Solving Linear–Quadratic Systems - Algebra 1

How many real solutions does the system have? (Solutions are intersection points of the line and the circle.) $\begin{cases} x^2 + y^2 = 9 \\ y = 3 \end{cases}$
Answer: 1 real solution This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (circle), finding the point(s) where they intersect. A linear-quadratic system has one equation that graphs as a straight line and one that graphs as a curve (parabola or circle): the solution(s) are the intersection point(s) where the line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line is tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities for linear-quadratic systems. After substituting the linear equation into the quadratic, we get $x^2 + 9 = 9$ , which simplifies to $x^2 = 0$ . The discriminant is $0^2 - 4(1)(0) = 0$ , which is zero, so one real solution—the line is tangent to the curve. Choice B correctly identifies the number of solutions as 1 real solution. Choice C claims 2 solutions, but actually the discriminant is zero, indicating 1 real solution(s). Check your algebra—the line and curve touch at exactly one point! The discriminant preview: after substitution and simplification, you'll have a quadratic equation $ax^2 + bx + c = 0$ . Before solving, check $b^2 - 4ac$ : if positive, you'll get 2 intersection points; if zero, 1 intersection (tangent); if negative, 0 intersections (line misses curve). This tells you what to expect and helps catch errors—if you get 3 solutions, something's wrong!
Verify whether $ (2,3) $ is a solution to the linear-quadratic system (i.e., an intersection point). Choose the correct statement. $\begin{cases} y = x^2 - 1 \\ y = 2x - 1 \end{cases}$
Answer: Yes; $ (2,3) $ satisfies both equations. This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola), finding the point(s) where they intersect. After solving, always verify: substitute each $ (x, y) $ pair into BOTH original equations. If both equations are satisfied (both give true statements), it's a valid solution. If even one equation isn't satisfied, something went wrong in your algebra. For linear-quadratic systems, this verification catches substitution errors and confirms you found actual intersection points! Checking if $ (2, 3) $ solves the system: Substitute into equation 1: $ 3 = $(2)^2$ - 1 $ → $ 3 = 4 - 1 $ → $ 3 = 3 $ true. Substitute into equation 2: $ 3 = 2*2 - 1 $ → $ 3 = 4 - 1 $ → $ 3 = 3 $ true. Both equations satisfied, so $ (2, 3) $ IS a solution! Choice A correctly verifies the solution satisfies both equations. Choice B claims it satisfies the linear only, but actually it satisfies both: checking the quadratic gives $ 4 - 1 = 3 $, which matches y=3. Don't forget to check both—verification is key! Common mistake: forgetting to back-substitute for the second variable. If you solve and get $ x = 3 $ and $ x = −2 $, you're not done! For each x, find the corresponding y by substituting into the linear equation. $ x = 3 $ might give $ y = 5 $, and $ x = −2 $ might give $ y = −4 $, so your solutions are $ (3, 5) $ and $ (−2, −4) $, not just 'x = 3 and x = −2.' Always complete the ordered pairs!
A ball's path is modeled by $y=-x^2+4x$ and a laser beam by $y=x$ . Where do they intersect? (Solve the system.) $\begin{cases} y = -x^2 + 4x \\ y = x \end{cases}$
Answer: $(0,0)$ and $(3,3)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually $y = mx + b$ is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system $\{ y = -x^2 + 4x, y = x \}$ by substitution: (1) The linear equation gives us $y = x$ . (2) Substitute into the quadratic equation: $x = -x^2 + 4x$ . (3) Simplify to standard form: $x^2 - 3x = 0$ . (4) Solve: $x(x - 3) = 0$ gives $x = 0$ or $x = 3$ . (5) Find corresponding y-values: for $x = 0$ , $y = 0$ ; for $x = 3$ , $y = 3$ . Solutions: $(0, 0)$ and $(3, 3)$ . Choice A correctly finds the intersection points as $(0,0)$ and $(3,3)$ through proper substitution and solving. Choice D only gives one solution when there are actually two: after substitution, we get $x^2 - 3x = 0$ with solutions $x = 0$ and $x = 3$ . Both are valid! When the line crosses a parabola, there are typically two intersection points—don't forget the second solution from the quadratic. Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often $y = mx + b$ is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both $(x, y)$ pairs in both original equations. This six-step process works every time!
Solve graphically and algebraically: the line $y=2x-4$ intersects the parabola $y=x^2-4$ . What are the intersection point(s)? $\begin{cases} y = 2x - 4 \\ y = x^2 - 4 \end{cases}$
Answer: $(0,-4)$ and $(2,0)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. Graphically, the solutions are where the line intersects the parabola (or circle): sketch both on the same axes and see where they meet. The graphical approach gives you a visual sense of how many solutions exist and approximately where they are, while the algebraic approach gives you exact coordinates. Using both methods together—graphing to see the big picture, algebra to get precise values—is powerful! Graphing both equations: the linear equation $y = 2x - 4$ graphs as a line with slope 2 and y-intercept -4, and the quadratic $y = x^2 - 4$ graphs as a parabola opening upward with vertex at $(0, -4)$ . Sketching shows intersections at $(0, -4)$ and $(2, 0)$ . Visually, we can see they intersect at these points. Solving algebraically confirms the exact coordinates are $(0, -4)$ and $(2, 0)$ . Choice A correctly finds the intersection points as $(0,-4)$ and $(2,0)$ through proper substitution and solving. Choice B only gives one solution when there are actually two: after substitution, we get $x^2 - 2x = 0$ with solutions $x = 0$ and $x = 2$ . Both are valid! When the line crosses a parabola, there are typically two intersection points—don't forget the second solution from the quadratic. Graphical verification tip: after solving algebraically, do a quick sketch: plot the line (easy: two points and connect), sketch the parabola or circle (use key features), see where they intersect. Do the intersection points roughly match your algebraic solutions? If your algebra gave $(2, 5)$ and $(-1, -2)$ , but your sketch shows intersections around $(5, 10)$ and $(3, 7)$ , recheck your algebra! The graph is a sanity check.
Find all points $(x,y)$ satisfying both equations (intersection points of the line and the parabola). $\begin{cases} y = 3x - 2 \\ y = x^2 - 2 \end{cases}$
Answer: $(0,-2)$ and $(3,7)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {y = 3x - 2, y = x² - 2} by substitution: (1) The linear equation gives us y = 3x - 2. (2) Substitute into the quadratic equation: 3x - 2 = x² - 2. (3) Simplify to standard form: x² - 3x = 0. (4) Solve: x(x - 3) = 0 gives x = 0 or x = 3. (5) Find corresponding y-values: for x = 0, y = 3(0) - 2 = -2; for x = 3, y = 3(3) - 2 = 7. Solutions: (0, -2) and (3, 7). Choice A correctly finds the intersection points as (0, -2) and (3, 7) through proper substitution and solving. Choice B makes an error in back-substitution: for x = 3, substituting into y = 3x - 2 gives y = 3(3) - 2 = 9 - 2 = 7, not 1. After finding x-values, don't forget to find the corresponding y-values carefully! Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often y = mx + b is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Solve algebraically (using substitution) and identify the intersection point(s) of the line and parabola: $\begin{cases}\n y = 2x + 1 \\ \n y = x^2 + 1\n\end{cases}$
Answer: $(0,1)$ and $(2,5)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually $y = mx + b$ is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system $\{ y = 2x + 1, y = x^2 + 1 \}$ by substitution: (1) The linear equation gives us $y = 2x + 1$ . (2) Substitute into the quadratic equation: $2x + 1 = x^2 + 1$ . (3) Simplify to standard form: $x^2 - 2x = 0$ . (4) Solve: $x(x - 2) = 0$ gives x = 0 or x = 2. (5) Find corresponding y-values: for x = 0, y = 2(0) + 1 = 1; for x = 2, y = 2(2) + 1 = 5. Solutions: $(0, 1)$ and $(2, 5)$ . Choice B correctly finds the intersection points as $(0,1)$ and $(2,5)$ through proper substitution and solving. Choice A only gives one solution when there are actually two: after substitution, we get $x^2 - 2x = 0$ with solutions x = 0 and x = 2. Both are valid! When the line crosses a parabola, there are typically two intersection points—don't forget the second solution from the quadratic. Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often $y = mx + b$ is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Solve the linear-quadratic system algebraically (substitution method). The solutions represent the intersection point(s) of the line and the parabola. $\begin{cases} y = 2x + 1 \\ y = x^2 + 1 \end{cases}$
Answer: $(0,1)$ and $(2,5)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {y = 2x + 1, y = x² + 1} by substitution: (1) The linear equation gives us y = 2x + 1. (2) Substitute into the quadratic equation: 2x + 1 = x² + 1. (3) Simplify to standard form: x² - 2x = 0. (4) Solve: x(x - 2) = 0 gives x = 0 or x = 2. (5) Find corresponding y-values: for x = 0, y = 2(0) + 1 = 1; for x = 2, y = 2(2) + 1 = 5. Solutions: (0, 1) and (2, 5). Choice C correctly finds the intersection points as (0, 1) and (2, 5) through proper substitution and solving. Choice B makes an error in back-substitution: for x = 1, substituting into y = 2x + 1 gives y = 2(1) + 1 = 3, but x = 1 isn't even a solution from our quadratic! After finding x-values, don't forget to find the corresponding y-values carefully! Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often y = mx + b is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Verify whether $(2,3)$ is a solution to the linear-quadratic system (it must satisfy both equations): $\begin{cases} y = x^2 - 1 \\ y = 2x - 1 \end{cases}$ Which statement is correct?
Answer: Yes; $(2,3)$ satisfies both equations. This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. After solving, always verify: substitute each (x, y) pair into BOTH original equations. If both equations are satisfied (both give true statements), it's a valid solution. If even one equation isn't satisfied, something went wrong in your algebra. For linear-quadratic systems, this verification catches substitution errors and confirms you found actual intersection points! Checking if (2, 3) solves the system: Substitute into equation 1 (y = x² - 1): 3 = 2² - 1 → 3 = 4 - 1 → 3 = 3 ✓ (true). Substitute into equation 2 (y = 2x - 1): 3 = 2(2) - 1 → 3 = 4 - 1 → 3 = 3 ✓ (true). Both equations satisfied, so (2, 3) IS a solution! Choice A correctly verifies the solution satisfies both equations. Choice B incorrectly claims it only satisfies the linear equation: checking (2, 3) in y = x² - 1 gives 3 = 4 - 1 = 3, which is true! For a system, BOTH equations must be satisfied. This point satisfies both equations, so it is a solution to the system! Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often y = mx + b is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Solve the linear-quadratic system algebraically (by substitution). The solutions correspond to the intersection point(s) of the line and the parabola: $\begin{cases} y = x + 2 \\ y = x^2 \end{cases}$ What are all solutions $(x,y)$ to the system?
Answer: $(2,4)$ and $(-1,1)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually $y = mx + b$ is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system $\begin{cases} y = x + 2 \\ y = x^2 \end{cases}$ by substitution: (1) The linear equation gives us $y = x + 2$ . (2) Substitute into the quadratic equation: $x + 2 = x^2$ . (3) Simplify to standard form: $x^2 - x - 2 = 0$ . (4) Solve: factoring gives $(x - 2)(x + 1) = 0$ , so $x = 2$ or $x = -1$ . (5) Find corresponding y-values: for $x = 2$ , $y = 2 + 2 = 4$ ; for $x = -1$ , $y = -1 + 2 = 1$ . Solutions: $(2, 4)$ and $(-1, 1)$ . Choice A correctly finds the intersection points as $(2, 4)$ and $(-1, 1)$ through proper substitution and solving. Choice D has the coordinates backwards: the solutions should be $(x, y)$ , not $(y, x)$ . After solving for x and then finding y, make sure you write them in the correct order: x-coordinate first, y-coordinate second! Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often $y = mx + b$ is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both $(x, y)$ pairs in both original equations. This six-step process works every time!
A system consists of the equations $y = x^2 - 4x + 3$ and $y = 2x - 5$ . If the system has two solutions, what is the sum of the x-coordinates of these solutions?
Answer: $6$ Setting the equations equal: $x^2 - 4x + 3 = 2x - 5$ . Rearranging: $x^2 - 6x + 8 = 0$ . Using Vieta's formulas, the sum of the roots equals the negative coefficient of x divided by the leading coefficient: $-(-6)/1 = 6$ . Choice B results from forgetting the negative sign in Vieta's formula. Choice C comes from incorrectly setting up the quadratic as $x^2 - 2x + 8 = 0$ . Choice D results from using $6/(-1)$ instead of $(-(-6))/1$ .
How many real solutions does the system have? (Think of solutions as intersection points.) $\begin{cases} x^2 + y^2 = 4 \\ y = 3 \end{cases}$
Answer: 0 real solutions This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. A linear-quadratic system has one equation that graphs as a straight line and one that graphs as a curve (parabola or circle): the solution(s) are the intersection point(s) where the line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line is tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities for linear-quadratic systems. Graphically, the solutions are where the line intersects the circle: the equation x² + y² = 4 represents a circle centered at origin with radius 2, and y = 3 is a horizontal line at height 3. Since the circle has radius 2 (reaching from y = -2 to y = 2), and the line is at y = 3, the line is completely above the circle and they don't intersect. Algebraically confirming: substituting y = 3 into x² + y² = 4 gives x² + 9 = 4, so x² = -5, which has no real solutions. Choice A correctly identifies 0 real solutions since the line y = 3 is above the circle of radius 2 (which only extends to y = 2). Choice C claims 2 real solutions, but the line doesn't intersect the circle at all—when we substitute, we get x² = -5, which has no real solutions. Check your visualization: the horizontal line y = 3 is completely above a circle of radius 2! Graphical verification tip: after solving algebraically, do a quick sketch: plot the line (easy: two points and connect), sketch the parabola or circle (use key features), see where they intersect. Do the intersection points roughly match your algebraic solutions? If your algebra gave no solutions but your sketch shows intersections, recheck both! The graph is a sanity check.
How many real solutions (intersection points) does the linear-quadratic system have? $\begin{cases} y = x + 3 \\ y = x^2 + 2x + 5 \end{cases}$
Answer: 0 real solutions This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. A linear-quadratic system has one equation that graphs as a straight line and one that graphs as a curve (parabola or circle): the solution(s) are the intersection point(s) where the line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line is tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities for linear-quadratic systems. After substituting the linear equation into the quadratic, we get the following: Starting with y = x + 3 and y = x² + 2x + 5, we substitute: x + 3 = x² + 2x + 5. Rearranging: 0 = x² + x + 2. The discriminant is b² - 4ac = 1² - 4(1)(2) = 1 - 8 = -7. This is negative, so no real solutions—the line doesn't intersect the curve at all. The discriminant tells us the number of intersections before we even solve! Choice A correctly identifies the number of solutions as 0 real solutions through proper analysis of the discriminant. Choice C claims 2 real solutions, but with a negative discriminant, there are no real solutions. After substitution, the discriminant is negative, indicating 0 real solutions. Check your algebra—the line and curve don't intersect! The discriminant preview: after substitution and simplification, you'll have a quadratic equation ax² + bx + c = 0. Before solving, check b² - 4ac: if positive, you'll get 2 intersection points; if zero, 1 intersection (tangent); if negative, 0 intersections (line misses curve). This tells you what to expect and helps catch errors—if you get 3 solutions, something's wrong!
Find all intersection points of the line and the parabola by solving the system: $\begin{cases} y = -x + 2 \\ y = x^2 \end{cases}$
Answer: $\{(-2,4),(1,1)\}$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. Solving the system {y = -x + 2, y = x²} by substitution: (1) The linear equation gives us y = -x + 2. (2) Substitute into the quadratic equation: -x + 2 = x². (3) Rearrange to standard form: x² + x - 2 = 0. (4) Factor: (x + 2)(x - 1) = 0, so x = -2 or x = 1. (5) Find corresponding y-values: for x = -2, y = (-2)² = 4; for x = 1, y = (1)² = 1. Solutions: (-2, 4) and (1, 1). Choice A correctly finds the intersection points as {(-2, 4), (1, 1)} through proper substitution and solving. Choice C makes an error in back-substitution: for x = -2, substituting into y = x² gives y = (-2)² = 4, not -4. After finding x-values, don't forget to find the corresponding y-values carefully! Common mistake: forgetting to back-substitute for the second variable. If you solve and get x = -2 and x = 1, you're not done! For each x, find the corresponding y by substituting into either equation (the quadratic y = x² is often easier). x = -2 gives y = 4, and x = 1 gives y = 1, so your solutions are (-2, 4) and (1, 1), not just 'x = -2 and x = 1.' Always complete the ordered pairs!
How many real solutions does this linear-quadratic system have (i.e., how many intersection points are there)? $\begin{cases} y = x + 3 \\ y = x^2 + 1 \end{cases}$
Answer: 2 This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. A linear-quadratic system has one equation that graphs as a straight line and one that graphs as a curve (parabola or circle): the solution(s) are the intersection point(s) where the line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line is tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities for linear-quadratic systems. After substituting the linear equation into the quadratic, we get $x^2 - x - 2 = 0$ . The discriminant is $b^2 - 4ac = 1 + 8 = 9$ . This is positive, so two real solutions—the line intersects the curve at two points. Choice C correctly identifies the number of solutions as 2. Choice A claims no solution, but actually there are solutions. After substitution, the discriminant is positive, indicating 2 real solution(s). Check your algebra—the line and curve do intersect! The discriminant preview: after substitution and simplification, you'll have a quadratic equation $ax^2 + bx + c = 0$ . Before solving, check $b^2 - 4ac$ : if positive, you'll get 2 intersection points; if zero, 1 intersection (tangent); if negative, 0 intersections (line misses curve). This tells you what to expect and helps catch errors—if you get 3 solutions, something's wrong!
How many real solutions does the system have? (Think of the number of intersection points.) $\begin{cases} y = x^2 + 4 \\ y = x + 1 \end{cases}$
Answer: 0 real solutions This question tests your ability to determine how many solutions a linear-quadratic system has by analyzing whether the line and parabola intersect. A linear-quadratic system has one equation that graphs as a straight line and one that graphs as a curve (parabola): the solution(s) are the intersection point(s) where the line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line is tangent), or 2 intersections (line passes through). After substituting the linear equation into the quadratic, we get x + 1 = x² + 4. Rearranging: x² - x + 3 = 0. The discriminant is b² - 4ac = (-1)² - 4(1)(3) = 1 - 12 = -11. This is negative, so no real solutions—the line doesn't intersect the curve at all. Choice A correctly identifies 0 real solutions based on the negative discriminant. Choice C might be tempting if you expect a line to always intersect a parabola twice, but when the parabola is positioned high enough (y = x² + 4 has vertex at (0, 4)) and the line has a gentle slope (y = x + 1), they can miss entirely. The discriminant preview: after substitution and simplification, you'll have a quadratic equation ax² + bx + c = 0. Before solving, check b² - 4ac: if positive, you'll get 2 intersection points; if zero, 1 intersection (tangent); if negative, 0 intersections (line misses curve). This tells you what to expect and helps catch errors—the negative discriminant confirms the line and parabola don't meet!
Find all intersection point(s) of the line and the parabola by solving algebraically: $\begin{cases} y = 2x \\ y = x^2 \end{cases}$
Answer: $(0,0)$ and $(2,4)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually $y = mx + b$ is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system { $y = 2x$ , $y = x^2$ } by substitution: (1) The linear equation gives us $y = 2x$ . (2) Substitute into the quadratic equation: $2x = x^2$ . (3) Simplify to standard form: $x^2 - 2x = 0$ . (4) Solve: factoring gives $x(x - 2) = 0$ , so $x = 0$ or $x = 2$ . (5) Find corresponding y-values: for $x = 0$ , $y = 0$ ; for $x = 2$ , $y = 4$ . Solutions: $(0, 0)$ and $(2, 4)$ . Choice A correctly finds the intersection points as $(0,0)$ and $(2,4)$ through proper substitution and solving. Choice C only gives one solution when there are actually two: after substitution, we get $x^2 - 2x = 0$ with solutions $x = 0$ and $x = 2$ . Both are valid! When the line crosses a parabola, there are typically two intersection points—don't forget the second solution from the quadratic. Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often $y = mx + b$ is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Solve the system algebraically. (The solution(s) are the intersection point(s) of the line and the circle.) $\begin{cases} x^2 + y^2 = 25 \\ y = 4 \end{cases}$
Answer: $(3,4)$ and $(-3,4)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {x² + y² = 25, y = 4} by substitution: (1) The linear equation gives us y = 4. (2) Substitute into the circle equation: x² + 4² = 25. (3) Simplify: x² + 16 = 25, so x² = 9. (4) Solve: x = ±3. (5) The y-value is already given as 4 for both points. Solutions: (3, 4) and (-3, 4). Choice A correctly finds the intersection points as (3, 4) and (-3, 4) through proper substitution and solving. Choice B has the coordinates backwards: the solutions should be (3, 4) and (-3, 4), not (4, 3) and (-4, 3). After solving for x and then finding y, make sure you write them in the correct order: x-coordinate first, y-coordinate second! Circle-line systems are special: when solving x² + y² = r² with a line, you often get solutions with radicals like (√5, √5). Don't be intimidated! These are exact answers. You can estimate decimals if asked, but the radical form is exact and preferred. Also, circle-line systems are symmetric—if (a, b) is a solution and the line passes through the origin or has special symmetry, there's often a matching solution (-a, -b) or similar.
Solve the system algebraically (substitution). The solutions are the intersection point(s) of the line and the circle: $\begin{cases} x^2 + y^2 = 10 \\ y = x \end{cases}$ What are all solutions $(x,y)$ ?
Answer: $(\sqrt{5},\sqrt{5})$ and $(-\sqrt{5},-\sqrt{5})$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually $y = mx + b$ is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system $\{x^2 + y^2 = 10, y = x\}$ by substitution: (1) The linear equation gives us $y = x$ . (2) Substitute into the circle equation: $x^2 + x^2 = 10$ . (3) Simplify: $2x^2 = 10$ , so $x^2 = 5$ . (4) Solve: $x = \pm \sqrt{5}$ . (5) Find corresponding y-values: since $y = x$ , for $x = \sqrt{5}$ , $y = \sqrt{5}$ ; for $x = -\sqrt{5}$ , $y = -\sqrt{5}$ . Solutions: $(\sqrt{5}, \sqrt{5})$ and $(-\sqrt{5}, -\sqrt{5})$ . Choice A correctly finds the intersection points as $(\sqrt{5}, \sqrt{5})$ and $(-\sqrt{5}, -\sqrt{5})$ through proper substitution and solving. Choice C gives $(\sqrt{5}, -\sqrt{5})$ and $(-\sqrt{5}, \sqrt{5})$ , but these don't satisfy $y = x$ : if $x = \sqrt{5}$ , then $y$ must equal $\sqrt{5}$ , not $-\sqrt{5}$ . For a system, BOTH equations must be satisfied. These points might solve the circle equation, but if they don't solve both, they're not solutions to the system! Circle-line systems are special: when solving $x^2 + y^2 = r^2$ with a line, you often get solutions with radicals like $(\sqrt{5}, \sqrt{5})$ . Don't be intimidated! These are exact answers. You can estimate decimals if asked, but the radical form is exact and preferred. Also, circle-line systems are symmetric—if $(a, b)$ is a solution and the line passes through the origin or has special symmetry, there's often a matching solution $(-a, -b)$ or similar.
Solve graphically and algebraically (substitution is fine). The solutions are the intersection point(s) of the line and the parabola: $\begin{cases} y = 2x - 1 \\ y = x^2 - 4 \end{cases}$
Answer: $(-1,-3)$ and $(3,5)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {y = 2x - 1, y = x² - 4} by substitution: (1) The linear equation gives us y = 2x - 1. (2) Substitute into the quadratic equation: 2x - 1 = x² - 4. (3) Simplify to standard form: 0 = x² - 2x - 3, which factors as 0 = (x - 3)(x + 1). (4) Solve: x = 3 or x = -1. (5) Find corresponding y-values: for x = 3, y = 2(3) - 1 = 5; for x = -1, y = 2(-1) - 1 = -3. Solutions: (-1, -3) and (3, 5). Choice B correctly finds the intersection points as (-1, -3) and (3, 5) through proper substitution and solving. Choice A makes an error with the first point: for x = 1, substituting into y = 2x - 1 gives y = 2(1) - 1 = 1, which would give (1, 1), but we need to check if this satisfies both equations. Substituting (1, 1) into y = x² - 4: 1 = 1² - 4 = -3, which is false. The correct x-values from solving the quadratic are -1 and 3, not 1 and 3! Linear-quadratic system solving recipe: (1) Solve the linear equation for y (or x, whichever is easier—often y = mx + b is already done), (2) Substitute that expression into the quadratic equation in place of that variable, (3) Simplify to get a quadratic equation in one variable, (4) Solve using factoring, quadratic formula, or other methods, (5) Back-substitute each solution into the linear equation to find the other coordinate, (6) Verify both (x, y) pairs in both original equations. This six-step process works every time!
Where does the line intersect the circle? Solve the linear-quadratic system. $\begin{cases} x^2 + y^2 = 13 \\ y = x \end{cases}$
Answer: $(\sqrt{26}/2,\sqrt{26}/2)$ and $(-\sqrt{26}/2,-\sqrt{26}/2)$ This question tests your ability to solve a system of equations consisting of one linear equation (straight line) and one quadratic equation (parabola or circle), finding the point(s) where they intersect. To solve a linear-quadratic system algebraically, we use substitution: (1) solve the linear equation for one variable (usually y = mx + b is already solved), (2) substitute that expression into the quadratic equation, giving you a quadratic equation in one variable, (3) solve that quadratic using factoring, quadratic formula, or other methods, (4) back-substitute each solution into the linear equation to find the corresponding other coordinate. This gives you all the intersection points! Solving the system {x² + y² = 13, y = x} by substitution: (1) The linear equation gives us y = x. (2) Substitute into the circle equation: x² + x² = 13. (3) Simplify: 2x² = 13, so x² = 13/2. (4) Solve: x = ±√(13/2) = ±√26/2. (5) Since y = x, we have y = ±√26/2 as well. Solutions: (√26/2, √26/2) and (-√26/2, -√26/2). Choice B correctly finds the intersection points as (√26/2, √26/2) and (-√26/2, -√26/2) through proper substitution and solving. Choice A makes an error: when substituting y = x into x² + y² = 13, we get 2x² = 13, not x² = 13. This gives x² = 13/2, so x = ±√(13/2) = ±√26/2, not ±√13. Careful substitution is crucial! Circle-line systems are special: when solving x² + y² = r² with a line, you often get solutions with radicals like (√5, √5). Don't be intimidated! These are exact answers. You can estimate decimals if asked, but the radical form is exact and preferred. Also, circle-line systems are symmetric—if (a, b) is a solution and the line passes through the origin or has special symmetry, there's often a matching solution (-a, -b) or similar.

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