Math II · A-REI.7

Solving Linear–Quadratic Systems Algebraically and Graphically

This objective teaches students how to find where two different rules are true at the same time. In real life, that is the mathematics of intersections: where a path meets a boundary, where a cost equals a revenue, where a straight-line plan meets a curved physical limit, or where two models make the same prediction.

Concept Algebra
Domain Reasoning with Equations and Inequalities
Read time 9 minutes

What this learning objective is really asking you to learn

A linear–quadratic system is a pair of equations in two variables where one equation is linear and the other contains a quadratic relationship. The linear equation usually graphs as a line. The quadratic equation may graph as a parabola, circle, or another simple second-degree curve, depending on the form of the equation. A solution to the system is an ordered pair \((x, y)\) that makes both equations true at the same time. Graphically, that means the point lies on both graphs. Algebraically, that means the same \(x\) and \(y\) values survive both symbolic tests.

The most important idea is simultaneous truth. A student is not merely solving a line, and not merely solving a quadratic. The student is asking, “Where do these two conditions overlap?” In Math I, students solved systems of two lines. Those systems could have one solution, no solution, or infinitely many solutions depending on whether the lines crossed, were parallel, or were the same line. In Math II, the second condition can curve. A line can cut a parabola twice, touch it once, or miss it completely. A line can cut a circle twice, touch it once as a tangent, or miss it. The visual picture is richer, and the algebra becomes a direct way to explain the picture.

A simple example is the system \(y = x + 1\) and \(y = x^2 - 3\). The first equation says \(y\) changes at a constant rate as \(x\) changes. The second says \(y\) changes quadratically. A solution must have a \(y\) value that is both \(x + 1\) and \(x^2 - 3\). Because both expressions equal \(y\), they must equal each other: \(x + 1 = x^2 - 3\). Rearranging gives \(x^2 - x - 4 = 0\). Solving that quadratic gives the \(x\)-coordinates of the intersection points. Substituting each \(x\) back into either original equation gives the corresponding \(y\)-coordinates. The graph shows the same story: the line and parabola meet where their heights are equal.

This is a major conceptual upgrade. In a linear system, substitution often leads to a linear equation. In a linear–quadratic system, substitution usually leads to a quadratic equation. That means students must bring together multiple parts of Math II: graphing, substitution, factoring, completing the square, the quadratic formula, and checking solutions. The objective is not a separate island. It is a meeting place.

The “why” is strong because many real situations involve a straight-line constraint and a curved relationship. A budget might be linear, while revenue or area is quadratic. A straight road might intersect a circular boundary. A planned constant-rate path might intersect the curved path of a projectile. A fixed price line might meet a demand or profit curve. A safety limit might be linear while a physical quantity grows with the square of speed. When students solve a linear–quadratic system, they are finding the moments or locations where two different descriptions of reality are both satisfied.

Students often ask, “When would I use this?” The honest answer is: any time two models compete or interact. If a ball follows a parabolic path and a wall is modeled by a line, the intersection tells whether the ball hits the wall and where. If a company’s cost is roughly linear but revenue is quadratic because price affects demand, intersections can represent break-even points. If a city designs a straight pipeline through a circular protected zone, intersections locate entry and exit points. If a camera’s line of sight crosses a curved path, intersections identify possible positions. The same algebraic structure keeps appearing because the world is full of constraints, paths, boundaries, and targets.

The historical machinery behind this idea

Linear–quadratic systems sit at the historical intersection of algebra and geometry. Ancient mathematicians solved practical quadratic problems long before modern symbolic notation existed. Babylonian tablets include procedures for problems involving areas and unknown lengths. Greek mathematicians studied conic sections, circles, and geometric intersections with extraordinary sophistication. But the modern power of this objective comes from a later breakthrough: the union of algebra and geometry.

René Descartes and Pierre de Fermat helped develop analytic geometry in the seventeenth century. Analytic geometry made it possible to describe geometric objects with equations and to solve geometric intersection problems with algebra. A line was no longer only a drawn object. It could be represented by an equation. A circle, parabola, or other conic could also be represented by an equation. Finding their intersection became a symbolic problem. This was revolutionary because it let people move between pictures and procedures. Geometry could guide intuition, while algebra could deliver exact answers.

This connection shaped modern science. Physics needed curves because motion under gravity is not usually linear. Astronomy needed conics because planetary paths and optical systems could be described by geometric curves. Engineering needed equations for loads, beams, projectiles, pressure, and design constraints. Economics later used equations to represent cost, revenue, demand, and equilibrium. In all of those areas, a solution often means an intersection: a moment when one model reaches another, a point where two conditions agree, or a boundary between possible and impossible.

The particular pairing of a line and a quadratic curve is also historically natural. Lines represent constant change and simple constraints. Quadratics represent area, acceleration, and squared distance. Circles arise from fixed distance from a center. Parabolas arise from projectile motion and focus-directrix geometry. These objects were studied for centuries because they are basic shapes of measurement and motion. A line meeting a quadratic curve is therefore not an artificial school problem. It is a simplified version of how people analyze the geometry of the physical world.

Technical execution: how to solve the system

The first method is graphing. Graph each equation on the same coordinate plane, using an appropriate scale. The solutions are the intersection points. Graphing helps students see how many solutions to expect. If the line crosses the quadratic curve in two places, expect two solutions. If it just touches the curve, expect one solution. If it misses, expect no real solutions. Graphing also helps students catch algebraic mistakes. If an algebraic answer says there are two intersections but the graph clearly shows none, something is wrong with the algebra, the graph, or both.

The second method is substitution. If the linear equation is written as \(y = mx + b\), substitute \(mx + b\) for \(y\) in the quadratic equation. If the quadratic is already \(y = ax^2 + bx + c\), set the two expressions equal: \(mx + b = ax^2 + bx + c\). Rearrange into standard quadratic form. Then solve the resulting quadratic equation. The solutions for \(x\) are not yet the full system solutions. They are the \(x\)-coordinates. Substitute each one back into one of the original equations, usually the simpler line equation, to find \(y\).

For example, solve \(y = 2x + 3\) and \(y = x^2 - 4x + 1\). Set the expressions equal: \(2x + 3 = x^2 - 4x + 1\). Move everything to one side: \(x^2 - 6x - 2 = 0\). This does not factor neatly, so use the quadratic formula: \(x = (6 ± \sqrt{36 + 8})/2 = (6 ± \sqrt{44})/2 = 3 ± \sqrt{11}\). Then compute \(y = 2x + 3\), giving \(y = 9 ± 2\sqrt{11}\). The two solutions are \((3 + \sqrt{11}, 9 + 2\sqrt{11})\) and \((3 - \sqrt{11}, 9 - 2\sqrt{11})\). A graph would show the line crossing the parabola twice.

Sometimes the quadratic equation is a circle, such as \(x^2 + y^2 = 25\), and the linear equation might be \(y = 3\). Substitution gives \(x^2 + 3^2 = 25\), so \(x^2 = 16\), and \(x = -4\) or \(x = 4\). The solutions are \((-4, 3)\) and \((4, 3)\). Graphically, the horizontal line \(y = 3\) cuts the circle at two points. If the line were \(y = 5\), substitution would give \(x^2 + 25 = 25\), so \(x = 0\); the line touches the circle at one point. If the line were \(y = 6\), substitution would give \(x^2 + 36 = 25\), so \(x^2 = -11\); there are no real intersection points. The algebra tells the same story as the graph.

The discriminant of the quadratic that appears after substitution gives useful information. If the discriminant is positive, the system has two real intersection points. If it is zero, the line is tangent to the quadratic curve and there is one real solution. If it is negative, there are no real intersections. Students do not need to treat the discriminant as a memorized gadget. It is a bridge between algebra and geometry: the sign of the discriminant predicts how the line and curve meet.

A careful solving routine has five steps. First, graph or sketch to predict the number of solutions. Second, substitute one equation into the other. Third, solve the resulting quadratic equation using the best available method. Fourth, substitute back to find the missing coordinate. Fifth, check each ordered pair in both original equations. Checking matters because a system solution must satisfy both conditions, not just the equation produced during substitution.

Why students should learn this math

A linear–quadratic system can represent an object moving along a straight path and another object or boundary described by a curve. Imagine a drone traveling along a straight route while a sensor boundary is circular. The intersection points are where the drone enters and exits the sensor zone. In a simplified sports setting, a ball’s height may be modeled by a quadratic function while a fence or target height is modeled by a line or constant value. The intersections tell whether the ball clears the target or hits it.

In business, linear cost and quadratic revenue can meet at break-even points. A small business might have costs that increase by a roughly constant amount for each unit produced, while revenue depends on both price and quantity, and quantity sold may change when price changes. Under simplified assumptions, the revenue model can become quadratic. The intersection of cost and revenue identifies sales levels where profit is zero. Between or beyond those points, the business may gain or lose money.

In design, a straight component may intersect a circular or parabolic boundary. Engineers, architects, and computer graphics programmers use intersection logic constantly. A ray intersects a surface. A line segment crosses a boundary. A tool path meets a curved edge. A camera view intersects a model. The exact equations may become more advanced, but this objective teaches the basic idea: solving systems is locating simultaneous constraints.

Where this fits in the big map of mathematics

This objective connects the systems work of Math I to the nonlinear world of Math II. Math I taught students that a system solution is an intersection of lines. Objective 066 extends that idea: a solution can be an intersection of different kinds of objects. That extension is the beginning of a much larger map. Later, students will solve systems involving circles, parabolas, rational functions, exponentials, logarithms, and trigonometric functions. They will also use technology to approximate intersections that cannot be solved neatly by hand.

The objective also prepares students for conic sections, coordinate proof, and calculus. In conics, equations describe circles, parabolas, ellipses, and hyperbolas. In calculus, intersections help solve optimization and area problems. In computer science and graphics, collision detection is often an intersection problem. In data modeling, two fitted models may be compared by finding where they agree or cross. Students who understand this objective are learning a general mathematical habit: when two rules must both be true, solve a system.

Common student traps and how to avoid them

One common trap is finding only \(x\) and forgetting \(y\). The quadratic equation after substitution gives possible \(x\)-coordinates, but a system solution is an ordered pair. Always substitute back.

A second trap is trusting a rough graph too much. Graphs are excellent for understanding and estimating, but algebra gives exact values when possible. A graph may hide intersections if the window or scale is poor.

A third trap is losing solutions while manipulating equations. Squaring, dividing, or simplifying without attention can remove or create candidates. Keep algebra organized and check final ordered pairs.

A fourth trap is assuming there must always be two solutions. A line can cross a curve twice, touch it once, or miss it. The number of real solutions depends on the geometry and on the discriminant of the resulting quadratic.

Problem Library

Problems in the App From This Objective

210 problems across 15 archetypes in the app.

substitute line equation into quadratic and solve.
15 problems Warmup Practice Mixed Review Assessment
Problem 1

Solve the linear-quadratic system by substitution: y=x^2, y=x+2.

Problem 2

Solve the linear-quadratic system by substitution: y=x^2-4, y=2x-1.

Open in simulator
Problem 3

Solve the linear-quadratic system by substitution: y=x^2+1, y=5.

Problem 4

Solve the linear-quadratic system by substitution: y=x^2-2x, y=x.

Problem 5

Solve the linear-quadratic system by substitution: y=x^2-3, y=x-1.

Problem 6

Solve the linear-quadratic system by substitution: y=x^2+2x+1, y=x+3.

Problem 7

Solve the linear-quadratic system by substitution: y=x^2-5x+6, y=2x.

Problem 8

Solve the linear-quadratic system by substitution: y=-x^2+4, y=x+2.

Problem 9

Solve the linear-quadratic system by substitution: y=x^2-6x+9, y=0.

Problem 10

Solve the linear-quadratic system by substitution: y=x^2+4x+4, y=x+2.

Problem 11

Solve the linear-quadratic system by substitution: y=x^2-9, y=0.

Problem 12

Solve the linear-quadratic system by substitution: y=x^2+3x, y=-x.

Problem 13

Solve the linear-quadratic system by substitution: y=2x^2, y=4x.

Problem 14

Solve the linear-quadratic system by substitution: y=x^2-2x+1, y=x-1.

Problem 15

Solve the linear-quadratic system by substitution: y=x^2+5, y=2x+5.

eliminate one variable and solve resulting quadratic.
15 problems Warmup Practice Mixed Review Assessment
Problem 16

Solve the standard-form linear-quadratic system by elimination or rearranging: x^2+y=6, y=x.

Problem 17

Solve the standard-form linear-quadratic system by elimination or rearranging: y-x^2=0, y-3x=0.

Problem 18

Solve the standard-form linear-quadratic system by elimination or rearranging: x^2+y=10, y=2x+2.

Problem 19

Solve the standard-form linear-quadratic system by elimination or rearranging: y+x=4, y=x^2.

Problem 20

Solve the standard-form linear-quadratic system by elimination or rearranging: y=x+1, y=x^2-5.

Problem 21

Solve the standard-form linear-quadratic system by elimination or rearranging: y=2x, y=x^2-3.

Problem 22

Solve the standard-form linear-quadratic system by elimination or rearranging: y=x-2, x^2+y=4.

Problem 23

Solve the standard-form linear-quadratic system by elimination or rearranging: y=-x+5, y=x^2-x+1.

Problem 24

Solve the standard-form linear-quadratic system by elimination or rearranging: y=2x+3, y=x^2.

Problem 25

Solve the standard-form linear-quadratic system by elimination or rearranging: y=-x, y=x^2-2x.

Problem 26

Solve the standard-form linear-quadratic system by elimination or rearranging: y=x+3, y=x^2+x-1.

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Problem 27

Solve the standard-form linear-quadratic system by elimination or rearranging: x=y+1, x=y^2-5.

Problem 28

Solve the standard-form linear-quadratic system by elimination or rearranging: y=4x, y=x^2+3x.

Problem 29

Solve the standard-form linear-quadratic system by elimination or rearranging: y=-2x+6, y=x^2-x.

Problem 30

Solve the standard-form linear-quadratic system by elimination or rearranging: y=x+6, x^2+y=12.

read intersection coordinates.
15 problems Warmup Practice Mixed Review Assessment
Problem 31

Estimate the intersection points from graph description Line y=2 crosses parabola y=x^2 near x=-1.4 and x=1.4.

Problem 32

Estimate the intersection points from graph description Line y=x+1 crosses y=x^2 at x about -0.6 and 1.6.

Problem 33

Estimate the intersection points from graph description Line y=4 crosses y=(x-1)^2 at x=-1 and x=3.

Problem 34

Estimate the intersection points from graph description Line y=3x touches y=x^2 at x=0 and crosses again near x=3.

Open in simulator
Problem 35

Estimate the intersection points from graph description Line y=x crosses parabola y=x^2 at x=0 and x=1.

Problem 36

Estimate the intersection points from graph description Line y=-x crosses parabola y=x^2 at x=0 and x=-1.

Problem 37

Estimate the intersection points from graph description Line y=x+2 crosses parabola y=x^2 at x=-1 and x=2.

Problem 38

Estimate the intersection points from graph description Line y=3 crosses parabola y=x^2-1 at x=-2 and x=2.

Problem 39

Estimate the intersection points from graph description Line y=-2x+1 crosses parabola y=x^2 near x=0.4 and x=-2.4.

Problem 40

Estimate the intersection points from graph description Line y=x+5 crosses parabola y=x^2+2x+1 near x=1.6 and x=-2.6.

Problem 41

Estimate the intersection points from graph description The x-axis crosses parabola y=(x-2)^2-1 at x=1 and x=3.

Problem 42

Estimate the intersection points from graph description Line y=-x+2 crosses parabola y=-x^2+4 at x=-1 and x=2.

Problem 43

Estimate the intersection points from graph description Line y=x-2 crosses parabola y=x^2-4x+2 at x=1 and x=4.

Problem 44

Estimate the intersection points from graph description Line y=x+3 crosses parabola y=x^2-x at x=-1 and x=3.

Problem 45

Estimate the intersection points from graph description Line y=5 crosses parabola y=2x^2 near x=-1.6 and x=1.6.

identify secant, tangent, or no intersection.
15 problems Warmup Practice Mixed Review Assessment
Problem 46

Determine the number of real intersections for y=x^2, y=4.

Problem 47

Determine the number of real intersections for y=x^2, y=0.

Problem 48

Determine the number of real intersections for y=x^2+3, y=1.

Problem 49

Determine the number of real intersections for y=(x-2)^2-1, y=-1.

Problem 50

Determine the number of real intersections for y=x^2-2, y=2.

Problem 51

Determine the number of real intersections for y=-x^2+5, y=1.

Problem 52

Determine the number of real intersections for y=x^2, y=x+2.

Problem 53

Determine the number of real intersections for y=(x+1)^2, y=0.

Open in simulator
Problem 54

Determine the number of real intersections for y=x^2, y=2x-1.

Problem 55

Determine the number of real intersections for y=x^2+1, y=0.

Problem 56

Determine the number of real intersections for y=-(x-3)^2-2, y=0.

Problem 57

Determine the number of real intersections for y=2x^2-3, y=5.

Problem 58

Determine the number of real intersections for y=-3x^2+1, y=1.

Problem 59

Determine the number of real intersections for y=x^2+5, y=x.

Problem 60

Determine the number of real intersections for y=-x^2+3, y=x.

explain equal outputs or meeting points.
12 problems Warmup Practice Mixed Review Assessment
Problem 61

Interpret the intersection points (2,20), (10,20) for context Revenue R=-p^2+12p and cost C=20 are both in dollars.

Problem 62

Interpret the intersection points (2,8), (4,8) for context Height h=-t^2+6t and platform height h=8.

Problem 63

Interpret the intersection points (5,25) for context Area A=x^2 and linear target A=25.

Problem 64

Interpret the intersection points (0,0), (8,0) for context Profit P=-x^2+8x and break-even line P=0.

Problem 65

Interpret the intersection points (0,0), (4,0) for context Height h=-16t^2+64t and ground h=0. t is time in seconds, h is height in feet.

Problem 66

Interpret the intersection points (3,9) for context Supply S=p^2 and demand D=-2p+15. p is price in dollars, S and D are quantities.

Open in simulator
Problem 67

Interpret the intersection points (1,6), (5,30) for context Population P_1=t^2+5 and P_2=6t. t is time in years, P is population in hundreds.

Problem 68

Interpret the intersection points (4,24), (6,24) for context Area A=x(10-x) and target area A=24. x is width in meters, A is area in square meters.

Problem 69

Interpret the intersection points (5-sqrt(10), 15), (5+sqrt(10), 15) for context Cost C=x^2-10x+30 and fixed cost C=15. x is units produced, C is cost in dollars.

Problem 70

Interpret the intersection points (3,9), (5,25) for context Distance d_1=t^2 and d_2=8t-15. t is time in hours, d is distance in miles.

Problem 71

Interpret the intersection points (10,50) for context Revenue R=-0.5x^2+10x and target revenue R=50. x is units sold, R is revenue in hundreds of dollars.

Problem 72

Interpret the intersection points (6,46) for context Temperature T=-t^2+12t+10 and target temperature T=46. t is hours after noon, T is temperature in degrees Celsius.

substitute constant x or y.
12 problems Warmup Practice Mixed Review Assessment
Problem 73

Solve the system with a horizontal or vertical line: y=x^2-1, y=3.

Open in simulator
Problem 74

Solve the system with a horizontal or vertical line: y=(x-1)^2, x=4.

Problem 75

Solve the system with a horizontal or vertical line: y=x^2+2x, y=-1.

Problem 76

Solve the system with a horizontal or vertical line: y=x^2, x=-3.

Problem 77

Solve the system with a horizontal or vertical line: y = x^2 + 1, y = 5.

Problem 78

Solve the system with a horizontal or vertical line: y = (x-2)^2, y = 0.

Problem 79

Solve the system with a horizontal or vertical line: y = -x^2 + 2, y = -2.

Problem 80

Solve the system with a horizontal or vertical line: y = -(x+3)^2, y = 0.

Problem 81

Solve the system with a horizontal or vertical line: x = y^2, x = 9.

Problem 82

Solve the system with a horizontal or vertical line: x = (y-2)^2, x = 1.

Problem 83

Solve the system with a horizontal or vertical line: x = y^2 + 4y + 4, x = 0.

Problem 84

Solve the system with a horizontal or vertical line: x = -y^2 + 5, x = 1.

connect discriminant to number of system solutions.
15 problems Warmup Practice Mixed Review Assessment
Problem 85

After substituting the line into the parabola, classify intersections using the discriminant for x^2-5x+6=0.

Problem 86

After substituting the line into the parabola, classify intersections using the discriminant for x^2-4x+4=0.

Problem 87

After substituting the line into the parabola, classify intersections using the discriminant for x^2+2x+5=0.

Problem 88

After substituting the line into the parabola, classify intersections using the discriminant for 2x^2-3x-2=0.

Problem 89

After substituting the line into the parabola, classify intersections using the discriminant for x^2-x-2=0.

Open in simulator
Problem 90

After substituting the line into the parabola, classify intersections using the discriminant for 3x^2+5x-2=0.

Problem 91

After substituting the line into the parabola, classify intersections using the discriminant for x^2-6x+9=0.

Problem 92

After substituting the line into the parabola, classify intersections using the discriminant for 4x^2+4x+1=0.

Problem 93

After substituting the line into the parabola, classify intersections using the discriminant for x^2+x+1=0.

Problem 94

After substituting the line into the parabola, classify intersections using the discriminant for 2x^2-x+1=0.

Problem 95

After substituting the line into the parabola, classify intersections using the discriminant for -x^2+3x+4=0.

Problem 96

After substituting the line into the parabola, classify intersections using the discriminant for -x^2+2x-1=0.

Problem 97

After substituting the line into the parabola, classify intersections using the discriminant for -x^2+x-2=0.

Problem 98

After substituting the line into the parabola, classify intersections using the discriminant for 5x^2-7x-6=0.

Problem 99

After substituting the line into the parabola, classify intersections using the discriminant for 9x^2-12x+4=0.

use shared points and linear equation.
12 problems Warmup Practice Mixed Review Assessment
Problem 100

Find a line that intersects parabola y=x^2 at specified point(s) (1,1) and (3,9).

Problem 101

Find a line that intersects parabola y=x^2-4 at specified point(s) (0,-4) and (2,0).

Problem 102

Find a line that intersects parabola y=(x-1)^2 at specified point(s) (1,0) and (3,4).

Problem 103

Find a line that intersects parabola y=x^2+1 at specified point(s) (-1,2) and (2,5).

Problem 104

Find a line that intersects parabola y=x^2+x at specified point(s) (0,0) and (1,2).

Problem 105

Find a line that intersects parabola y=-x^2 at specified point(s) (-1,-1) and (1,-1).

Problem 106

Find a line that intersects parabola y=x^2-2x+1 at specified point(s) (0,1) and (2,1).

Problem 107

Find a line that intersects parabola y=2x^2 at specified point(s) (-2,8) and (0,0).

Problem 108

Find a line that intersects parabola y=x^2+2x+3 at specified point(s) (-3,6) and (-1,2).

Problem 109

Find a line that intersects parabola y=-x^2+5 at specified point(s) (1,4) and (2,1).

Problem 110

Find a line that intersects parabola y=(x+2)^2 at specified point(s) (-2,0) and (0,4).

Open in simulator
Problem 111

Find a line that intersects parabola y=3x^2-x at specified point(s) (1,2) and (2,10).

substitute ordered pairs into both equations.
15 problems Warmup Practice Mixed Review Assessment
Problem 112

Verify whether (2,4) solves the system y=x^2 and y=x+2.

Problem 113

Verify whether (1,3) solves the system y=x^2 and y=x+2.

Problem 114

Verify whether (3,5) solves the system y=x^2-4 and y=2x-1.

Problem 115

Verify whether (3,4) solves the system y=(x-1)^2 and y=4.

Problem 116

Verify whether (2,3) solves the system y=x^2-1 and y=x+1.

Problem 117

Verify whether (-1,0) solves the system y=x^2-1 and y=x+1.

Problem 118

Verify whether (2,4) solves the system y=x^2 and y=x+1.

Problem 119

Verify whether (0,3) solves the system y=x^2+1 and y=x+3.

Problem 120

Verify whether (-2,0) solves the system y=x^2+2x and y=x+2.

Problem 121

Verify whether (1,3) solves the system y=x^2+2x and y=x+2.

Problem 122

Verify whether (0,0) solves the system y=x^2-3x+2 and y=x-1.

Problem 123

Verify whether (2,0) solves the system y=x^2-5x+6 and y=x-2.

Problem 124

Verify whether (4,2) solves the system y=x^2-5x+6 and y=x-2.

Problem 125

Verify whether (0,2) solves the system y=-x^2+4 and y=x+2.

Problem 126

Verify whether (2,1) solves the system y=x^2-2x+1 and y=x+1.

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decide based on exact versus approximate need.
12 problems Warmup Practice Mixed Review Assessment
Problem 127

Choose an algebraic or graphical method for y=x^2 and y=x+2, exact solutions needed and justify the choice.

Open in simulator
Problem 128

Choose an algebraic or graphical method for A graph shows a line and parabola crossing near noninteger points, estimate only and justify the choice.

Problem 129

Choose an algebraic or graphical method for y=(x-3)^2+1 and y=1 and justify the choice.

Problem 130

Choose an algebraic or graphical method for Real-world plotted data line and fitted parabola, nearest tenth requested and justify the choice.

Problem 131

Choose an algebraic or graphical method for y = x^2 - 4x + 3 and y = -x + 3, exact solutions required and justify the choice.

Problem 132

Choose an algebraic or graphical method for The intersection of y = x^3 - 2x + 1 and y = -x^2 + 3, estimate the positive x-intercept to one decimal place. and justify the choice.

Problem 133

Choose an algebraic or graphical method for y = x^2 - 5 and y = -x^2 + 3, find all exact points of intersection. and justify the choice.

Problem 134

Choose an algebraic or graphical method for Given a graph of f(x) and g(x), estimate the x-values where f(x) = g(x) to the nearest half-unit. and justify the choice.

Problem 135

Choose an algebraic or graphical method for 2x + y = 5 and x - y = 1, find the exact solution. and justify the choice.

Problem 136

Choose an algebraic or graphical method for y = 2^x and y = x + 3, approximate the intersection points to two decimal places. and justify the choice.

Problem 137

Choose an algebraic or graphical method for (x+1)^2 + y^2 = 25 and y = 0, find exact x-intercepts. and justify the choice.

Problem 138

Choose an algebraic or graphical method for Intersection of a sine wave and a parabola, estimate all real solutions from a provided plot. and justify the choice.

substitute and solve efficiently.
15 problems Warmup Practice Mixed Review Assessment
Problem 139

Solve the system using the factored quadratic form: y=(x-1)(x-4), y=0.

Problem 140

Solve the system using the factored quadratic form: y=(x+2)(x-3), y=x+2.

Problem 141

Solve the system using the factored quadratic form: y=2(x-1)(x+3), y=0.

Problem 142

Solve the system using the factored quadratic form: y=(x-2)(x+5), y=6.

Problem 143

Solve the system using the factored quadratic form: y=(x+1)(x-5), y=0.

Problem 144

Solve the system using the factored quadratic form: y=-1(x-2)(x+4), y=0.

Problem 145

Solve the system using the factored quadratic form: y=(x+3)(x-1), y=5.

Problem 146

Solve the system using the factored quadratic form: y=2(x-3)(x+1), y=-8.

Open in simulator
Problem 147

Solve the system using the factored quadratic form: y=(x-1)(x-3), y=x-3.

Problem 148

Solve the system using the factored quadratic form: y=-1(x+1)(x-3), y=x+1.

Problem 149

Solve the system using the factored quadratic form: y=(x-3)^2, y=4.

Problem 150

Solve the system using the factored quadratic form: y=(x+1)^2, y=2x+2.

Problem 151

Solve the system using the factored quadratic form: y=(x-5)(x+2), y=x+2.

Problem 152

Solve the system using the factored quadratic form: y=-(x-4)(x+1), y=2x-2.

Problem 153

Solve the system using the factored quadratic form: y=x(x-4), y=-4.

substitute and handle squared binomial.
15 problems Warmup Practice Mixed Review Assessment
Problem 154

Solve the system with a vertex-form quadratic: y=(x-2)^2+1, y=5.

Problem 155

Solve the system with a vertex-form quadratic: y=-(x+1)^2+6, y=2.

Problem 156

Solve the system with a vertex-form quadratic: y=(x-3)^2-4, y=x-4.

Problem 157

Solve the system with a vertex-form quadratic: y=2(x-1)^2-3, y=5.

Problem 158

Solve the system with a vertex-form quadratic: y=(x-1)^2+2, y=6.

Problem 159

Solve the system with a vertex-form quadratic: y=-(x+2)^2+5, y=1.

Problem 160

Solve the system with a vertex-form quadratic: y=(x-3)^2+1, y=1.

Problem 161

Solve the system with a vertex-form quadratic: y=(x-1)^2-3, y=x-2.

Problem 162

Solve the system with a vertex-form quadratic: y=(x-2)^2-1, y=-3x+3.

Problem 163

Solve the system with a vertex-form quadratic: y=2(x-1)^2-1, y=2x+1.

Problem 164

Solve the system with a vertex-form quadratic: y=(x-2)^2+3, y=2x-2.

Problem 165

Solve the system with a vertex-form quadratic: y=-(x+1)^2+5, y=-2x+4.

Problem 166

Solve the system with a vertex-form quadratic: y=(x+5)^2-3, y=1.

Problem 167

Solve the system with a vertex-form quadratic: y=-2(x-3)^2+8, y=4x-4.

Open in simulator
Problem 168

Solve the system with a vertex-form quadratic: y=3(x+1)^2-2, y=10.

catch substitution, quadratic solving, and graph-reading errors.
15 problems Warmup Practice Mixed Review Assessment
Problem 169

Identify and correct the error in solving this linear-quadratic system: For y=x^2 and y=x+2, x^2=x+2 gives x=2 only.

Open in simulator
Problem 170

Identify and correct the error in solving this linear-quadratic system: For y=(x-1)^2 and y=4, x-1=4 so x=5.

Problem 171

Identify and correct the error in solving this linear-quadratic system: For y=x^2-4 and y=2x-1, roots x=1,3 so points are (1,3),(3,5).

Problem 172

Identify and correct the error in solving this linear-quadratic system: For y=x^2 and y=-1, x^2=-1 so there is one solution x=-1.

Problem 173

Identify and correct the error in solving this linear-quadratic system: For y=x^2-2x and y=x, x^2-2x=x gives x^2-x=0, so x(x-1)=0, roots x=0,1.

Problem 174

Identify and correct the error in solving this linear-quadratic system: For y=x^2+1 and y=5, x^2+1=5 leads to x^2=4, so x=2.

Problem 175

Identify and correct the error in solving this linear-quadratic system: For y=x^2-x and y=2, x^2-x-2=0 gives x=2, x=-1. The points are (2,2) and (-1,-2).

Problem 176

Identify and correct the error in solving this linear-quadratic system: For y=x^2 and y=2(x-1)+2, x^2=2x-1+2, so x^2=2x+1. This gives x^2-2x-1=0.

Problem 177

Identify and correct the error in solving this linear-quadratic system: For y=x^2+3x and y=-1, x^2+3x+1=0. Using the quadratic formula, the discriminant is 3^2-4(1)(1)=5, so x=(-3 +/- sqrt(9))/2.

Problem 178

Identify and correct the error in solving this linear-quadratic system: For y=x^2 and y=2x-1, the graphs touch at one point, so there are no solutions.

Problem 179

Identify and correct the error in solving this linear-quadratic system: For y=x^2+2x+1 and y=x+3, x^2+2x+1 = x+3. Moving terms gives x^2+3x-2=0.

Problem 180

Identify and correct the error in solving this linear-quadratic system: For y=x^2-4x+5 and y=2, x^2-4x+3=0 gives x=1, x=3. The solutions are (1,2) and (3,?).

Problem 181

Identify and correct the error in solving this linear-quadratic system: For y=x^2-5x+6 and y=0, (x-2)(x-3)=0 so x=-2, x=-3.

Problem 182

Identify and correct the error in solving this linear-quadratic system: For y=x^2-4x+4 and y=0, (x-2)^2=0. Since there's only one x-value, there are no real solutions.

Problem 183

Identify and correct the error in solving this linear-quadratic system: For y=x^2+3x and y=4, x^2+3x=4. Factoring gives x(x+3)=4, so x=4 or x+3=4 (x=1).

build both equations and solve intersections.
15 problems Warmup Practice Mixed Review Assessment
Problem 184

Build and solve a linear-quadratic system for context: A ball has height h=-t^2+6t and a platform has height h=8.

Open in simulator
Problem 185

Build and solve a linear-quadratic system for context: Revenue is R=-p^2+12p and cost is C=20.

Problem 186

Build and solve a linear-quadratic system for context: A square area A=x^2 equals a linear target A=3x+10.

Problem 187

Build and solve a linear-quadratic system for context: A path height h=(t-1)^2 and a sensor line h=4.

Problem 188

Build and solve a linear-quadratic system for context: The trajectory of a projectile is y=x^2-4x+5 and a laser beam follows y=x+1.

Problem 189

Build and solve a linear-quadratic system for context: A parabolic arch is described by y=-x^2+8x-10 and a support beam is y=2x-2.

Problem 190

Build and solve a linear-quadratic system for context: The path of a bouncing ball is y=x^2+2x+1 and the ground is y=0.

Problem 191

Build and solve a linear-quadratic system for context: A parabolic antenna has a cross-section y=x^2-6x+9 and a signal is detected at height y=4.

Problem 192

Build and solve a linear-quadratic system for context: The profit of a company is P=-x^2+4x and a break-even line is P=x.

Problem 193

Build and solve a linear-quadratic system for context: The path of a bird is y=x^2-4x+6 and a power line is y=x+2.

Problem 194

Build and solve a linear-quadratic system for context: The height of a rocket is h=-t^2+7t-6 and a drone flies at h=t+2.

Problem 195

Build and solve a linear-quadratic system for context: A diver's path is y=x^2+5x+4 and the water surface is y=0.

Problem 196

Build and solve a linear-quadratic system for context: The path of a thrown object is y=-2x^2+10x-8 and the ground is y=0.

Problem 197

Build and solve a linear-quadratic system for context: A parabolic mirror has a shape y=x^2-8x+16 and a light ray follows y=2x-8.

Problem 198

Build and solve a linear-quadratic system for context: A suspension bridge cable forms y=(x+2)^2-1 and a horizontal support beam is at y=3.

compute corresponding y-values and interpret both coordinates.
12 problems Warmup Practice Mixed Review Assessment
Problem 199

Explain why solving y=x^2 and y=x+2 requires ordered pairs, not just x-values.

Problem 200

Explain why solving y=(x-3)^2+1 and y=5 requires ordered pairs, not just x-values.

Open in simulator
Problem 201

Explain why solving y=x^2-4 and y=2x-1 requires ordered pairs, not just x-values.

Problem 202

Explain why solving y=x^2 and x=-2 requires ordered pairs, not just x-values.

Problem 203

Explain why solving y = 2x + 1 and y = -x + 4 requires ordered pairs, not just x-values.

Problem 204

Explain why solving x^2 + y^2 = 5 and y = x + 1 requires ordered pairs, not just x-values.

Problem 205

Explain why solving y = x^2 - 2x + 1 and y = 0 requires ordered pairs, not just x-values.

Problem 206

Explain why solving y = x^2 and y = -x^2 + 4 requires ordered pairs, not just x-values.

Problem 207

Explain why solving y = |x| and y = 2 requires ordered pairs, not just x-values.

Problem 208

Explain why solving y = 2^x and y = 4 requires ordered pairs, not just x-values.

Problem 209

Explain why solving y = 1/x and y = 1 requires ordered pairs, not just x-values.

Problem 210

Explain why solving y = x^2 + 2x and x = -3 requires ordered pairs, not just x-values.