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.

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!

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!

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. This gives you all the intersection points! Solving the system $\begin{cases} y = 2x + 1 \\ y = x^2 + 1 \end{cases}$ 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 A correctly finds the intersection points as $(0,1)$ and $(2,5)$ 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!

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!

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 + 1 = x² + 2x + 5, which rearranges to x² + 3x + 4 = 0. The discriminant is b² - 4ac = 3² - 4(1)(4) = 9 - 16 = -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 actually the discriminant is negative, indicating no real solutions. After substitution, the discriminant is -7 < 0, 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!

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: $y = x^2 - 1$ → $3 = 2^2 - 1$ → $3 = 4 - 1$ → $3 = 3$ ✓ (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 that (2, 3) satisfies both equations. Choice B includes a point that doesn't satisfy both equations: checking the linear equation shows $3 = 2(2) - 1 = 3$, which is true. For a system, BOTH equations must be satisfied. This point satisfies both equations, so it IS a solution to the system! 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, 3), and your sketch shows an intersection around that point, you're on track! The graph is a sanity check.

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, y = x² - 4} by substitution: (1) The linear equation gives us y = x - 2. (2) Substitute into the quadratic equation: x - 2 = x² - 4. (3) Simplify to standard form: x² - 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 = 0; for x = -1, y = -3. Solutions: (2, 0) and (-1, -3). Choice A correctly finds the intersection points as (2,0) and (-1,-3) through proper substitution and solving. Choice B only gives one solution when there are actually two: after substitution, we get x² - x - 2 = 0 with solutions x = 2 and x = -1. 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!

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. 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 $x^2 + 1 = 2$, which simplifies to $x^2 - 1 = 0$. The discriminant is $0^2 - 4(1)(-1) = 4$, which is positive, so two real solutions—the line intersects the curve at two points. Choice C correctly identifies the number of solutions as 2 real solutions. Choice A claims no solution, but actually there are two: after substitution, the discriminant is 4, which 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!

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!