of any horizontal line is of the form y=ky = k, where kk is a constant equal to the y-coordinate of every point on the line. Since the line must pass through (−7,4)(-7, 4), its y-coordinate must be 4. Therefore, the equation of the line is y=4y = 4.

Question 20

A line has a positive y-intercept and a positive x-intercept. The y-intercept is twice the x-intercept. If the line passes through the point (2,4)(2, 4), what is its equation?

  1. y=−x+6y = -x + 6
  2. y=−12x+5y = -\frac{1}{2}x + 5
  3. y=2xy = 2x
  4. y=−2x+8y = -2x + 8 (correct answer)
Explanation: Let the x-intercept be at (a,0)(a, 0) and the y-intercept be at (0,b)(0, b). The problem states b=2ab = 2a. The slope of the line is m=b−00−a=2a−a=−2m = \frac{b - 0}{0 - a} = \frac{2a}{-a} = -2. The equation of the line is in the form y=mx+by = mx + b, so y=−2x+by = -2x + b. Since the line passes through (2,4)(2, 4), we can substitute these values to find bb: 4=−2(2)+b  ⟹  4=−4+b  ⟹  b=84 = -2(2) + b \implies 4 = -4 + b \implies b = 8. Thus, the equation of the line is y=−2x+8y = -2x + 8.