College Algebra Quiz: Solve Systems Graphically And Interpret Intersections

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.

Question 1

1. Intersect at exactly one point, making the system consistent and independent
2. Be parallel with no intersection points, making the system inconsistent (correct answer)
3. Intersect at infinitely many points, making the system consistent and dependent
4. 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

1. $t = 8$ months, $1400; Account A grows faster afterward
2. $t = 8$ months, $1400; Account B grows faster afterward (correct answer)
3. $t = 10$ months, $1500; both grow at same rate afterward
4. $t = 6$ 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

1. $(25, 25)$ - a single square configuration that maximizes area
2. $(10, 40)$ and $(40, 10)$ - representing minimum and maximum length configurations
3. $(20, 30)$ and $(30, 20)$ - two possible rectangular configurations (correct answer)
4. 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 $x = 30$ or $x = 20$. 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

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

1. 250 minutes; this is where the slopes of the two cost functions are equal
2. 150 minutes; this is the maximum number of minutes before Plan A becomes more expensive
3. 200 minutes; this represents the point where Plan B reaches its minimum possible cost
4. 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.