Systems of Polynomial Equations - SAT Math

A revenue curve is $y=-x^2+10x-21$ and a break-even line is $y=x-1$ . The break-even points are intersections. How many solutions does the system have? $\begin{cases}y=-x^2+10x-21\\y=x-1\end{cases}$
Answer: 2 We need to find how many break-even points exist where revenue equals the break-even line. The system has $y = -x^2 + 10x - 21$ and $y = x - 1$ , which we solve by substitution. Setting them equal: $-x^2 + 10x - 21 = x - 1$ , which simplifies to $-x^2 + 9x - 20 = 0$ , or $x^2 - 9x + 20 = 0$ . This factors as $(x - 4)(x - 5) = 0$ , giving $x = 4$ or $x = 5$ , so there are 2 solutions. These represent two different production levels where the company breaks even. A common mistake is errors in sign when rearranging equations—always be careful with negative coefficients. Remember that finding the number of solutions often requires only the discriminant or factoring, not the actual solution values.
A projectile's path is $y=-x^2+2x+3$ . A sensor beam is the horizontal line $y=4$ . How many solutions does the system have? $\begin{cases}y=-x^2+2x+3\\y=4\end{cases}$
Answer: 1 We need to find how many times the projectile's parabolic path reaches the horizontal sensor beam height. This system has a quadratic $y = -x^2 + 2x + 3$ and a horizontal line $y = 4$ , solvable by substitution. Setting them equal: $-x^2 + 2x + 3 = 4$ , which simplifies to $-x^2 + 2x - 1 = 0$ or $x^2 - 2x + 1 = 0$ . This factors as $(x - 1)^2 = 0$ , giving $x = 1$ as a repeated root, meaning the parabola just touches the line at one point. A common error is assuming that because a quadratic equation is involved, there must be 2 solutions—but when the discriminant equals 0, there's only 1 solution (tangent point). Remember: parabola and line can intersect at 0, 1, or 2 points.
A water fountain's arc is modeled by $y=-x^2+4x$ and a walkway edge is $y=2$ . The wet spots occur where the stream meets the walkway. If $(x_1,y_1)$ and $(x_2,y_2)$ are the intersection points, what is $x_1+x_2$ ? $\begin{cases}y=-x^2+4x\\y=2\end{cases}$
Answer: $4$ We need to find where the fountain's parabolic arc meets the horizontal walkway, then sum the x-coordinates. This system has a quadratic $y = -x^2 + 4x$ and a horizontal line $y = 2$ , solvable by substitution. Setting them equal: $-x^2 + 4x = 2$ , which rearranges to $x^2 - 4x + 2 = 0$ . Using the quadratic formula or Vieta's formulas, for $ax^2 + bx + c = 0$ , the sum of roots equals $-b/a = -(-4)/1 = 4$ . A common error is trying to find individual x-values when only their sum is needed—Vieta's formulas provide a shortcut. For any quadratic equation, the sum of roots $x_1 + x_2 = -b/a$ , saving calculation time on the SAT.
A ball's height (in meters) after $t$ seconds is modeled by $h=-t^2+6t+1$ . A drone flies along the line $h=2t+1$ . The ball and drone are at the same height when the system is satisfied: $\begin{cases} h=-t^2+6t+1\\ h=2t+1 \end{cases}$ How many real solutions does this system have?
Answer: 2 This problem asks for the number of times a ball and drone are at the same height. The system consists of a parabola ( $h = -t^2 + 6t + 1$ ) and a line ( $h = 2t + 1$ ), which is a linear-nonlinear system. To solve, we substitute: $-t^2 + 6t + 1 = 2t + 1$ , which simplifies to $-t^2 + 4t = 0$ , or $t(-t + 4) = 0$ , giving $t = 0$ or $t = 4$ . Since we get two distinct real values of t, the system has 2 real solutions. A common error is forgetting that a parabola and line can intersect at 0, 1, or 2 points depending on the discriminant.
A flashlight beam is modeled by the line $y=\tfrac{1}{2}x+2$ and a curved mirror edge by $y=x^2-2x+2$ . The bright spots occur at intersections. What are all solutions $(x,y)$ ? $\begin{cases}y=\tfrac{1}{2}x+2\\y=x^2-2x+2\end{cases}$
Answer: $(0,2)$ and $(\tfrac{5}{2},\tfrac{13}{4})$ We need to find where the flashlight beam line intersects the curved mirror edge. The system has $y = \frac{1}{2}x + 2$ and $y = x^2 - 2x + 2$ , solvable by substitution. Setting them equal: $\frac{1}{2}x + 2 = x^2 - 2x + 2$ , which simplifies to $\frac{1}{2}x = x^2 - 2x$ , then $0 = x^2 - 2x - \frac{1}{2}x = x^2 - \frac{5}{2}x$ . Factoring: $x(x - \frac{5}{2}) = 0$ , so $x = 0$ or $x = \frac{5}{2}$ . When $x = 0$ , $y = \frac{1}{2}(0) + 2 = 2$ ; when $x = \frac{5}{2}$ , $y = \frac{1}{2}(\frac{5}{2}) + 2 = \frac{5}{4} + 2 = \frac{13}{4}$ . The solutions are $(0, 2)$ and $(\frac{5}{2}, \frac{13}{4})$ . A common error with fractions is arithmetic mistakes—always double-check calculations involving fractions on the SAT.
On the coordinate plane, the line $y=-2x+1$ intersects the parabola $y=x^2+1$ . What are the intersection points?
Answer: $(-2,5)$ and $(0,1)$ We need to find where the line y = -2x + 1 intersects the parabola y = x² + 1. This system combines a linear and quadratic equation, solved by substitution. Setting -2x + 1 = x² + 1 gives -2x = x², or x² + 2x = 0, which factors to x(x + 2) = 0. This yields x = 0 and x = -2, giving intersection points (0, 1) and (-2, 5). We can verify: at x = 0, both equations give y = 1; at x = -2, line gives y = 4 + 1 = 5 and parabola gives y = 4 + 1 = 5. The key observation is that when both equations have the same constant term, one intersection occurs at that y-intercept.
A ball's height $y$ (in meters) after $x$ seconds is modeled by $y=-x^2+6x+1$ . A nearby platform has height given by the linear model $y=2x+1$ . Solving the system gives the times when the ball is exactly at platform height. What are all solutions $(x,y)$ to the system?
Answer: $(0,1)$ and $(4,9)$ The question asks for all solutions (x, y) to the system of equations modeling the ball's height and the platform's height, representing the times when the ball is exactly at the platform height. This system consists of a quadratic equation (downward-opening parabola) and a linear equation (line), which can intersect at 0, 1, or 2 points, and we solve by substitution, setting the y-expressions equal. Set -x² + 6x + 1 = 2x + 1, then subtract 2x + 1 from both sides to get -x² + 4x = 0; multiply by -1 for x² - 4x = 0, and factor as x(x - 4) = 0, yielding x = 0 or x = 4. Substitute into the line: for x = 0, y = 1; for x = 4, y = 9, so solutions are (0, 1) and (4, 9). A key error might be failing to multiply by -1, leading to incorrect roots, or forgetting to compute y-values. Another mistake is assuming only positive x-values are valid without context. For test-taking, since a line and parabola intersect at most twice, verify both solutions satisfy the original equations to confirm.
A projectile's path is modeled by $y=-\tfrac12 x^2+3x+2$ , and a safety net is modeled by the line $y=x+2$ . The system's solutions represent where the projectile hits the net. How many solutions does the system have?
Answer: 2 The question asks for the number of solutions to the system, representing where the projectile hits the safety net. This linear-quadratic system can have 0, 1, or 2 solutions, found by setting equations equal. Set - (1/2) x² + 3x + 2 = x + 2, subtract x + 2 to get - (1/2) x² + 2x = 0, multiply by -2 for x² - 4x = 0, factor as x(x - 4) = 0, yielding x = 0 and x = 4. Both are real, confirming two solutions. A key error is mishandling the fraction, leading to scaling mistakes. Another is assuming tangency without checking discriminant (here, 16 > 0). For tests, simplify by multiplying out fractions early and remember maximum two hits between line and parabola.
Solve the system: $\begin{cases} y=3x+2 \\ y=x^2+2 \end{cases}$ What are all solutions $(x,y)$ ?
Answer: $(0,2)$ and $(3,11)$ We need to solve $y = 3x + 2$ and $y = x^2 + 2$ , a system with linear and quadratic equations. Setting $3x + 2 = x^2 + 2$ gives $3x = x^2$ , or $x^2 - 3x = 0$ , which factors to $x(x - 3) = 0$ . This yields $x = 0$ and $x = 3$ , giving solutions $(0, 2)$ and $(3, 11)$ . We can verify: at $x = 0$ , both equations give $y = 2$ ; at $x = 3$ , line gives $y = 9 + 2 = 11$ and parabola gives $y = 9 + 2 = 11$ . Notice that both equations have the same constant term (+2), which guarantees one intersection at $x = 0$ . The factoring pattern $x^2 - bx = x(x - b)$ is useful when there's no constant term after simplification.
A small business compares two demand models: $y=x^2-6x+8$ and $y=0$ , where $x$ is price (in dollars) and $y$ is demand (in hundreds). How many solutions does the system have? $\begin{cases} y=x^2-6x+8 \\ y=0 \end{cases}$
Answer: 2 We need to find how many price points result in zero demand. The system has $y = x^2 - 6x + 8$ and $y = 0$ , so we solve $x^2 - 6x + 8 = 0$ . This factors as $(x - 2)(x - 4) = 0$ , giving $x = 2$ or $x = 4$ , which means there are 2 solutions: $(2, 0)$ and $(4, 0)$ . These represent the two prices where demand drops to zero. A common mistake is confusing the number of solutions with the actual solution values—the question asks "how many," which is 2. Remember that a parabola can cross the x-axis ( $y = 0$ ) at 0, 1, or 2 points, depending on the discriminant.
On a coordinate plane, the graphs of $y=x^2-4x+1$ and $y=3x-5$ represent two design constraints for a ramp. How many solutions does the system have (that is, how many intersection points are there)?
Answer: 2 The question asks for the number of intersection points between the parabola y = x² - 4x + 1 and the line y = 3x - 5. This is a system of one quadratic and one linear equation, which we solve by substitution. Setting equal: x² - 4x + 1 = 3x - 5, which becomes x² - 7x + 6 = 0. Using the discriminant b² - 4ac = (-7)² - 4(1)(6) = 49 - 24 = 25 > 0, we know there are two real solutions. The positive discriminant tells us the parabola and line intersect at exactly 2 points. When solving systems graphically or algebraically, remember that a parabola and line can intersect 0, 1, or 2 times depending on their relative positions.
A drone's planned altitude is $y=x^2-2x+2$ and a safety ceiling is $y=2x-2$ , where $x$ is time in seconds and $y$ is meters. If $(x_1,y_1)$ and $(x_2,y_2)$ are the intersection points, what is $x_1+x_2$ ?
Answer: $4$ The question asks for the sum x₁ + x₂ of intersection points between a drone's quadratic altitude and a linear safety ceiling. This quadratic-linear system is solved by substitution, recalling possible intersections of 0, 1, or 2 for a parabola and line. Set x² - 2x + 2 = 2x - 2, simplify to x² - 4x + 4 = 0, or (x - 2)² = 0, with double root x = 2. The sum of roots from the quadratic coefficients is 4, treating the tangent case as repeated roots. Errors often occur by counting only one root and ignoring multiplicity for sums. A strategy is to use Vieta's formulas directly for sums without solving fully.
On the coordinate plane, the line $y=x+1$ and the parabola $y=x^2-1$ are graphed. How many solutions does the system have (that is, how many intersection points are there)?
Answer: 2 The question asks for the number of solutions to the system, which corresponds to the number of intersection points between the line and the parabola on the coordinate plane. This is a system of a linear equation and a quadratic equation, capable of having 0, 1, or 2 intersections, solved by setting the equations equal and solving the resulting quadratic. Set x + 1 = x² - 1, rearrange to x² - x - 2 = 0, and factor as (x - 2)(x + 1) = 0, giving x = 2 and x = -1. Both roots are real and distinct, confirming two intersection points. A common error is misarranging the equation, like forgetting to move all terms to one side, resulting in a wrong quadratic. Students might also confuse the number of roots with the discriminant without calculating it (here, 1 + 8 = 9 > 0, two real roots). A useful strategy is to quickly compute the discriminant of the resulting quadratic to determine the number of solutions without fully solving.
A parabola $y=-x^2+4x$ and a line $y=x$ are used to model two possible routes on a map grid. The routes meet where the equations are both true. What are all solutions $(x,y)$ to the system?
Answer: $(0,0)$ and $(3,3)$ To find where the parabola y = -x² + 4x meets the line y = x, we solve this system using substitution. Setting equal: -x² + 4x = x, which simplifies to -x² + 3x = 0, then -x(x - 3) = 0. This gives x = 0 or x = 3. When x = 0, y = 0; when x = 3, y = 3. The solutions are (0,0) and (3,3). A common error is forgetting that x = 0 is a valid solution when factoring out x - always include it when you factor out a variable.
A cyclist's distance from home (in miles) is modeled by $d=t^2+2t$ , and a friend's estimate is $d=6t$ . At what times $t$ do the two models agree? Give all solutions for $t$ .
Answer: $t=0$ and $t=4$ This problem asks when two distance models agree: d = t² + 2t (cyclist) and d = 6t (friend's estimate). We solve this system by setting the equations equal, giving us one quadratic and one linear equation. Setting t² + 2t = 6t simplifies to t² - 4t = 0, which factors as t(t - 4) = 0. This gives t = 0 (at the start) and t = 4 (after 4 time units). We can verify: at t = 0, both models give d = 0; at t = 4, cyclist's model gives d = 16 + 8 = 24 and friend's gives d = 24. The key insight is recognizing that factoring is often faster than the quadratic formula when the equation has no constant term.
Solve the system and select all solutions: $\begin{cases} y=x^2+2x-8 \\ y=0 \end{cases}$
Answer: $(-4,0)$ and $(2,0)$ This problem asks for solutions to $y = x^2 + 2x - 8$ and $y = 0$ , which means finding where the parabola crosses the x-axis. Setting $x^2 + 2x - 8 = 0$ , we can factor by finding two numbers that multiply to -8 and add to 2: these are 4 and -2. So $(x + 4)(x - 2) = 0$ , giving $x = -4$ and $x = 2$ . The solutions are $(-4, 0)$ and $(2, 0)$ , representing the x-intercepts of the parabola. A common error is to confuse the signs when factoring; remember that if $(x + 4) = 0$ , then $x = -4$ . When $y = 0$ , we're specifically looking for x-intercepts, which is a special case of a parabola-line system.
A designer compares $y=-x^2+4x+4$ with the line $y=4$ . The intersections show where the design reaches a constant height. What are all solutions $(x,y)$ to the system?
Answer: $(0,4)$ and $(4,4)$ We need to find where the parabola y = -x² + 4x + 4 intersects the horizontal line y = 4. This system has a quadratic and a constant linear equation, solved by substitution. Setting equal: -x² + 4x + 4 = 4, which simplifies to -x² + 4x = 0, then -x(x - 4) = 0. This gives x = 0 or x = 4. Since y = 4 for both solutions (given by the line equation), our intersection points are (0,4) and (4,4). A common error is thinking a horizontal line can only intersect a parabola once, but it can intersect 0, 1, or 2 times depending on the parabola's position and shape.
A ride's height is modeled by $y=-x^2+2x+8$ , and a minimum clearance line is $y=4$ . How many solutions does the system have, and what does that mean for clearance points?
Answer: 2 This problem asks how many times the ride height $y = -x^2 + 2x + 8$ meets the clearance line $y = 4$ . Setting $-x^2 + 2x + 8 = 4$ gives $-x^2 + 2x + 4 = 0$ , or $x^2 - 2x - 4 = 0$ after multiplying by -1. Using the discriminant: $b^2 - 4ac = 4 - 4(1)(-4) = 4 + 16 = 20 > 0$ . Since the discriminant is positive, there are two real solutions, meaning the ride crosses the clearance height at two points. This makes physical sense: the parabola opens downward (negative coefficient of $x^2$ ), so it rises above y = 4, reaches a maximum, then falls back below y = 4. For safety analysis, two clearance points typically represent entry and exit from a safe zone.
A company's profit (in thousands of dollars) is $y=x^2-4x+3$ , where $x$ is the number of ads (in tens). A consultant claims profit will follow $y=2x-1$ . How many solutions does the system have? $\begin{cases}y=x^2-4x+3\\y=2x-1\end{cases}$
Answer: 2 This asks for the number of solutions when a parabola meets a line. We have a quadratic equation $y = x^2 - 4x + 3$ and a linear equation $y = 2x - 1$ , so we use substitution. Setting them equal: $x^2 - 4x + 3 = 2x - 1$ , which simplifies to $x^2 - 6x + 4 = 0$ . Using the discriminant $b^2 - 4ac = (-6)^2 - 4(1)(4) = 36 - 16 = 20$ , which is positive, indicating 2 real solutions. A common mistake is assuming all parabola-line systems have 2 solutions, but the discriminant determines whether there are 0, 1, or 2 intersection points. When the discriminant is positive (as here), there are exactly 2 solutions.
A ball's height (in meters) after $x$ seconds is modeled by $y=-x^2+6x-5$ . A drone follows the straight path $y=x+1$ at the same time. The times when the drone is exactly at the ball's height correspond to the solutions of the system. What are all solutions $(x,y)$ to the system? $\begin{cases}y=-x^2+6x-5\\y=x+1\end{cases}$
Answer: $(1,2)$ and $(4,5)$ We need to find where the ball's parabolic path intersects with the drone's linear path. This is a system with one quadratic equation (parabola) and one linear equation, which can be solved by substitution. Since both equations equal y, we set them equal: $-x^2 + 6x - 5 = x + 1$ , which simplifies to $-x^2 + 5x - 6 = 0$ or $x^2 - 5x + 6 = 0$ . Factoring gives $(x - 1)(x - 4) = 0$ , so $x = 1$ or $x = 4$ . When $x = 1$ , $y = 1 + 1 = 2$ ; when $x = 4$ , $y = 4 + 1 = 5$ . A common error is to stop after finding x-values without calculating the corresponding y-values. Remember that a parabola and line can intersect at 0, 1, or 2 points—here we have 2 intersection points.

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