This question tests your ability to solve systems of a line and circle to find all intersection points. A linear-quadratic system with a circle can have 0, 1, or 2 intersections based on the line's position. For y = x and $x^2 + y^2 = 8$, substitute: $x^2 + x^2 = 8$, $2x^2 = 8$, $x^2 = 4$, $x = \pm 2$; $y = \pm 2$, giving $(2,2)$ and $(-2,-2)$, both valid. Choice A correctly identifies the points from proper substitution. Choice B might result from $\sqrt{8}$ instead of $\sqrt{4}$, but solve the quadratic accurately after combining terms! Strategy: substitute linear into circle, simplify, solve for x, find y, verify; discriminant previews the count. Wonderful effort—this will sharpen your geometry-algebra skills!

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 like a circle: the solutions are the intersection points, which can be 0, 1, or 2 depending on whether the line misses, touches, or crosses the circle. For x² + y² = 25 and y = -3x, substitute: x² + (-3x)² = 25 becomes 10x² = 25, so x² = 2.5, x = ±√(5/2) = ±5/√10 after simplifying; then y = -3x gives (5/√10, -15/√10) and (-5/√10, 15/√10), both satisfying the equations. Choice A correctly finds both points with proper substitution and radical simplification. Choice C might appeal if you incorrectly simplify √(25/10) to 1 and -1, but remember to handle the square root accurately—x² = 25/10 means x = ±5/√10! Always substitute the linear into the quadratic, solve for the variable, back-substitute, and check the discriminant for the number of solutions. Keep up the excellent work—this approach will help you tackle any linear-circle system confidently!

This question tests your ability to solve linear-quadratic systems to find intersection points of a line and parabola. Such systems can yield 0, 1, or 2 solutions depending on intersections. For $y = -x + 2$ and $y = x^2$, set equal: $x^2 = -x + 2$, $x^2 + x - 2 = 0$; discriminant $1 + 8 = 9$, $x = \frac{-1 \pm 3}{2} = 1$ or $-2$; $y = 1$ and $4$, giving $(1,1)$ and $(-2,4)$, both checked. Choice A correctly lists the points. Choice B might come from sign errors in quadratic, but track coefficients carefully! Use the method: substitute, form quadratic, solve with formula or factoring, back-substitute, verify. Keep going—you're building expertise step by step!

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!

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 without fully solving, focusing on intersection points of a parabola and line. A linear-quadratic system can have 0, 1, or 2 real intersections based on the discriminant of the resulting quadratic after substitution. For y = x² + 2x - 3 and y = -x + 1, set equal: x² + 2x - 3 = -x + 1, so x² + 3x - 4 = 0; discriminant 9 + 16 = 25 > 0, indicating two real solutions. Choice C correctly identifies two real solutions from the positive discriminant. Choice A might tempt if you miscalculate the discriminant as negative, but double-check arithmetic—b² - 4ac is key! Use the strategy: substitute to form ax² + bx + c = 0, compute discriminant to predict solutions (positive: 2, zero: 1, negative: 0), then solve if needed. Great job exploring this—you'll ace predicting solution counts in no time!

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 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 like a parabola: the solutions are the intersection points where the line and parabola meet, and there can be 0, 1, or 2 real solutions depending on how they intersect. For the system $y = x + 1$ and $y = x^2 - 2x + 1$, substitute the linear 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 $x = 3$; then $y = 1$ and $y = 4$, giving points $(0, 1)$ and $(3, 4)$, which both satisfy the original equations. Choice C correctly identifies both intersection points through accurate substitution and back-substitution. Choice A might tempt if you mistakenly discard $x = 3$, but remember quadratics often yield two solutions—check both! To master linear-quadratic systems, always substitute the linear expression into the quadratic, solve the resulting quadratic equation, back-substitute to find y-values, and verify each point in both equations. You're doing great—practicing this method will make solving these systems second nature!

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!