This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points, applied to a real-world scenario like paths intersecting. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. For $y = -x^2 + 6x$ and $y = 2x$, set equal: $-x^2 + 6x = 2x$, $-x^2 + 4x = 0$, $-x(x - 4) = 0$, $x = 0$ or $4$; $y = 0$ and $8$, giving $(0,0)$ and $(4,8)$. Choice A correctly finds both intersection points through factoring. A common error is missing the negative sign, but careful rearrangement prevents that—verify by plugging back in. Excellent effort: this shows how math models intersections in physics, like trajectories!

This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in $y = mx + b$ form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system $y = x + 1$ and $y = x^2 - 2x + 1$, substitute the linear expression into the quadratic: $x + 1 = x^2 - 2x + 1$, rearrange to $x^2 - 3x = 0$, factor as $x(x - 3) = 0$, so $x = 0$ or $x = 3$; back-substitute to get $y = 1$ and $y = 4$, giving points $(0, 1)$ and $(3, 4)$. Choice B correctly identifies both intersection points through proper substitution and solving. Remember the transferable strategy: always check the discriminant of the resulting quadratic to predict the number of solutions, and verify each point in both equations to ensure accuracy—you've got this!

This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (circle) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: substitute y = -3x into $x^2 + y^2 = 25$, yielding $x^2 + 9x^2 = 25$, so $10x^2 = 25$, $x^2 = 2.5$, $x = \pm \sqrt{\frac{5}{2}} = \pm \frac{5}{\sqrt{10}}$; then y = -3x gives $(\frac{5}{\sqrt{10}}, -\frac{15}{\sqrt{10}})$ and $(-\frac{5}{\sqrt{10}}, \frac{15}{\sqrt{10}})$. Choice A correctly finds both intersection points with accurate substitution and rationalized forms. A common mistake is forgetting to rationalize or mixing signs in y-values, but double-checking with the circle equation ensures correctness. Keep practicing: visualize the line through the origin with slope -3 crossing the circle centered at origin with radius 5—great job verifying graphically too!

This question tests your ability to determine the number of real solutions in a linear-quadratic system, corresponding to intersection points of a line and parabola. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. For $y = x^2 + 4$ and $y = 2x + 1$, set equal: $x^2 + 4 = 2x + 1$, $x^2 - 2x + 3 = 0$; discriminant $4 - 12 = -8 < 0$, so no real solutions. Choice A correctly indicates zero real solutions based on the negative discriminant. If you miscounted terms, recheck the setup—the parabola $y = x^2 + 4$ is always above $y = 4$, and the line doesn't reach it. Awesome insight: discriminants quickly reveal if intersections occur without graphing!

This question tests your ability to determine the number of real solutions in a linear-quadratic system, corresponding to intersection points of a line and parabola. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. For $y = x^2 + 2x - 3$ and $y = -x + 1$, set equal: $x^2 + 2x - 3 = -x + 1$, $x^2 + 3x - 4 = 0$; discriminant $9 + 16 = 25 > 0$, so two real solutions. Choice C correctly indicates two real solutions based on the positive discriminant. If you got a different count, recheck your algebra—substitution errors are common, but the discriminant is your reliable guide. Keep up the great work: this method predicts without fully solving, saving time on tests!

This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for $y$ (often already in $y = mx + b$ form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system $y = x + 1$ and $y = x^2 - 2x + 1$, substitute: $x + 1 = x^2 - 2x + 1$, rearrange to $0 = x^2 - 3x$, factor as $x(x - 3) = 0$, so $x = 0$ or $x = 3$; then $y = 1$ and $y = 4$, giving $(0, 1)$ and $(3, 4)$. Choice A correctly identifies both intersection points through accurate substitution and solving. Keep up the good work—remember to verify each point in both equations to confirm!

This question tests your ability to verify if a given point satisfies both equations in a linear-quadratic system, meaning it's an intersection point. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution, but for verification, plug the point into both equations and check if they hold true. For the point $ (2, 3) $ in $ y = x^2 - 1 $ and $ y = 3x - 3 $: first equation $ 2^2 - 1 = 3 $, second $ 3(2) - 3 = 3 $, so yes, it satisfies both. Choice A correctly confirms it works for both. You're awesome—verification is a key step to catch errors after solving!

This question tests your ability to determine the number of real solutions in a linear-quadratic system, corresponding to intersection points of a line and parabola. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in y = mx + b form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system y = -x + 1 and y = x² + 2x - 3, substitute: -x + 1 = x² + 2x - 3, rearrange to x² + 3x - 4 = 0, discriminant 9 + 16 = 25 > 0, so two real solutions (x = 1 and x = -4). Choice C correctly identifies there are 2 real solutions based on the positive discriminant. Excellent—using the discriminant is a quick way to preview the number without full solving!

This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (parabola) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: solve the linear equation for y (often already in y = mx + b form), substitute into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system y = -x² + 6x and y = 2x, substitute: -x² + 6x = 2x, rearrange to -x² + 4x = 0, factor -x(x - 4) = 0, x = 0 or 4, y = 0 or 8, giving (0, 0) and (4, 8). Choice A correctly finds both points with proper steps. Fantastic—modeling real-world scenarios like this builds intuition!

This question tests your ability to solve systems consisting of one linear equation (line) and one quadratic equation (circle) to find their intersection points. A linear-quadratic system has one equation graphing as a straight line and one as a curve (parabola or circle): the solutions are the intersection points where line and curve meet. Depending on how the line crosses the curve, there can be 0 intersections (line misses), 1 intersection (line tangent to curve), or 2 intersections (line passes through curve)—these are the three possibilities. To solve algebraically, use substitution: substitute the linear expression into the quadratic equation, giving a quadratic in one variable, then solve and back-substitute for the other coordinate. For the system x² + y² = 25 and y = -3x, substitute: x² + (-3x)² = 25, so 10x² = 25, x² = 2.5, x = ±√(5/2) = ±5/√10 (rationalized), y = -3x giving (5/√10, -15/√10) and (-5/√10, 15/√10). Choice C correctly finds both intersection points using proper substitution. Great job—circles follow the same method as parabolas!