Systems of Linear and Non-Linear Equations Practice Questions — ACT

Question 1Question →

The elevation $y$ (in meters) of a roller coaster track is modeled by the equation $y = (x - 3)^2 - 4$ , where $x$ represents the horizontal distance (in meters) from the start of the ride. A straight support beam is designed such that the height of the track is 7 meters less than 7 times the horizontal distance. The support beam connects to the roller coaster track at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the value of $x_1 y_1 + x_2 y_2$ ?

126

-42

924

336

2664

Show answer & explanation

Answer: 924

Answer

The value of the expression is 924, which corresponds to the coordinates of the two connection points

(1, 0)

and

(12, 77)

.

By translating the verbal relationship to the equation

y = 7x - 7

, we can equate it to the track's equation:

(x - 3)^2 - 4 = 7x - 7

. Expanding and simplifying yields

x^2 - 13x + 12 = 0

, which factors as

(x - 1)(x - 12) = 0

. This gives the horizontal intersection coordinates

x_1 = 1

and

x_2 = 12

. Substituting these coordinates back into the linear equation gives the vertical coordinates

y_1 = 0

and

y_2 = 77

. Evaluating the final expression yields

(1)(0) + (12)(77) = 924

.

Step-by-Step Solution

1

Translate the verbal description of the support beam's height into an algebraic equation.

y = 7x - 7

The description states that the height (

y

) is '7 meters less than 7 times the horizontal distance (

x

)', which translates to

7x - 7

.

2

Set the two equations for

y

equal to each other to find the

x

-coordinates of the intersection points.

(x - 3)^2 - 4 = 7x - 7 \Rightarrow x^2 - 6x + 5 = 7x - 7 \Rightarrow x^2 - 13x + 12 = 0

At the points of intersection, the elevation of the track and the support beam must be equal.

3

Solve the quadratic equation for

x

by factoring.

(x - 1)(x - 12) = 0 \Rightarrow x_1 = 1

and

x_2 = 12

Factoring the quadratic expression allows us to identify its roots directly.

4

Substitute the

x

-values back into the linear equation to find the corresponding

y

-coordinates.

For

x_1 = 1

,

y_1 = 7(1) - 7 = 0

. For

x_2 = 12

,

y_2 = 7(12) - 7 = 77

. The intersection points are

(1, 0)

and

(12, 77)

.

The intersection points must satisfy both equations in the system.

5

Calculate the required expression

x_1 y_1 + x_2 y_2

.

(1)(0) + (12)(77) = 0 + 924 = 924

This evaluates the target expression using the coordinates found in the previous steps.

Key Concept

Solving systems of linear and non-linear (quadratic) equations using substitution and factoring.

Question 2Question →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 12y + 27 = 0$ . A parabola that opens downward has its vertex at $(0, k)$ and is defined by the equation $y = -x^2 + k$ . If the system of equations consisting of this circle and parabola has exactly three distinct real solution points, what is the value of $k$ ?

Show answer & explanation

Answer: 9

Answer

The value of

k

is 9.

The correct value of

k

is 9 because when

k=9

, the system of equations reduces to a quadratic in

y

with roots

y=9

and

y=4

. Both roots satisfy the real-number constraint

y \leq 9

for the parabola

x^2 = 9-y

, producing three distinct real solutions:

(0, 9)

,

(\sqrt{5}, 4)

, and

(-\sqrt{5}, 4)

.

Step-by-Step Solution

1

Complete the square for the circle's equation.

x^2 + (y-6)^2 = 9

To identify the circle's center at

(0, 6)

and radius

R=3

for geometric interpretation.

2

Express

x^2

in terms of

y

using the parabola's equation.

x^2 = k - y

To substitute into the circle's equation and eliminate the

x

variable.

3

Substitute

x^2

into the circle's equation and simplify.

y^2 - 13y + (k+27) = 0

To create a quadratic equation in

y

representing the y-coordinates of the intersection points.

4

Set

y = k

in the quadratic equation.

k^2 - 12k + 27 = 0

, which factors as

(k-3)(k-9) = 0

An intersection must lie on the y-axis (

x=0

, which means

y=k

) to yield an odd number of intersection points.

5

Verify which candidate value of

k

yields exactly three real solutions.

For

k=3

, the solutions are restricted because

y=10

gives no real

x

value, resulting in only 1 solution. For

k=9

, the roots

y=9

and

y=4

both yield real

x

values, resulting in exactly 3 solutions:

(0, 9)

,

(\sqrt{5}, 4)

, and

(-\sqrt{5}, 4)

.

The algebraic condition for real

x

coordinates is

x^2 = k - y \geq 0

, so we must verify that the roots

y

satisfy

y \leq k

.

Key Concept

Solving systems of non-linear equations algebraically and analyzing the number of real intersection points under coordinate constraints.

Alternative Method

Geometrically, a parabola opening downward with its vertex on the y-axis will intersect a circle centered on the y-axis in exactly three points if and only if its vertex is at the top of the circle and its curvature is less than that of the circle at that point. Completing the square for the circle

x^2 + y^2 - 12y + 27 = 0

gives

x^2 + (y-6)^2 = 9

, which shows the top point of the circle is

(0, 9)

. Thus, the vertex of the downward-opening parabola must be at

(0, 9)

, meaning

k = 9

. We then algebraically verify that this curvature indeed allows two other real intersections.

Estimated Time:3m 0s

Question 3Question →

A system of equations consists of the linear equation $y = 2x + 1$ and the quadratic equation $y = x^2 - 2$ . If $(x, y)$ is a solution to this system such that $x > 0$ , what is the value of $y$ ?

Show answer & explanation

Answer: 7

Answer

The correct value of

y

is 7.

Substituting

x = 3

into either equation yields the

y

-value of 7.

Step-by-Step Solution

1

Set the two expressions for

y

equal to each other.

x^2 - 2 = 2x + 1

Since both equations define

y

in terms of

x

, equating them allows us to find the

x

-coordinates of the intersection points.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 2x - 3 = 0

Subtracting

2x

and

1

from both sides collects all terms on one side of the equation.

3

Factor the quadratic equation.

(x - 3)(x + 1) = 0

Finding two numbers that multiply to

-3

and add to

-2

gives

-3

and

1

, allowing the quadratic to be factored.

4

Solve for

x

and apply the constraint

x > 0

.

x = 3

The factored equation yields solutions of

x = 3

and

x = -1

. The constraint that

x

must be greater than zero means we select

x = 3

.

5

Substitute the

x

-value back into one of the original equations to solve for

y

.

y = 7

Plugging

x = 3

into the linear equation

y = 2x + 1

yields

y = 2(3) + 1 = 7

. Substituting into

y = x^2 - 2

yields

y = 3^2 - 2 = 7

, which confirms the solution.

Key Concept

Solving a system of linear and quadratic equations using substitution.

Question 4Question →

A line with a positive slope passes through the point $(0, -4)$ and is tangent to the circle $x^2 + y^2 = 4$ . If this same line is also tangent to the parabola $y = x^2 + k$ , what is the value of the constant $k$ ?

Show answer & explanation

Answer: -3.25

Answer

The constant

k

must be

-3.25

.

By writing the equation of the line passing through

(0, -4)

as

y = mx - 4

and applying the condition that it is tangent to the circle

x^2 + y^2 = 4

, we find the positive slope is

m = \sqrt{3}

. Substituting this tangent line

y = \sqrt{3}x - 4

into the parabola equation

y = x^2 + k

yields the quadratic equation

x^2 - \sqrt{3}x + (k + 4) = 0

. For the line to be tangent to the parabola, the discriminant of this equation must be zero, which gives

3 - 4(k + 4) = 0

, leading to the final value

k = -3.25

.

Step-by-Step Solution

1

Set up the equation of the line passing through

(0, -4)

and substitute it into the equation of the circle.

x^2 + (mx - 4)^2 = 4 \implies (1 + m^2)x^2 - 8mx + 12 = 0

This establishes the quadratic equation representing the intersection points of the line and the circle.

2

Solve for the slope

m

by setting the discriminant of the intersection equation to zero.

(-8m)^2 - 4(1 + m^2)(12) = 0 \implies 64m^2 - 48 - 48m^2 = 0 \implies 16m^2 = 48 \implies m = \sqrt{3}

Since the line is tangent to the circle, the intersection equation must have exactly one real solution, meaning its discriminant is zero. The positive root is selected because the slope is specified as positive.

3

Equate the line

y = \sqrt{3}x - 4

and the parabola

y = x^2 + k

to analyze their intersection.

x^2 - \sqrt{3}x + (k + 4) = 0

This sets up the quadratic equation for the intersection points between the tangent line and the parabola.

4

Solve for

k

by setting the discriminant of this second quadratic equation to zero.

(-\sqrt{3})^2 - 4(1)(k + 4) = 0 \implies 3 - 4k - 16 = 0 \implies -13 - 4k = 0 \implies k = -3.25

For the line to be tangent to the parabola, the system must have exactly one solution, meaning the discriminant of their intersection equation must be zero.

Key Concept

Systems of Linear and Non-Linear Equations

Question 5Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + y^2 = 10$ . The line $y = 3x$ intersects the circle at a point $(x, y)$ in the first quadrant. What is the value of $x + y$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

Substituting

y = 3x

into the equation

x^2 + y^2 = 10

yields

x^2 + (3x)^2 = 10

. Simplifying this expression gives

x^2 + 9x^2 = 10

, which combines to

10x^2 = 10

. Dividing both sides by 10 results in

x^2 = 1

. Since the point of intersection lies in the first quadrant, we take the positive square root, so

x = 1

. Substituting

x = 1

back into

y = 3x

gives

y = 3(1) = 3

. The sum of the coordinates is

x + y = 1 + 3 = 4

.

Step-by-Step Solution

1

Substitute the linear expression for

y

into the equation of the circle.

x^2 + (3x)^2 = 10

Substituting one equation into another allows us to solve for a single variable.

2

Simplify the equation and solve for

x

.

10x^2 = 10 \Rightarrow x^2 = 1 \Rightarrow x = 1

Combining like terms gives

10x^2 = 10

, which simplifies to

x^2 = 1

. Since the intersection point is in the first quadrant,

x

must be positive.

3

Substitute the value of

x

back into the linear equation to find

y

.

y = 3(1) = 3

Using the value of

x

in the linear equation determines the corresponding

y

-coordinate.

4

Calculate the sum of the coordinates

x + y

.

1 + 3 = 4

The question asks for the value of the sum of the coordinates of the intersection point.

Key Concept

Solving a system of linear and non-linear (circular) equations using substitution.

Question 6Question →

A line and a parabola intersect at a point $(x, y)$ in the first quadrant of the standard $(x, y)$ coordinate plane. If the equation of the line is $y = x + 1$ and the equation of the parabola is $y = (x - 1)^2$ , what is the value of $x + y$ ?

3

5

1

7

9

Show answer & explanation

Answer: 7

Answer

7

To find the intersection point, we set the two equations equal to each other:

(x - 1)^2 = x + 1

. Expanding the left side gives

x^2 - 2x + 1 = x + 1

. Subtracting

x + 1

from both sides yields

x^2 - 3x = 0

. Factoring gives

x(x - 3) = 0

, so

x = 0

or

x = 3

. The first quadrant requires positive coordinates, so we choose

x = 3

. Substituting this back into either equation gives

y = 3 + 1 = 4

. Thus, the intersection point is

(3, 4)

, and the sum of the coordinates is

3 + 4 = 7

.

Step-by-Step Solution

1

Set the two equations equal to find the x-coordinates of the intersection points.

(x - 1)^2 = x + 1

Since both equations are equal to

y

, their right-hand sides must be equal at the points of intersection.

2

Expand the quadratic expression and simplify the equation.

x^2 - 2x + 1 = x + 1 \implies x^2 - 3x = 0

Expanding the binomial

(x-1)^2

yields

x^2 - 2x + 1

. Subtracting

x

and

1

from both sides simplifies the equation to standard quadratic form.

3

Factor the quadratic equation to solve for

x

.

x(x - 3) = 0 \implies x = 0

or

x = 3

Factoring out the greatest common factor

x

allows us to find the roots of the equation.

4

Determine which solution lies in the first quadrant and calculate the corresponding y-coordinate.

x = 3 \implies y = 3 + 1 = 4

The first quadrant requires both coordinates to be strictly positive. The solution

x = 0

gives

(0, 1)

, which is on the y-axis. Thus, we must use

x = 3

, which gives

(3, 4)

.

5

Calculate the sum

x + y

.

3 + 4 = 7

The question asks for the value of

x + y

for the first-quadrant intersection point.

Key Concept

Solving systems of linear and non-linear equations by substitution, factoring quadratic equations, and applying coordinate plane quadrant constraints.

Estimated Time:1m 0s

Question 7Question →

Consider the system of equations below:

\begin{aligned} x^2 + y^2 &= 9 \\ y &= x - 3 \end{aligned}

If $(x, y)$ is a solution to the system such that $x > 0$ , what is the value of $x + y$ ?

3

-3

0

6

9

Show answer & explanation

Answer: 3

Answer

3

The correct answer is

3

. Substituting

y = x - 3

into

x^2 + y^2 = 9

gives

x^2 + (x - 3)^2 = 9

. Expanding this yields

2x^2 - 6x = 0

, which factors as

2x(x - 3) = 0

. Since

x > 0

, the

x

-coordinate must be

3

. Substituting

x = 3

back into the linear equation gives

y = 3 - 3 = 0

. Therefore,

x + y = 3 + 0 = 3

.

Step-by-Step Solution

1

Substitute the linear equation into the quadratic equation.

x^2 + (x - 3)^2 = 9

This reduces the system of two variables to a single quadratic equation in terms of

x

.

2

Expand the squared binomial and simplify the equation.

x^2 + (x^2 - 6x + 9) = 9 \implies 2x^2 - 6x = 0

Expanding

(x - 3)^2

yields

x^2 - 6x + 9

. Subtracting

9

from both sides simplifies the equation.

3

Factor the quadratic equation to solve for

x

.

2x(x - 3) = 0 \implies x = 0

or

x = 3

Factoring out the greatest common factor

2x

reveals the roots of the quadratic equation.

4

Apply the constraint

x > 0

to find the correct

x

-coordinate.

x = 3

The problem specifies that the solution

(x, y)

must have

x > 0

.

5

Find the corresponding

y

-coordinate and calculate

x + y

.

y = 3 - 3 = 0 \implies x + y = 3 + 0 = 3

Substituting

x = 3

into the linear equation gives

y = 0

, and summing them gives the required value.

Key Concept

Solving systems of linear and quadratic equations by substitution

Question 8Question →

A system of equations consists of a parabola with the equation $y = x^2$ and a line with the equation $y = 3x - 2$ . The line and the parabola intersect at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the product of the $y$ -coordinates, $y_1 \cdot y_2$ , of these two points?

Show answer & explanation

Answer: 4

Answer

The product of the y-coordinates of the intersection points is 4.

Solving the system of equations by setting

x^2 = 3x - 2

leads to the quadratic equation

x^2 - 3x + 2 = 0

, which factors as

(x - 1)(x - 2) = 0

. The x-coordinates of the intersection points are

x = 1

and

x = 2

. Substituting these into

y = x^2

gives the corresponding y-coordinates as

y = 1^2 = 1

and

y = 2^2 = 4

. The product of these y-coordinates is

1 \cdot 4 = 4

.

Step-by-Step Solution

1

Equate the two equations to find the x-coordinates of the intersection points.

x^2 = 3x - 2

Since both equations are solved for y, setting them equal to each other allows us to find the x-coordinates where they intersect.

2

Move all terms to one side to set the quadratic equation to zero.

x^2 - 3x + 2 = 0

This puts the equation in the standard form

ax^2 + bx + c = 0

, which can be solved by factoring.

3

Factor the quadratic equation.

(x - 1)(x - 2) = 0

, so

x_1 = 1

and

x_2 = 2

Factoring shows that the values of x that satisfy the equation are the roots of the quadratic.

4

Substitute the x-coordinates back into the equation of the parabola to find the y-coordinates.

For

x_1 = 1

,

y_1 = 1^2 = 1

. For

x_2 = 2

,

y_2 = 2^2 = 4

.

Substituting x into either of the original equations gives the corresponding y-coordinates of the intersection points.

5

Calculate the product of the y-coordinates.

1 \cdot 4 = 4

The question asks for the product of the y-coordinates, so we multiply

y_1

and

y_2

.

Key Concept

Solving systems of linear and quadratic equations by substitution.

Question 9Question →

In the standard $(x, y)$ coordinate plane, the line defined by the equation $3x - 4y = k$ is tangent to the circle defined by the equation $x^2 + y^2 - 2x - 4y = 4$ . If $k > 0$ , what is the value of $k$ ?

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10. Completing the square for the circle's equation gives

(x - 1)^2 + (y - 2)^2 = 9

, showing the center is

(1, 2)

and the radius is

3

. The distance from

(1, 2)

to the line

3x - 4y - k = 0

is

\frac{|3(1) - 4(2) - k|}{\sqrt{3^2 + (-4)^2}} = \frac{|k + 5|}{5}

. For tangency, this distance must equal the radius:

\frac{|k + 5|}{5} = 3

, which gives

|k + 5| = 15

. Solving this absolute value equation gives

k = 10

or

k = -20

. Since

k

must be positive, the value is 10.

Step-by-Step Solution

1

Complete the square for the circle's equation

x^2 + y^2 - 2x - 4y = 4

.

(x - 1)^2 + (y - 2)^2 = 9

, which represents a circle with center

(1, 2)

and radius

R = 3

.

To find the center and radius of the circle, which are needed to use the distance formula.

2

Express the distance

d

from the center

(1, 2)

to the line

3x - 4y - k = 0

using the formula

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

.

d = \frac{|3(1) - 4(2) - k|}{\sqrt{3^2 + (-4)^2}} = \frac{|-5 - k|}{5} = \frac{|k + 5|}{5}

.

A line is tangent to a circle if and only if the distance from the center of the circle to the line equals the radius.

3

Set the distance equal to the radius (

3

) and solve for

k

.

\frac{|k + 5|}{5} = 3 \Rightarrow |k + 5| = 15

, which yields

k + 5 = 15 \Rightarrow k = 10

, or

k + 5 = -15 \Rightarrow k = -20

.

To find the values of

k

that make the line tangent to the circle.

4

Select the positive value of

k

.

k = 10

.

The problem specifies that

k > 0

.

Key Concept

Systems of Linear and Non-Linear Equations

Estimated Time:2m 30s

Question 10Question →

A system of equations is given below:

\begin{aligned} y &= (x - 1)^2 - 6 \\ y &= x - 1 \end{aligned}

If $(x, y)$ is a solution to this system in the first quadrant, what is the value of $x + y$ ?

5

-3

7

4

9

Show answer & explanation

Answer: 7

Answer

The sum of the coordinates of the first quadrant solution is 7.

To solve the system, substitute the expression for

y

from the linear equation into the quadratic equation to get

x - 1 = (x - 1)^2 - 6

. Substituting

u = x - 1

yields

u^2 - u - 6 = 0

. Factoring the quadratic expression gives

(u - 3)(u + 2) = 0

, so

u = 3

or

u = -2

. Since

u = x - 1

, this means

x = 4

or

x = -1

. Evaluating both cases gives the intersection points

(4, 3)

and

(-1, -2)

. Only

(4, 3)

is in the first quadrant. Adding these coordinates together yields

4 + 3 = 7

.

Step-by-Step Solution

1

Substitute

y = x - 1

into the quadratic equation to set up an equation in terms of

x

.

x - 1 = (x - 1)^2 - 6

To find the

x

-coordinates of the intersection points.

2

Solve the equation for

x

by substituting

u = x - 1

.

u = u^2 - 6 \Rightarrow u^2 - u - 6 = 0 \Rightarrow (u - 3)(u + 2) = 0

. This gives

u = 3

or

u = -2

.

To find the values of the substituted variable

u

.

3

Find the corresponding

x

and

y

values for both cases.

Case 1:

x - 1 = 3 \Rightarrow x = 4

, which gives

y = 3

. Point is

(4, 3)

. Case 2:

x - 1 = -2 \Rightarrow x = -1

, which gives

y = -2

. Point is

(-1, -2)

.

To determine the full coordinates of all intersection points.

4

Identify the first quadrant solution and calculate the sum of its coordinates.

The first quadrant solution is

(4, 3)

. The sum of the coordinates is

4 + 3 = 7

.

To answer the question's requirement for the sum of coordinates in the first quadrant.

Key Concept

Solving systems of linear and quadratic equations by substitution and identifying quadrant-specific solutions.

Question 11Question →

The equation of a circle is $x^2 + y^2 = 25$ , and the equation of a line is $2x - y = 5$ . If the line intersects the circle at the points $P$ and $Q$ , what is the sum of the $y$ -coordinates of $P$ and $Q$ ?

-18

-6

-2

2

4

Show answer & explanation

Answer: -2

Answer

The sum of the

y

-coordinates is

-2

.

Substituting the linear equation into the circle equation yields a quadratic equation with roots

x = 0

and

x = 4

. Evaluating the linear equation at these values gives the

y

-coordinates

-5

and

3

. Adding these

y

-coordinates results in

-2

.

Step-by-Step Solution

1

Rearrange the linear equation to express

y

in terms of

x

.

y = 2x - 5

Expressing one variable in terms of the other allows for substitution into the quadratic circle equation.

2

Substitute

y = 2x - 5

into the circle equation

x^2 + y^2 = 25

and simplify.

x^2 + (2x - 5)^2 = 25 \implies x^2 + 4x^2 - 20x + 25 = 25 \implies 5x^2 - 20x = 0

This substitution reduces the system of equations to a single quadratic equation in terms of

x

.

3

Solve the quadratic equation

5x^2 - 20x = 0

for

x

.

5x(x - 4) = 0 \implies x_1 = 0

and

x_2 = 4

Finding the roots of this quadratic equation gives the

x

-coordinates of the intersection points.

4

Substitute the

x

-values back into the linear equation

y = 2x - 5

to find the corresponding

y

-coordinates.

For

x_1 = 0

,

y_1 = 2(0) - 5 = -5

. For

x_2 = 4

,

y_2 = 2(4) - 5 = 3

. The intersection points are

P(0, -5)

and

Q(4, 3)

.

This step determines the coordinates of the two points of intersection.

5

Calculate the sum of the

y

-coordinates of the points

P

and

Q

.

y_1 + y_2 = -5 + 3 = -2

This yields the final value requested by the question.

Key Concept

Solving a system of linear and circular equations by substitution

Estimated Time:2m 0s

Question 12Question →

A parabola is defined by the equation $y = 2x^2 - 5x + 1$ and a line is defined by the equation $y = x - 3$ . If the parabola and the line intersect at the points $(x_1, y_1)$ and $(x_2, y_2)$ , what is the value of $y_1 + y_2$ ?

Show answer & explanation

Answer: -3

Answer

The sum of the y-coordinates of the intersection points is -3.

By setting the two equations equal to each other, we obtain a quadratic equation in terms of

x

. Solving this equation gives the x-coordinates of the intersection points. Substituting these x-values back into the linear equation yields the corresponding y-coordinates. Summing these y-coordinates gives the final value of

-3

.

Step-by-Step Solution

1

Equate the equations for the line and the parabola to find the x-coordinates of their intersection points.

2x^2 - 5x + 1 = x - 3

Since both equations are equal to

y

, their right-hand sides must be equal at the points of intersection.

2

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

2x^2 - 6x + 4 = 0

Grouping all terms on one side of the equation allows us to solve for $x.

3

Divide the entire equation by 2 to simplify.

x^2 - 3x + 2 = 0

Simplifying the quadratic equation makes factoring easier.

4

Factor the quadratic equation to solve for

x

.

(x - 1)(x - 2) = 0

, yielding

x = 1

and

x = 2

.

Finding the roots of the quadratic equation gives the x-coordinates of the intersection points.

5

Substitute each x-coordinate back into the linear equation

y = x - 3

to find the corresponding y-coordinates.

For

x_1 = 1

:

y_1 = 1 - 3 = -2

. For

x_2 = 2

:

y_2 = 2 - 3 = -1

.

The intersection points must satisfy both equations in the system.

6

Calculate the sum of the two y-coordinates,

y_1 + y_2

.

y_1 + y_2 = -2 + (-1) = -3

The question asks for the value of the sum of the y-coordinates.

Key Concept

Solving systems of linear and quadratic equations by substitution

Question 13Question →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 = 17$ and a line is defined by the equation $y = x - 3$ . If $(x, y)$ represents the intersection point of the circle and the line that lies in the first quadrant, what is the value of $x + y$ ?

1

3

5

7

-5

Show answer & explanation

Answer: 5

Answer

The sum of the coordinates of the first-quadrant intersection point is 5.

The correct answer is the sum of the coordinates of the first-quadrant intersection point. By substituting

y = x - 3

into the circle's equation, we get

x^2 + (x - 3)^2 = 17

. Expanding the squared term gives

x^2 + x^2 - 6x + 9 = 17

, which simplifies to

2x^2 - 6x - 8 = 0

. Dividing the entire equation by 2 yields

x^2 - 3x - 4 = 0

. Factoring this quadratic equation gives

(x - 4)(x + 1) = 0

, which has solutions

x = 4

and

x = -1

. Because the intersection point must lie in the first quadrant, the

x

-coordinate must be positive, so we choose

x = 4

. Substituting

x = 4

back into the linear equation gives

y = 4 - 3 = 1

. The sum of these coordinates is

4 + 1 = 5

.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the circle's equation.

x^2 + (x - 3)^2 = 17

This eliminates

y

so that we can solve for

x

.

2

Expand the squared binomial and combine like terms to write the equation in standard quadratic form.

2x^2 - 6x - 8 = 0

Expanding

(x - 3)^2

gives

x^2 - 6x + 9

, and adding

x^2

and subtracting 17 from both sides gives the quadratic equation.

3

Simplify the quadratic equation by dividing all terms by 2, then factor the resulting quadratic expression.

(x - 4)(x + 1) = 0

Dividing by 2 gives

x^2 - 3x - 4 = 0

, which factors into

(x - 4)(x + 1) = 0

since

(-4) \times 1 = -4

and

-4 + 1 = -3

.

4

Solve for

x

and select the positive solution since the point lies in the first quadrant.

x = 4

The solutions are

x = 4

and

x = -1

. In the first quadrant, both coordinates must be positive, so we choose

x = 4

.

5

Substitute

x = 4

back into the linear equation to find the corresponding

y

-coordinate.

y = 1

Using

y = x - 3

with

x = 4

gives

y = 4 - 3 = 1

.

6

Calculate the sum of the coordinates

x + y

.

4 + 1 = 5

The question asks for the value of

x + y

.

Key Concept

Solving a system consisting of a linear equation and a quadratic circle equation by substitution, factoring the resulting quadratic equation, and applying quadrant constraints.

Question 14Question →

A line intersects a parabola at two distinct points in the standard $(x, y)$ coordinate plane. The system of equations representing these curves is given by:

\begin{aligned} x^2 - 7x + 3y &= 6 \\ 3y - 4x &= 6 \end{aligned}

What is the distance between the two intersection points?

Show answer & explanation

Answer: 5

Answer

5

Solving the system of equations yields the intersection points

(0, 2)

and

(3, 6)

. The distance between these points is computed using the distance formula:

\sqrt{(3 - 0)^2 + (6 - 2)^2} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Isolate the

3y

term in the linear equation.

3y = 4x + 6

This allows for a direct substitution into the quadratic equation without introducing fractional coefficients.

2

Substitute

3y

into the first equation and simplify.

x^2 - 3x = 0

Substituting

4x + 6

for

3y

yields

x^2 - 7x + (4x + 6) = 6

. Subtracting 6 from both sides and combining like terms simplifies this to a basic quadratic equation.

3

Solve the quadratic equation for

x

and determine the corresponding

y

-coordinates.

The intersection points are

(0, 2)

and

(3, 6)

.

Factoring gives

x(x - 3) = 0

, so the

x

-coordinates are

0

and

3

. Substituting

x = 0

into the linear relation gives

3y = 6 \implies y = 2

. Substituting

x = 3

gives

3y = 18 \implies y = 6

.

4

Calculate the distance between the two coordinates.

5

Using the distance formula:

d = \sqrt{(3 - 0)^2 + (6 - 2)^2} = \sqrt{9 + 16} = 5

.

Key Concept

Systems of Linear and Non-Linear Equations

Question 15Question →

A system of equations consists of the circle defined by $x^2 + (y - 4)^2 = 10$ and the line defined by $y = 2x - 1$ . The two points of intersection of this system and the origin, $(0, 0)$ , form the vertices of a triangle in the standard $(x, y)$ coordinate plane. What is the area of this triangle?

2

\sqrt{7}

1

4

5

Show answer & explanation

Answer: 1

Answer

The correct area of the triangle is 1.

The correct answer of 1 is found by substituting the linear equation into the circle's equation, solving the resulting quadratic equation to find the coordinates of the two intersection points,

(1, 1)

and

(3, 5)

, and then applying the coordinate area formula for a triangle with a vertex at the origin.

Step-by-Step Solution

1

Substitute the linear equation into the circle's equation to form a quadratic equation in terms of

x

.

x^2 + (2x - 5)^2 = 10

This allows us to solve for the x-coordinates of the intersection points by replacing

y

with

2x - 1

in

(y - 4)^2

to get

(2x - 5)^2

.

2

Expand the squared binomial and simplify the quadratic equation.

5x^2 - 20x + 15 = 0

Expanding

(2x - 5)^2

yields

4x^2 - 20x + 25

. Adding

x^2

and subtracting

10

from both sides simplifies the equation to standard quadratic form.

3

Divide the quadratic equation by 5 and solve for

x

by factoring.

x = 1

and

x = 3

Dividing the equation by 5 yields

x^2 - 4x + 3 = 0

, which factors as

(x - 1)(x - 3) = 0

.

4

Find the corresponding y-coordinates by substituting the

x

-values back into the linear equation

y = 2x - 1

.

The intersection points are

(1, 1)

and

(3, 5)

.

For

x = 1

,

y = 2(1) - 1 = 1

. For

x = 3

,

y = 2(3) - 1 = 5

.

5

Calculate the area of the triangle with vertices at the origin

(0, 0)

and the two intersection points

(1, 1)

and

(3, 5)

.

Area = 1

Using the coordinate area formula for a triangle with one vertex at the origin,

\text{Area} = \frac{1}{2} |x_1 y_2 - x_2 y_1| = \frac{1}{2} |1(5) - 3(1)| = \frac{1}{2} |2| = 1

.

Key Concept

Solving a system of linear and non-linear equations and using coordinate geometry to find the area of a triangle.

Alternative Method

Instead of using the coordinate area formula, we can find the distance between the two intersection points to serve as the base of the triangle (

b = \sqrt{(3-1)^2 + (5-1)^2} = 2\sqrt{5}

), and find the perpendicular distance from the origin to the line

2x - y - 1 = 0

to serve as the height (

h = \frac{|-1|}{\sqrt{2^2 + (-1)^2}} = \frac{1}{\sqrt{5}}

). The area is then

\frac{1}{2} \times 2\sqrt{5} \times \frac{1}{\sqrt{5}} = 1

.

Estimated Time:2m 30s

Question 16Question →

A line with a positive slope passes through the point $(0, -5)$ and is tangent to the parabola $y = x^2 - 6x + 11$ . What is the slope of this line?

Show answer & explanation

Answer: 2

Answer

The slope of the line is 2.

The correct slope is 2. Representing the line as

y = mx - 5

and setting it equal to the parabola

y = x^2 - 6x + 11

results in the quadratic equation

x^2 - (6 + m)x + 16 = 0

. For the line to be tangent, this equation must have exactly one real solution, meaning its discriminant must equal zero:

(6 + m)^2 - 4(1)(16) = 0

. Solving this gives

6 + m = 8

or

6 + m = -8

, which results in

m = 2

or

m = -14

. Since the problem specifies that the slope is positive, the value of

m

must be 2.

Step-by-Step Solution

1

Write the equation of the line in slope-intercept form.

y = mx - 5

, where

m > 0

.

The line passes through

(0, -5)

, which represents the y-intercept of the line.

2

Equate the line and the parabola to set up the system of equations.

x^2 - 6x + 11 = mx - 5

Setting the two expressions equal allows us to find the x-coordinates of any intersection points.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - (6 + m)x + 16 = 0

Writing the equation in standard form identifies the coefficients

a = 1

,

b = -(6 + m)

, and

c = 16

needed for the discriminant.

4

Apply the condition for tangency by setting the discriminant to zero.

b^2 - 4ac = (-(6 + m))^2 - 4(1)(16) = 0 \implies (6 + m)^2 - 64 = 0

A line is tangent to a parabola if they touch at exactly one point, meaning the quadratic equation has exactly one real root (discriminant equals zero).

5

Solve the quadratic equation for the slope

m

and filter for the positive value.

(6+m)^2 = 64 \implies 6+m = \pm 8

, yielding

m = 2

or

m = -14

. Since the slope is positive,

m = 2

.

Solving the equation gives two possible slope values for tangent lines, and we select the positive slope as specified in the problem statement.

Key Concept

Determining tangency between a linear and quadratic equation by setting the discriminant of the intersection equation to zero.

Question 17Question →

A system of equations consists of the line $2x - y = 5$ and the parabola $y = x^2 - 4x + c$ , where $c$ is a constant. If the line and the parabola intersect at two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ such that the positive difference between their $x$ -coordinates is 2, what is the value of $c$ ?

0

3

5

8

11

Show answer & explanation

Answer: 3

Answer

3

To find the constant

c

, we equate the line and the parabola equations:

x^2 - 4x + c = 2x - 5

. Bringing all terms to one side gives the quadratic equation

x^2 - 6x + (c + 5) = 0

. Applying the quadratic formula, the

x

-coordinates of the intersection points are

x = 3 \pm \sqrt{4 - c}

. The positive difference between these coordinates is

(3 + \sqrt{4 - c}) - (3 - \sqrt{4 - c}) = 2\sqrt{4 - c}

. Setting this difference equal to the given value of 2 gives

2\sqrt{4 - c} = 2

, which simplifies to

\sqrt{4 - c} = 1

. Squaring both sides yields

4 - c = 1

, which gives

c = 3

. This corresponds to the correct option.

Step-by-Step Solution

1

Equate the equations for the line and the parabola to set up the equation for their intersection points.

x^2 - 4x + c = 2x - 5

The intersection points of the system of equations occur where the

y

-values are equal.

2

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 - 6x + (c + 5) = 0

Grouping the terms allows us to identify the coefficients:

a = 1

,

b = -6

, and the constant term is

c + 5

.

3

Apply the quadratic formula to find the

x

-coordinates of the intersection points in terms of

c

.

x = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(c + 5)}}{2} = 3 \pm \sqrt{4 - c}

The quadratic formula yields the roots

x_1 = 3 + \sqrt{4 - c}

and

x_2 = 3 - \sqrt{4 - c}

.

4

Set up an equation representing the positive difference between the

x

-coordinates and solve for

c

.

(3 + \sqrt{4 - c}) - (3 - \sqrt{4 - c}) = 2 \implies 2\sqrt{4 - c} = 2 \implies \sqrt{4 - c} = 1 \implies 4 - c = 1 \implies c = 3

The difference between the two coordinates is given as 2, which allows us to isolate and solve for

c

.

Key Concept

Solving systems of linear and non-linear equations by finding the intersection of a line and a parabola and using root properties to determine unknown constants

Alternative Method

Instead of using the quadratic formula, you can apply Vieta's formulas. Let the roots of

x^2 - 6x + (c + 5) = 0

be

x_1

and

x_2

. Vieta's formulas state that

x_1 + x_2 = 6

and

x_1 x_2 = c + 5

. We are given that

|x_1 - x_2| = 2

. Squaring this equation gives

(x_1 - x_2)^2 = 4

. Since

(x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1 x_2

, we substitute the known values:

6^2 - 4(c + 5) = 4 \implies 36 - 4c - 20 = 4 \implies 16 - 4c = 4 \implies 4c = 12 \implies c = 3

.

Estimated Time:2m 0s

Question 18Question →

A line is defined by the equation $y = 3x + k$ , where $k$ is a constant. This line intersects the parabola $y = x^2 - x + 2$ at two distinct points, $P$ and $Q$ . If the midpoint of the line segment $PQ$ lies on the line $y = 2x + 7$ , what is the value of $k$ ?

Show answer & explanation

Answer: 5

Answer

The value of

k

is

5

.

Equating the equations of the line and the parabola gives a quadratic equation

x^2 - 4x + (2 - k) = 0

. The average of the roots of this quadratic equation gives the

x

-coordinate of the midpoint,

x_m = 2

. Substituting this into the first line's equation gives the

y

-coordinate of the midpoint,

y_m = 6 + k

. Since the midpoint

(2, 6+k)

lies on the line

y = 2x + 7

, we substitute these coordinates to get

6 + k = 11

, which simplifies to

k = 5

.

Step-by-Step Solution

1

Equate the equations of the line and the parabola.

x^2 - 4x + (2 - k) = 0

To find the

x

-coordinates of the intersection points

P

and

Q

.

2

Determine the sum of the

x

-coordinates and find the midpoint's

x

-coordinate.

x_m = 2

By Vieta's formulas, the sum of the roots of the quadratic equation is

4

. The

x

-coordinate of the midpoint is the average of these roots:

4 / 2 = 2

.

3

Find the

y

-coordinate of the midpoint.

y_m = 6 + k

Because the midpoint lies on the line segment

PQ

, its coordinates must satisfy the equation of the line passing through

P

and

Q

, which is

y = 3x + k

.

4

Substitute the midpoint coordinates into the second line's equation and solve for

k

.

k = 5

We are given that the midpoint lies on the line

y = 2x + 7

.

Key Concept

Systems of Linear and Quadratic Equations and Midpoint Properties

Question 19Question →

In the standard $(x, y)$ coordinate plane, a parabola is defined by the equation $y = x^2 + 2x + 7$ and a line is defined by the equation $y = mx + 3$ , where $m$ is a constant. If the system of equations consisting of this parabola and line has exactly one real solution, and this solution lies in the first quadrant, what is the value of $m$ ?

-2

-6

6

2\sqrt{3}

2 + 2\sqrt{7}

Show answer & explanation

Answer: 6

Answer

6

The correct answer is

6

. When setting the equations of the parabola and line equal, we get

x^2 + (2-m)x + 4 = 0

. Setting the discriminant to zero yields

(2-m)^2 - 16 = 0

, which gives

m = 6

or

m = -2

. Substituting

m = 6

back gives a single intersection point of

(2, 15)

, which is in the first quadrant since both coordinates are positive. The other value,

m = -2

, gives an intersection point of

(-2, 7)

, which is in the second quadrant.

Step-by-Step Solution

1

Set the equations of the parabola and the line equal to each other to find their intersection points.

x^2 + 2x + 7 = mx + 3

To find the coordinates where the two graphs intersect.

2

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 + (2 - m)x + 4 = 0

To express the intersection condition as a single quadratic equation where we can analyze the number of solutions.

3

Set the discriminant of the quadratic equation to zero.

\Delta = (2 - m)^2 - 4(1)(4) = 0 \implies (2 - m)^2 - 16 = 0

For the system to have exactly one real solution, the quadratic equation must have a discriminant of zero (tangency).

4

Solve for the possible values of

m

.

(2 - m)^2 = 16 \implies 2 - m = \pm 4

, giving

m = -2

or

m = 6

.

To find all values of

m

that result in exactly one intersection point.

5

Find the intersection point for each value of

m

and determine which lies in the first quadrant.

For

m = -2

,

(x+2)^2 = 0 \implies x = -2

and

y = -2(-2)+3 = 7

, yielding

(-2, 7)

(Quadrant II). For

m = 6

,

(x-2)^2 = 0 \implies x = 2

and

y = 6(2)+3 = 15

, yielding

(2, 15)

(Quadrant I).

To satisfy the condition that the single solution must lie in the first quadrant (where both

x > 0

and

y > 0

).

Key Concept

Solving systems of linear and quadratic equations and applying the discriminant to find conditions for tangency.

Estimated Time:2m 0s

Question 20Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 1)^2 + (y - 2)^2 = 10$ , and a line is defined by the equation $y = 3x - 1$ . The line intersects the circle at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the sum of the $y$ -coordinates of these two points of intersection?

Show answer & explanation

Answer: 4

Answer

The sum of the

y

-coordinates of the intersection points is 4.

Substituting

y = 3x - 1

into the circle's equation gives

(x-1)^2 + (3x-3)^2 = 10

. Factoring out 3 from the second term yields

(x-1)^2 + 9(x-1)^2 = 10

, which simplifies to

10(x-1)^2 = 10

, or

(x-1)^2 = 1

. Solving for

x

gives

x = 0

and

x = 2

. Substituting these values into the linear equation gives the corresponding

y

-coordinates:

y = -1

when

x = 0

, and

y = 5

when

x = 2

. The sum of these

y

-coordinates is

5 + (-1) = 4

.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the circle's equation.

(x-1)^2 + (3x-3)^2 = 10

To reduce the system of two equations to a single equation in terms of

x

.

2

Factor out 3 from the term

(3x-3)

and simplify the equation.

10(x-1)^2 = 10

, which simplifies to

(x-1)^2 = 1

To solve for the

x

-coordinates of the intersection points.

3

Solve the simplified quadratic equation for

x

.

x_1 = 2

and

x_2 = 0

To find the

x

-coordinates of the two intersection points.

4

Substitute the

x

-values back into the linear equation

y = 3x - 1

to find the corresponding

y

-coordinates.

y_1 = 3(2) - 1 = 5

and

y_2 = 3(0) - 1 = -1

To determine the

y

-coordinates of the intersection points

(2, 5)

and

(0, -1)

.

5

Add the two

y

-coordinates together.

5 + (-1) = 4

To find the sum of the

y

-coordinates as requested by the question.

Key Concept

Solving systems of linear and circular equations by substitution

Alternative Method

Instead of finding the individual coordinates, substitute

y_1 = 3x_1 - 1

and

y_2 = 3x_2 - 1

to write the sum as

y_1 + y_2 = 3(x_1 + x_2) - 2

. Expanding the substitution equation gives

10x^2 - 20x = 0

. By Vieta's formulas, the sum of the roots

x_1 + x_2 = -(-20)/10 = 2

. Substituting this back gives

y_1 + y_2 = 3(2) - 2 = 4

.

Estimated Time:1m 30s

Systems of Linear and Non-Linear Equations

Answer

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