College Algebra Quiz: Solve Systems Graphically And Interpret Intersections

5 questions · exam conditions

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Solve Systems Graphically And Interpret IntersectionsQuestion 1 of 5

A system of equations has lines with slopes $m_1 = 3$ and $m_2 = 3$ , but different y-intercepts. When graphed, these lines will:

Intersect at exactly one point, making the system consistent and independent

Be parallel with no intersection points, making the system inconsistent

Intersect at infinitely many points, making the system consistent and dependent

Intersect at exactly two points due to their identical slopes

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College Algebra Quiz

College Algebra Quiz: Solve Systems Graphically And Interpret Intersections

Practice Solve Systems Graphically And Interpret Intersections in College Algebra 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 Solve Systems Graphically And Interpret Intersections, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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 system of equations has lines with slopes $m_1 = 3$ and $m_2 = 3$ , but different y-intercepts. When graphed, these lines will:

Intersect at exactly one point, making the system consistent and independent
Be parallel with no intersection points, making the system inconsistent (correct answer)
Intersect at infinitely many points, making the system consistent and dependent
Intersect at exactly two points due to their identical slopes

Explanation: Lines with identical slopes (

m_1 = m_2 = 3

) but different y-intercepts are parallel and never intersect. This creates an inconsistent system with no solution. Choice A would be correct if the slopes were different. Choice C would be correct if the lines were identical (same slope AND same y-intercept). Choice D is impossible since two lines can intersect at most once.

Question 2

Two investment accounts grow linearly over time. Account A has value $V_A = 1000 + 50t$ and Account B has value $V_B = 800 + 75t$ , where $t$ is time in months and values are in dollars. When do the accounts have equal value, and what happens after this point?

$t = 8$ months, $1400; Account A grows faster afterward
$t = 8$ months, $1400; Account B grows faster afterward (correct answer)
$t = 10$ months, $1500; both grow at same rate afterward
$t = 6$ months, $1300; Account B decreases afterward

Explanation: Setting

V_A = V_B

1000 + 50t = 800 + 75t

. Solving:

200 = 25t

, so

t = 8

months. At

t = 8

V_A = V_B = 1400

. After this point, Account B grows at $75 per month vs Account A's $50 per month, so B grows faster. Choice A has wrong growth comparison. Choice C has wrong intersection time. Choice D has wrong time and incorrect behavior.

Question 3

A farmer has 100 feet of fencing to create a rectangular pen. If the length is $y$ feet and the width is $x$ feet, the constraint equation is $2x + 2y = 100$ . If the farmer also wants the area $A = xy$ to equal 600 square feet, at what dimensions do these constraints intersect?

$(25, 25)$ - a single square configuration that maximizes area
$(10, 40)$ and $(40, 10)$ - representing minimum and maximum length configurations
$(20, 30)$ and $(30, 20)$ - two possible rectangular configurations (correct answer)
No intersection exists because the area constraint cannot be met with 100 feet of fencing

Explanation: When you encounter problems involving multiple constraints, you're looking for points where two equations intersect - this is a classic system of equations problem. You have two constraints: the perimeter constraint

2x + 2y = 100

(which simplifies to

x + y = 50

) and the area constraint

xy = 600

. To find where these intersect, substitute the first equation into the second. From

x + y = 50

, you get

y = 50 - x

. Substituting into

xy = 600

x(50 - x) = 600

50x - x^2 = 600

x^2 - 50x + 600 = 0

Using the quadratic formula:

x = \frac{50 \pm \sqrt{2500 - 2400}}{2} = \frac{50 \pm 10}{2}

This gives

x = 30

x = 20

. When

x = 20

y = 30

; when

x = 30

y = 20

. So the intersections are

(20, 30)

and

(30, 20)

. Answer A is wrong because

(25, 25)

gives an area of only 625 square feet, not 600. Answer B is incorrect because

(10, 40)

and

(40, 10)

satisfy the perimeter constraint but give areas of 400 square feet, not 600. Answer D is wrong because we just proved intersections do exist. Study tip: When solving constraint problems, always check that your solutions satisfy both original equations. Systems with a linear and quadratic equation typically yield zero, one, or two intersection points.

Question 4

A company's revenue $R$ (in thousands of dollars) is given by $R = 8x$ , where $x$ is the number of units sold (in hundreds). The company's costs $C$ (in thousands of dollars) follow $C = 3x + 15$ . At what production level do the revenue and cost functions intersect, and what is the economic interpretation?

$x = 3$ hundred units; break-even point where revenue equals cost at $24,000 (correct answer)
$x = 5$ hundred units; break-even point where revenue equals cost at $40,000
$x = 3$ hundred units; maximum profit point where costs are minimized
$x = 5$ hundred units; point where marginal revenue equals marginal cost

Explanation: To find the intersection, set

R = C

8x = 3x + 15

. Solving:

5x = 15

, so

x = 3

. At

x = 3

R = 8(3) = 24

and

C = 3(3) + 15 = 24

. Both equal 24 thousand dollars, making this the break-even point. Choice B has wrong calculation. Choice C misinterprets intersection as optimization. Choice D confuses intersection with marginal analysis.

Question 5

Two cell phone plans are compared: Plan A charges $25 per month plus $0.10 per minute, while Plan B charges $15 per month plus $0.15 per minute. If the total monthly costs are equal, how many minutes were used, and what is the significance of this point graphically?

250 minutes; this is where the slopes of the two cost functions are equal
150 minutes; this is the maximum number of minutes before Plan A becomes more expensive
200 minutes; this represents the point where Plan B reaches its minimum possible cost
200 minutes; this is where the two linear cost functions intersect, representing equal total cost (correct answer)

Explanation: When you encounter problems comparing two linear cost functions, you're looking for the intersection point where both plans cost the same amount. This requires setting up equations for each plan and solving for when they're equal. Let's set up the cost functions. Plan A:

C_A = 25 + 0.10m

, and Plan B:

C_B = 15 + 0.15m

, where

m

is minutes used. To find when costs are equal, set the equations equal:

25 + 0.10m = 15 + 0.15m

. Subtracting

0.10m

from both sides gives

25 = 15 + 0.05m

, then

10 = 0.05m

, so

m = 200

minutes. Graphically, this represents the intersection point of two lines, where both plans cost exactly the same amount. Answer A incorrectly suggests this point relates to equal slopes, but the slopes (

0.10

and

0.15

) are the per-minute rates and never change. Answer B gives the wrong number of minutes and misinterprets the significance—after 200 minutes, Plan B becomes more expensive than Plan A, not at 150 minutes. Answer C correctly identifies 200 minutes but incorrectly describes this as Plan B's minimum cost, when actually Plan B's minimum occurs at zero minutes (just the base fee). Answer D correctly identifies both the 200-minute solution and its graphical meaning: the intersection point of two linear functions where total costs are equal. Study tip: In cost comparison problems, always set up equations for each option and solve for equality. The intersection point tells you exactly when switching between options makes financial sense.