Nonlinear Systems of Equations Practice Questions — SAT — Page 2

Question 21Question →

In the $xy$ -plane, a circle has center $(6, 8)$ and is tangent to the $y$ -axis. A line with a positive slope $m$ passes through the origin and is tangent to the circle. If $m$ is written as a fraction in simplest form, $\frac{p}{q}$ , what is the value of $p + q$ ?

Show answer & explanation

Answer: 31

Answer

The value of

p + q

is 31.

The radius of the circle is 6 since the center is

(6, 8)

and it is tangent to the

y

-axis. Substituting the line

y = mx

into the circle's equation

(x-6)^2 + (y-8)^2 = 36

yields a quadratic equation in

x

:

(1+m^2)x^2 - (12+16m)x + 64 = 0

. Since the line is tangent to the circle, there is exactly one solution, meaning the discriminant of this quadratic must equal zero:

(12+16m)^2 - 4(1+m^2)(64) = 0

. Simplifying this equation yields

24m - 7 = 0

, which gives a slope of

m = \frac{7}{24}

. Since the fraction is in simplest form,

p=7

and

q=24

, and their sum is

31

.

Step-by-Step Solution

1

Determine the radius of the circle.

r = 6

Since the center is at

(6, 8)

and the circle is tangent to the

y

-axis, the radius is the horizontal distance from the center to the

y

-axis.

2

Write the equation of the circle.

(x-6)^2 + (y-8)^2 = 36

A circle with center

(h, k)

and radius

r

has the equation

(x-h)^2 + (y-k)^2 = r^2

.

3

Substitute

y = mx

into the circle's equation.

(1+m^2)x^2 - (12+16m)x + 64 = 0

Substituting the equation of the line

y = mx

into the circle's equation and expanding gives a quadratic equation in terms of

x

.

4

Set the discriminant of the quadratic equation to zero.

\Delta = (12+16m)^2 - 256(1+m^2) = 0

For the line to be tangent to the circle, the system must have exactly one real solution, meaning the quadratic equation must have a discriminant of zero.

5

Solve for the positive slope

m

.

m = \frac{7}{24}

Expanding and simplifying the discriminant equation yields

24m - 7 = 0

, which gives

m = \frac{7}{24}

.

6

Calculate

p + q

using the simplest form fraction.

31

Since

m = \frac{7}{24}

is in simplest form,

p = 7

and

q = 24

. Therefore,

p + q = 7 + 24 = 31

.

Key Concept

Nonlinear Systems of Equations

Question 22Question →

For a real solution $(x, y)$ to the system of equations below, which of the following is a possible value of $x + y$ ?

\begin{aligned} x^2 + 3y^2 &= 21 \\ x^2 - 2y &= 5 \end{aligned}

-5

-3

1

5

Show answer & explanation

Answer: 5

Answer

The correct answer is 5, which is the sum of the coordinates of the real solution (3, 2) of the system.

Substituting

x^2 = 2y + 5

into the first equation yields the quadratic equation

3y^2 + 2y - 16 = 0

. Solving this quadratic equation gives

y = 2

or

y = -8/3

. Since

x^2

must be non-negative for real solutions,

y = -8/3

yields no real solutions because

x^2 = 2(-8/3) + 5 = -1/3 < 0

. For

y = 2

, we have

x^2 = 2(2) + 5 = 9

, which gives

x = 3

or

x = -3

. The real solutions to the system are

(3, 2)

and

(-3, 2)

. The possible values for

x + y

are

3 + 2 = 5

and

-3 + 2 = -1

. Therefore, 5 is a possible value of

x + y

.

Step-by-Step Solution

1

Rearrange the second equation to express

x^2

in terms of

y

.

x^2 = 2y + 5

This allows for direct substitution into the first equation to eliminate the

x

variable.

2

Substitute the expression for

x^2

into the first equation and simplify the resulting quadratic equation in terms of

y

.

3y^2 + 2y - 16 = 0

Substituting

2y + 5

for

x^2

in

x^2 + 3y^2 = 21

gives

(2y + 5) + 3y^2 = 21

, which simplifies to

3y^2 + 2y - 16 = 0

.

3

Factor or use the quadratic formula to solve for

y

.

y = 2

or

y = -8/3

The quadratic equation factors as

(3y + 8)(y - 2) = 0

, giving the roots

y = 2

and

y = -8/3

.

4

Find the corresponding values of

x

for each

y

and determine the real solutions.

For

y = 2

,

x^2 = 2(2) + 5 = 9 \implies x = \pm 3

. For

y = -8/3

,

x^2 = 2(-8/3) + 5 = -1/3

, which has no real solutions.

A real number squared cannot be negative, so only

y = 2

yields real solutions:

(3, 2)

and

(-3, 2)

.

5

Calculate the possible values of

x + y

for the real solutions.

For

(3, 2)

,

x + y = 3 + 2 = 5

. For

(-3, 2)

,

x + y = -3 + 2 = -1

.

Evaluating the target expression for both valid real solutions gives 5 and -1. Since 5 is the only value present in the options, it is the correct answer.

Key Concept

Solving systems of nonlinear equations algebraically using substitution and analyzing real solutions.

Question 23Question →

A system of two equations is given:

y = x^2 + kx + 4

y = 4x - k

In these equations, $k$ represents a positive constant. If the system has a single real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 12

Answer

12

Setting the two equations equal yields the quadratic equation

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

. For the system to have exactly one real solution, the discriminant of this quadratic equation must be equal to

0

. The discriminant is

(k - 4)^2 - 4(1)(k + 4) = k^2 - 12k

. Solving

k^2 - 12k = 0

yields

k = 0

or

k = 12

. Since

k

must be positive, the value of

k

is

12

.

Step-by-Step Solution

1

Set the two equations equal to each other to form a single quadratic equation in terms of

x

.

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

Equating the expressions for

y

allows us to find the

x

-coordinates where the graphs of the two equations intersect.

2

Set the discriminant of the quadratic equation to zero.

(k - 4)^2 - 4(1)(k + 4) = 0

A system of equations consisting of a line and a parabola has a single real solution if and only if the line is tangent to the parabola, which corresponds to a quadratic equation with a discriminant of zero.

3

Expand and simplify the equation for

k

.

k^2 - 12k = 0

Expanding

(k - 4)^2

yields

k^2 - 8k + 16

, and distributing

-4

yields

-4k - 16

. Combining like terms simplifies the relation.

4

Solve for

k

and apply the constraint that

k > 0

.

k = 12

Factoring

k(k - 12) = 0

gives

k = 0

or

k = 12

. Since

k

is specified to be positive,

k = 12

is the correct value.

Key Concept

Solving nonlinear systems of equations by substitution and using the discriminant to find conditions for a single real solution.

Question 24Question →

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

\begin{aligned} x^2 + y^2 &= 50 \\ x - y &= 6 \end{aligned}

Show answer & explanation

Answer: 8

Answer

8

Solving the linear equation for

x

gives

x = y + 6

. Substituting this into the circle equation yields

(y + 6)^2 + y^2 = 50

, which simplifies to

2y^2 + 12y - 14 = 0

. Dividing by

2

gives

y^2 + 6y - 7 = 0

, which factors as

(y + 7)(y - 1) = 0

. This yields

y = 1

(with

x = 7

) or

y = -7

(with

x = -1

). The constraint

x > 0

requires selecting the solution

(7, 1)

. The sum of

x

and

y

for this solution is

7 + 1 = 8

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation.

x = y + 6

This allows for substitution into the second equation to reduce the system to a single variable.

2

Substitute the expression for

x

into the equation of the circle and expand.

2y^2 + 12y - 14 = 0

Substituting

x = y + 6

into

x^2 + y^2 = 50

yields

(y + 6)^2 + y^2 = 50

. Expanding gives

y^2 + 12y + 36 + y^2 = 50

.

3

Simplify and solve the quadratic equation by factoring.

y = 1

or

y = -7

Dividing the equation by 2 gives

y^2 + 6y - 7 = 0

, which factors as

(y + 7)(y - 1) = 0

.

4

Calculate the corresponding values of

x

for each

y

.

For

y = 1

,

x = 7

; for

y = -7

,

x = -1

Using the relation

x = y + 6

yields the coordinates of the two intersection points.

5

Apply the constraint

x > 0

and calculate

x + y

.

x + y = 8

Since

x > 0

, the only valid solution is

(7, 1)

. The sum of the coordinates is

7 + 1 = 8

.

Key Concept

Solving a nonlinear system of equations representing a circle and a line using substitution and factoring.

Question 25Question →

In the $xy$ -plane, the graphs of the linear function $3x + y = 6$ and the quadratic function $y = 3x^2 - 5x - 2$ intersect at two points. What is the sum of the $y$ -coordinates of these two intersection points?

10

\frac{2}{3}

14

2

Show answer & explanation

Answer: 10

Answer

10

The correct answer is 10. By rewriting the linear equation as

y = 6 - 3x

and substituting it into the quadratic equation, we get

6 - 3x = 3x^2 - 5x - 2

. Collecting all terms on one side yields the quadratic equation

3x^2 - 2x - 8 = 0

. Factoring this equation gives

(3x + 4)(x - 2) = 0

, which results in

x = 2

and

x = -\frac{4}{3}

. Substituting these

x

-values back into the linear equation gives the

y

-coordinates:

y = 6 - 3(2) = 0

and

y = 6 - 3(-\frac{4}{3}) = 10

. The sum of these

y

-coordinates is

0 + 10 = 10

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation.

y = 6 - 3x

Isolating

y

makes substitution into the quadratic equation straightforward.

2

Substitute the expression for

y

into the quadratic equation and set the equation to zero.

6 - 3x = 3x^2 - 5x - 2

, which simplifies to

3x^2 - 2x - 8 = 0

.

This forms a single quadratic equation in terms of

x

to find the

x

-coordinates of the intersection points.

3

Solve the quadratic equation by factoring.

(3x + 4)(x - 2) = 0

, yielding

x = 2

and

x = -\frac{4}{3}

.

Factoring determines the values of

x

at the points of intersection.

4

Substitute the

x

-values back into the linear equation

y = 6 - 3x

to find the corresponding

y

-coordinates.

For

x = 2

:

y = 6 - 3(2) = 0

. For

x = -\frac{4}{3}

:

y = 6 - 3(-\frac{4}{3}) = 10

.

This identifies the coordinates of the two intersection points as

(2, 0)

and

(-\frac{4}{3}, 10)

.

5

Calculate the sum of the

y

-coordinates.

0 + 10 = 10

The question asks for the sum of the

y

-coordinates of the two intersection points.

Key Concept

Solving a system of linear and quadratic equations using substitution.

Estimated Time:1m 30s

Question 26Question →

In the $xy$ -plane, the graphs of the linear equation $y = 3x + k$ , where $k$ is a constant, and the quadratic equation $y = x^2 - 2x + 7$ intersect at two points. If the distance between the two points of intersection is $\sqrt{130}$ , what is the value of $k$ ?

1

-4

4

10

Show answer & explanation

Answer: 4

Answer

4

The correct value of

k

is

4

. Equating the two equations yields the quadratic equation

x^2 - 5x + (7 - k) = 0

. Since the slope of the line is

3

, the difference in the y-coordinates of the intersection points is

3

times the difference in the x-coordinates. Applying the distance formula shows that the distance is equal to

\sqrt{10(x_2 - x_1)^2}

. Setting this equal to the given distance of

\sqrt{130}

yields

(x_2 - x_1)^2 = 13

. Substituting Vieta's relations (

x_1 + x_2 = 5

and

x_1 x_2 = 7 - k

) into the identity

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

results in

13 = 25 - 4(7 - k)

, which simplifies to

k = 4

.

Step-by-Step Solution

1

Equate the linear and quadratic equations to set up the intersection condition.

x^2 - 2x + 7 = 3x + k \dots \implies x^2 - 5x + (7 - k) = 0

The x-coordinates of the intersection points are the roots of this quadratic equation.

2

Express the distance between the two points

(x_1, y_1)

and

(x_2, y_2)

in terms of the difference of their x-coordinates.

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + (3(x_2 - x_1))^2} = \sqrt{10(x_2 - x_1)^2}

Since both points lie on the line

y = 3x + k

, the difference in their y-coordinates is

y_2 - y_1 = 3(x_2 - x_1)

.

3

Use the given distance to find the value of

(x_2 - x_1)^2

.

10(x_2 - x_1)^2 = 130 \implies (x_2 - x_1)^2 = 13

The distance between the intersection points is given as

\sqrt{130}

.

4

Apply Vieta's formulas and the algebraic identity for the squared difference of the roots to solve for

k

.

x_1 + x_2 = 5

and

x_1 x_2 = 7 - k

. Substituting these into

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

gives

13 = 5^2 - 4(7 - k) \implies 13 = 25 - 28 + 4k \implies 4k = 16 \implies k = 4

.

Vieta's formulas relate the coefficients of the quadratic equation to the sum and product of its roots.

Key Concept

Solving systems of linear and quadratic equations and finding constants based on geometric constraints.

Question 27Question →

In the system of equations below, $k$ is a constant.

\begin{aligned} y &= 3x^2 - 12x + 7 \\ y &= kx - 5 \end{aligned}

If the system has two distinct real solutions, $(x_1, y_1)$ and $(x_2, y_2)$ , such that the average of the $x$ -coordinates of the solutions is $3$ , what is the value of $k$ ?

-6

-3

6

0

Show answer & explanation

Answer: 6

Answer

6

To find the value of

k

, we set the two equations equal to each other to solve for the

x

-coordinates of the intersection points:

3x^2 - 12x + 7 = kx - 5

. Moving all terms to one side gives the quadratic equation

3x^2 - (12+k)x + 12 = 0

. The sum of the roots

x_1 + x_2

is given by

-\frac{b}{a} = \frac{12+k}{3}

. The average of the roots is

\frac{x_1 + x_2}{2} = \frac{12+k}{6}

. Since the average of the

x

-coordinates is

3

, we set

\frac{12+k}{6} = 3

, which gives

12+k = 18

, and solving for

k

yields

6

. Evaluating the discriminant with

k = 6

gives

(-18)^2 - 4(3)(12) = 324 - 144 = 180 > 0

, confirming two distinct real solutions.

Step-by-Step Solution

1

Equate the two equations to find the

x

-coordinates of the intersection points.

3x^2 - 12x + 7 = kx - 5

The intersection points of the system occur where the

y

-values of the two functions are equal.

2

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

.

3x^2 - (12+k)x + 12 = 0

Grouping the like terms allows us to identify the quadratic coefficients

a = 3

,

b = -(12+k)

, and

c = 12

.

3

Express the sum and the average of the solutions in terms of

k

using Vieta's formulas.

Sum of solutions

x_1 + x_2 = -\frac{b}{a} = \frac{12+k}{3}

, so the average is

\frac{x_1 + x_2}{2} = \frac{12+k}{6}

.

Vieta's formulas state that the sum of the roots of a quadratic equation is

-\frac{b}{a}

. The average is half of the sum.

4

Set the average equal to

3

and solve for

k

.

\frac{12+k}{6} = 3 \implies 12+k = 18 \implies k = 6

.

The problem states that the average of the

x

-coordinates of the solutions is

3

.

5

Verify that

k = 6

yields two distinct real solutions by checking the discriminant.

For

k = 6

, the equation is

3x^2 - 18x + 12 = 0

. The discriminant is

D = (-18)^2 - 4(3)(12) = 324 - 144 = 180

. Since

D > 0

, two distinct real solutions exist.

A positive discriminant ensures that there are two distinct real solutions, satisfying the problem's condition.

Key Concept

Solving nonlinear systems of equations using quadratic properties and Vieta's formulas.

Question 28Question →

For a constant $k$ , the circle $(x - 5)^2 + (y - 5)^2 = 18$ and the line $y = kx$ are graphed in the $xy$ -plane. For how many integer values of $k$ will the circle and the line intersect at exactly two points?

5

6

7

8

Show answer & explanation

Answer: 6

Answer

6

The correct answer is 6. By substituting the line equation

y = kx

into the circle equation, we obtain the quadratic equation

(k^2 + 1)x^2 - 10(k + 1)x + 32 = 0

. For the circle and line to intersect at exactly two points, this equation must have two distinct real roots, meaning its discriminant must be positive. This leads to the inequality

7k^2 - 50k + 7 < 0

, which factors as

(7k - 1)(k - 7) < 0

. The solution is the interval

\frac{1}{7} < k < 7

. The integer values of

k

in this range are

1, 2, 3, 4, 5,

and

6

, giving a total of 6 integers.

Step-by-Step Solution

1

Substitute the line equation

y = kx

into the circle equation

(x - 5)^2 + (y - 5)^2 = 18

.

(x - 5)^2 + (kx - 5)^2 = 18

Substitution reduces the system of equations to a single quadratic equation in

x

, representing the

x

-coordinates of the intersection points.

2

Expand the terms and write the equation in standard quadratic form

Ax^2 + Bx + C = 0

.

(k^2 + 1)x^2 - 10(k + 1)x + 32 = 0

Putting the equation in standard form is necessary to analyze its discriminant.

3

Set the discriminant

\Delta = b^2 - 4ac

strictly greater than zero to ensure exactly two distinct real solutions.

[-10(k+1)]^2 - 4(k^2 + 1)(32) > 0 \Rightarrow 7k^2 - 50k + 7 < 0

A quadratic equation has two distinct real roots if and only if its discriminant is positive.

4

Solve the quadratic inequality

7k^2 - 50k + 7 < 0

by factoring the quadratic expression.

\frac{1}{7} < k < 7

The roots of

(7k - 1)(k - 7) = 0

are

k = \frac{1}{7}

and

k = 7

. Since the leading coefficient is positive, the expression is negative between these roots.

5

Identify and count the integer values of

k

that lie within the interval

(\frac{1}{7}, 7)

.

The integers are

1, 2, 3, 4, 5, 6

, which gives a total of 6 values.

The integers that are strictly greater than

\frac{1}{7}

and strictly less than

7

are the whole numbers from

1

to

6

.

Key Concept

Solving systems consisting of a circle and a line algebraically by utilizing the quadratic discriminant to find the condition for two real intersection points.

Question 29Question →

The system of equations below consists of a linear equation and a quadratic equation:

\begin{aligned} y - 2x &= 5 \\ y &= x^2 - 3x - 1 \end{aligned}

Let

(x_1, y_1)

and

(x_2, y_2)

represent the two distinct real coordinate points where the graphs of these equations intersect. If

y_1 > y_2

, what is the value of

x_1 - x_2

?

Show answer & explanation

Answer: 7

Answer

7

Solving the system algebraically by expressing

y

as

2x + 5

and substituting it into the quadratic equation leads to

x^2 - 5x - 6 = 0

. Factoring this equation yields

x = 6

and

x = -1

. Evaluating these values in the linear equation gives the coordinates

(6, 17)

and

(-1, 3)

. Since

17 > 3

, the point

(6, 17)

corresponds to

(x_1, y_1)

and

(-1, 3)

corresponds to

(x_2, y_2)

. The difference

x_1 - x_2

is equal to

6 - (-1) = 7

.

Step-by-Step Solution

1

Solve for

y

in the linear equation to express it in terms of

x

.

y = 2x + 5

This allows for substitution into the quadratic equation to eliminate one variable.

2

Substitute

2x + 5

for

y

in the quadratic equation.

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

To create a single-variable quadratic equation in terms of

x

.

3

Rearrange the equation by subtracting

2x

and

5

from both sides to set it equal to zero.

x^2 - 5x - 6 = 0

Setting the quadratic equation to zero is the standard first step to find its roots.

4

Factor the quadratic trinomial.

(x - 6)(x + 1) = 0

, which gives

x = 6

and

x = -1

.

Factoring finds the

x

-coordinates of the points of intersection.

5

Substitute the

x

-values back into the linear equation

y = 2x + 5

to determine the corresponding

y

-values.

For

x = 6

,

y = 2(6) + 5 = 17

. For

x = -1

,

y = 2(-1) + 5 = 3

. The intersection points are

(6, 17)

and

(-1, 3)

.

This defines the full coordinate pairs of the system's solutions.

6

Apply the condition

y_1 > y_2

to identify

(x_1, y_1)

and

(x_2, y_2)

.

Since

17 > 3

, the first solution is

(x_1, y_1) = (6, 17)

and the second solution is

(x_2, y_2) = (-1, 3)

.

This aligns the coordinate values with the variables defined in the problem constraint.

7

Calculate the value of

x_1 - x_2

.

6 - (-1) = 7

Subtracting a negative value is equivalent to adding its absolute value, yielding the final required answer.

Key Concept

Nonlinear Systems of Equations

Question 30Question →

In the system of equations below,

k

is a constant.

\begin{aligned} x^2 + y^2 + 8y &= 9 \\ x^2 + y &= k \end{aligned}

If the system has exactly 3 distinct real solutions

(x, y)

, what is the value of

k

?

Show answer & explanation

Answer: 1

Answer

The value of

k

is 1.

Substituting

x^2 = k - y

into

x^2 + y^2 + 8y = 9

gives

y^2 + 7y + k - 9 = 0

. For the system to have exactly 3 distinct real solutions, the vertex of the parabola

y = -x^2 + k

must lie on the circle, which corresponds to the root

y = k

. Substituting

y = k

into the quadratic equation yields

k^2 + 8k - 9 = 0

, which gives

k = 1

or

k = -9

. For

k = 1

, the roots of the quadratic are

y = 1

and

y = -8

. The root

y = 1

yields 1 real solution,

(0, 1)

, and the root

y = -8

yields 2 real solutions,

(3, -8)

and

(-3, -8)

, for a total of 3 real solutions. For

k = -9

, the roots are

y = -9

and

y = 2

. The root

y = 2

does not yield any real solutions for

x

because

2 > -9

, so the system has only 1 real solution. Thus,

k = 1

.

Step-by-Step Solution

1

Express

x^2

in terms of

y

and

k

from the second equation.

x^2 = k - y

(with constraint

y \le k

for real

x

)

To prepare for substitution into the first equation and establish the boundary condition for real solutions.

2

Substitute

x^2 = k - y

into the first equation.

y^2 + 7y + k - 9 = 0

To create a single quadratic equation in terms of

y

.

3

Analyze the conditions on the roots of the quadratic equation to get exactly 3 distinct real solutions.

One root must equal

k

and the other must be less than

k

.

A root

y = k

yields 1 real solution for

x

(

x = 0

), while a root

y < k

yields 2 real solutions (

x = \pm\sqrt{k-y}

).

4

Find the candidate values of

k

by setting

y = k

in the quadratic equation.

k^2 + 8k - 9 = 0 \implies k = 1

or

k = -9

To find the values of

k

where the parabola's vertex lies on the circle.

5

Verify which candidate value of

k

satisfies all conditions.

For

k = 1

, the roots are

y = 1

and

y = -8 < 1

(3 solutions). For

k = -9

, the roots are

y = -9

and

y = 2 > -9

(1 solution). Therefore,

k = 1

.

To ensure the second root is strictly less than

k

, guaranteeing exactly 3 solutions.

Key Concept

Analyzing the number of solutions in a nonlinear system of equations using algebraic substitution and boundary constraints.

Estimated Time:3m 0s

Question 31Question →

In the $xy$ -plane, the graph of the linear equation $3x + 4y = k$ , where $k$ is a positive constant, is tangent to the circle with equation $x^2 + y^2 - 2x - 4y = 20$ . What is the value of $k$ ?

Show answer & explanation

Answer: 36

Answer

The value of the constant

k

is

36

.

To find the value of

k

where the line

3x + 4y = k

is tangent to the circle, we first rewrite the circle's equation

x^2 + y^2 - 2x - 4y = 20

in standard form by completing the square:

(x^2 - 2x + 1) + (y^2 - 4y + 4) = 20 + 1 + 4

, which simplifies to

(x - 1)^2 + (y - 2)^2 = 25

. This is a circle centered at

(1, 2)

with a radius of

5

. A line is tangent to a circle if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. Using the distance formula

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

for the point

(1, 2)

and the line

3x + 4y - k = 0

, we set up the equation:

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

. This simplifies to

\frac{|11 - k|}{5} = 5

, or

|11 - k| = 25

. Solving the absolute value equation gives

11 - k = 25 \Rightarrow k = -14

and

11 - k = -25 \Rightarrow k = 36

. Since the problem specifies that

k

is a positive constant, the correct answer is

36

.

Step-by-Step Solution

1

Rewrite the circle's equation by completing the square for both variables.

(x - 1)^2 + (y - 2)^2 = 25

To identify the center and radius of the circle.

2

Identify the center and radius from the standard form of the circle's equation.

Center is

(1, 2)

and radius is

5

.

The standard form of a circle is

(x-h)^2 + (y-k)^2 = r^2

.

3

Apply the condition for tangency: the perpendicular distance from the center

(1, 2)

to the line

3x + 4y - k = 0

must equal the radius

5

.

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

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

4

Simplify the distance equation and solve the resulting absolute value equation for

k

.

|11 - k| = 25

, yielding

k = -14

or

k = 36

.

To find all mathematically possible values of the constant

k

.

5

Select the positive value for

k

as specified by the problem constraints.

k = 36

The problem states that

k

is a positive constant.

Key Concept

The relationship between a line and a circle in a nonlinear system, specifically using the distance from the center to a tangent line to solve for an unknown constant.

Question 32Question →

A circle in the $xy$ -plane is defined by the equation $(x - 3)^2 + (y - 2)^2 = 13$ . The line $y = 3x$ intersects the circle at the origin $(0, 0)$ and at a second point $P$ . What is the $y$ -coordinate of point $P$ ?

\frac{9}{5}

\frac{11}{5}

\frac{27}{5}

-\frac{9}{5}

Show answer & explanation

Answer: $\frac{27}{5}$

Answer

The y-coordinate of point P is

\frac{27}{5}

.

Substituting the linear relationship

y = 3x

into the circle equation yields a quadratic equation in terms of

x

. Solving this equation gives the

x

-coordinates of the intersection points as

x = 0

and

x = \frac{9}{5}

. Since point

P

is not the origin, its

x

-coordinate is

\frac{9}{5}

. Substituting this back into

y = 3x

gives the

y

-coordinate of point

P

as

\frac{27}{5}

.

Step-by-Step Solution

1

Substitute the linear equation

y = 3x

into the equation of the circle.

(x - 3)^2 + (3x - 2)^2 = 13

Substitution reduces the system of two equations with two variables to a single quadratic equation with one variable.

2

Expand and simplify the equation.

10x^2 - 18x = 0

Expanding

(x - 3)^2

to

x^2 - 6x + 9

and

(3x - 2)^2

to

9x^2 - 12x + 4

, then combining like terms and subtracting 13 from both sides, isolates the quadratic terms.

3

Solve for

x

by factoring.

2x(5x - 9) = 0

, which gives

x = 0

or

x = \frac{9}{5}

.

The roots of the factored equation represent the x-coordinates of the two intersection points.

4

Find the y-coordinate of the second point

P

using

y = 3x

.

y = 3 \left(\frac{9}{5}\right) = \frac{27}{5}

Since point

P

is not the origin, its x-coordinate must be

\frac{9}{5}

. Substituting this into the linear equation gives its corresponding y-coordinate.

Key Concept

Solving nonlinear systems of equations involving a circle and a line by substitution.

Question 33Question →

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

If $(x, y)$ is a solution to the system of equations above and $x < 0$ , what is the value of $x + y$ ?

18

-3

-2

0

Show answer & explanation

Answer: -3

Answer

The correct answer is -3.

The correct answer is

-3

. To find this, substitute the expression for

y

from the first equation into the second equation, which yields

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

. Distributing the negative sign and combining like terms gives

-x^2 + 3x + 7 = -3

. Setting the equation to zero results in

x^2 - 3x - 10 = 0

. Factoring the quadratic expression gives

(x - 5)(x + 2) = 0

, which means

x = 5

or

x = -2

. Since the question specifies that

x < 0

, we select

x = -2

. Substituting

x = -2

back into the linear equation gives

y = 2(-2) + 3 = -1

. Finally, calculating

x + y

gives

-2 + (-1) = -3

.

Step-by-Step Solution

1

Substitute the expression for

y

from the first equation into the second equation.

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

This allows us to eliminate

y

and solve for

x

in a single variable quadratic equation.

2

Distribute the negative sign and simplify the equation.

-x^2 + 3x + 7 = -3

Expanding the parentheses correctly preserves the signs of the quadratic terms.

3

Rearrange the terms to set the quadratic equation to zero.

x^2 - 3x - 10 = 0

Writing the equation in standard form (

ax^2 + bx + c = 0

) allows us to factor it.

4

Factor the quadratic expression.

(x - 5)(x + 2) = 0

Finding factors that multiply to

-10

and add to

-3

gives us the solutions for

x

.

5

Solve for

x

and apply the constraint

x < 0

.

x = -2

The equation yields

x = 5

or

x = -2

. The constraint

x < 0

means we must choose

x = -2

.

6

Substitute

x = -2

back into one of the original equations to solve for

y

.

y = 2(-2) + 3 = -1

We need the corresponding

y

-value to find the value of

x + y

.

7

Calculate the value of

x + y

.

x + y = -2 + (-1) = -3

This provides the final required value specified in the question.

Key Concept

Solving a system of one linear equation and one quadratic equation using substitution.

Question 34Question →

\begin{aligned} x - y &= 3 \\ y &= x^2 - 7x + 12 \end{aligned}

If $(x, y)$ is a solution to the system of equations above and $y > 0$ , what is the value of $xy$ ?

Show answer & explanation

Answer: 10

Answer

10

By substituting the linear equation

y = x - 3

into the quadratic equation, we obtain a single quadratic equation

x^2 - 8x + 15 = 0

. Factoring this equation yields the solutions

x = 3

and

x = 5

. The corresponding

y

-coordinates are

y = 0

and

y = 2

, respectively. Since the problem specifies that

y > 0

, we choose the solution

(5, 2)

. The product of the coordinates is

5 \cdot 2 = 10

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation.

y = x - 3

This allows for substitution into the second equation.

2

Substitute

y = x - 3

into the quadratic equation.

x - 3 = x^2 - 7x + 12

To create a single quadratic equation in terms of

x

.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - 8x + 15 = 0

Subtracting

x

and adding

3

to both sides enables factoring.

4

Factor the quadratic expression.

(x - 3)(x - 5) = 0

To find the roots of the equation, which are

x = 3

and

x = 5

.

5

Find the corresponding

y

-values for each

x

-value.

For

x = 3

,

y = 3 - 3 = 0

. For

x = 5

,

y = 5 - 3 = 2

.

To obtain the complete coordinate pairs of the intersection points.

6

Apply the constraint

y > 0

to select the correct solution pair.

The solution

(5, 2)

is selected since its

y

-coordinate is greater than

0

.

The other solution,

(3, 0)

, has

y = 0

, which does not satisfy the constraint

y > 0

.

7

Calculate the value of

xy

for the chosen solution.

xy = 5 \cdot 2 = 10

To find the final requested value.

Key Concept

Solving a nonlinear system of equations containing a linear equation and a quadratic equation by substitution.

Estimated Time:1m 30s

Question 35Question →

\begin{aligned} (x - 4)^2 + (y + 1)^2 &= 18 \\ x + y &= k \end{aligned}

In the system of equations above, $k$ is a constant. If the system has exactly one real solution, what is the sum of all possible values of $k$ ?

-6

2

6

10

Show answer & explanation

Answer: 6

Answer

The sum of all possible values of

k

is

6

.

The system of equations has exactly one real solution when the line is tangent to the circle. The circle

(x - 4)^2 + (y + 1)^2 = 18

has center

(4, -1)

and radius

\sqrt{18} = 3\sqrt{2}

. The distance from the center of the circle to the line

x + y - k = 0

is given by

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

. Setting this distance equal to the radius yields

\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6

. Solving this equation gives

k = 9

and

k = -3

. The sum of these possible values is

9 + (-3) = 6

.

Step-by-Step Solution

1

Relate the algebraic condition of having exactly one solution to the geometric relationship between the circle and the line.

The line must be tangent to the circle, meaning the perpendicular distance from the center of the circle to the line must equal the radius of the circle.

A system consisting of a circle and a line has exactly one real solution if and only if the line is tangent to the circle.

2

Identify the center and radius of the circle from the equation

(x - 4)^2 + (y + 1)^2 = 18

.

The center of the circle is

(4, -1)

and the radius is

r = \sqrt{18} = 3\sqrt{2}

.

The standard form of a circle equation is

(x - h)^2 + (y - k)^2 = r^2

, where

(h, k)

is the center and

r

is the radius.

3

Express the line equation in standard form

Ax + By + C = 0

and set up the perpendicular distance formula.

The line equation is

x + y - k = 0

. The distance from the center

(4, -1)

to the line is

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

.

The perpendicular distance from a point

(x_0, y_0)

to a line

Ax + By + C = 0

is given by

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

.

4

Set the distance equal to the radius of the circle and solve the absolute value equation for

k

.

\frac{|3 - k|}{\sqrt{2}} = 3\sqrt{2} \implies |3 - k| = 6

. This yields two equations:

3 - k = 6 \implies k = -3

, and

3 - k = -6 \implies k = 9$.

Setting the distance equal to the radius ensures that the line is tangent to the circle, resulting in exactly one point of intersection.

5

Calculate the sum of all possible values of

k

.

Sum =

-3 + 9 = 6

.

The question asks for the sum of all values of

k

that satisfy the tangency condition.

Key Concept

Determining the relationship between a line and a circle in a system of equations by setting the perpendicular distance from the center of the circle equal to its radius.

Alternative Method

Substitute

y = k - x

into the circle equation to obtain a quadratic equation in

x

:

(x - 4)^2 + (k - x + 1)^2 = 18

. Expanding and combining like terms yields

2x^2 - 2(k + 5)x + (k^2 + 2k - 1) = 0

. For the system to have exactly one real solution, the discriminant

\Delta

of this quadratic equation must be zero:

\Delta = [-2(k + 5)]^2 - 4(2)(k^2 + 2k - 1) = 0

. Simplifying this equation leads to

4(k^2 + 10k + 25) - 8(k^2 + 2k - 1) = 0 \implies k^2 - 6k - 27 = 0

. By Vieta's formulas, the sum of the solutions to this quadratic equation for

k

is given by

-\frac{b}{a} = -\frac{-6}{1} = 6

.

Estimated Time:2m 30s

Question 36Question →

A system of equations consists of the quadratic function $f(x) = -x^2 + 6x - 2$ and the linear function $g(x) = 2x + 1$ . If the graphs of $f$ and $g$ intersect at the points $(x_1, y_1)$ and $(x_2, y_2)$ in the $xy$ -plane, what is the value of $y_1 + y_2$ ?

Show answer & explanation

Answer: 10

Answer

10

By setting the two equations equal to each other, we obtain

-x^2 + 6x - 2 = 2x + 1

. Rearranging terms to one side gives

x^2 - 4x + 3 = 0

, which factors as

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

. The x-coordinates of the intersection points are

x = 1

and

x = 3

. Substituting these into the linear equation gives the y-coordinates: for

x = 1

,

y = 2(1) + 1 = 3

; for

x = 3

,

y = 2(3) + 1 = 7

. Summing these y-coordinates gives

3 + 7 = 10

.

Step-by-Step Solution

1

Set the quadratic function equal to the linear function to find the x-coordinates of the intersection points.

-x^2 + 6x - 2 = 2x + 1

At the points of intersection, the y-values of both functions must be equal.

2

Rearrange the equation into standard quadratic form and solve for

x

by factoring.

x^2 - 4x + 3 = 0 \Rightarrow (x - 1)(x - 3) = 0 \Rightarrow x = 1 \text{ or } x = 3

Factoring the quadratic equation gives the x-coordinates of the intersection points.

3

Substitute the x-coordinates back into the linear equation to find the corresponding y-coordinates.

For

x_1 = 1

,

y_1 = 2(1) + 1 = 3

. For

x_2 = 3

,

y_2 = 2(3) + 1 = 7

.

The linear equation

g(x) = 2x + 1

is simpler to use for finding the y-coordinates than the quadratic equation.

4

Sum the two y-coordinates.

y_1 + y_2 = 3 + 7 = 10

The question asks for the sum of the y-coordinates of the intersection points.

Key Concept

Solving a system consisting of a linear equation and a quadratic equation by substitution or equating them.

Question 37Question →

\begin{aligned} y &= 3x^2 - 4x - 5 \\ y - 2x &= 4 \end{aligned}

If $(x_1, y_1)$ and $(x_2, y_2)$ are the solutions to the system of equations above, and $y_1 > y_2$ , what is the value of $x_1 + y_2$ ?

-3

-1

5

1

Show answer & explanation

Answer: 5

Answer

5

To find the solutions to the system of equations, substitute the expression for

y

from the second equation into the first equation. First, rewrite the second equation as

y = 2x + 4

. Substituting this into the first equation yields

2x + 4 = 3x^2 - 4x - 5

. Rearranging terms to set the equation to zero gives

3x^2 - 6x - 9 = 0

. Dividing the entire equation by 3 simplifies it to

x^2 - 2x - 3 = 0

. Factoring this quadratic equation gives

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

, which yields the solutions

x_1 = 3

and

x_2 = -1

. Next, find the corresponding

y

-coordinates by substituting these

x

-values back into the linear equation

y = 2x + 4

. For

x = 3

,

y_1 = 2(3) + 4 = 10

. For

x = -1

,

y_2 = 2(-1) + 4 = 2

. We are given that

y_1 > y_2

, which confirms that

(x_1, y_1) = (3, 10)

and

(x_2, y_2) = (-1, 2)

. Finally, calculate

x_1 + y_2

, which is

3 + 2 = 5

.

Step-by-Step Solution

1

Rewrite the linear equation to express

y

in terms of

x

.

y = 2x + 4

This allows for substitution into the quadratic equation.

2

Substitute

y = 2x + 4

into the quadratic equation and set to zero.

3x^2 - 6x - 9 = 0

To find the

x

-coordinates of the intersection points.

3

Simplify and factor the quadratic equation.

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

, giving

x = 3

and

x = -1

.

To solve for the

x

-values of the intersection points.

4

Substitute the

x

-values back into the linear equation to find the corresponding

y

-values.

For

x = 3

,

y = 10

. For

x = -1

,

y = 2

. The intersection points are

(3, 10)

and

(-1, 2)

.

To find the complete coordinates of the intersection points.

5

Identify

(x_1, y_1)

and

(x_2, y_2)

using the condition

y_1 > y_2

, and compute

x_1 + y_2

.

Since

10 > 2

,

y_1 = 10

(with

x_1 = 3

) and

y_2 = 2

(with

x_2 = -1

). Then,

x_1 + y_2 = 3 + 2 = 5

.

To compute the required target expression.

Key Concept

Solving a nonlinear system of equations by substituting a linear expression into a quadratic equation and solving the resulting quadratic equation.

Question 38Question →

\begin{aligned} y &= 2x^2 - kx + 8 \\ y &= 4x - 10 \end{aligned}

In the system of equations above, $k$ is a constant. If the system has exactly one real solution, what is the product of all possible values of $k$ ?

-48

-128

-20

128

Show answer & explanation

Answer: -128

Answer

-128

To find the number of solutions to the system, set the two equations equal to each other:

2x^2 - kx + 8 = 4x - 10

. Moving all terms to one side gives

2x^2 - (k + 4)x + 18 = 0

. For this quadratic equation to have exactly one real solution, its discriminant,

\Delta = B^2 - 4AC

, must equal zero. Substituting the coefficients

A = 2

,

B = -(k + 4)

, and

C = 18

into the discriminant formula gives

(-(k + 4))^2 - 4(2)(18) = 0

, which simplifies to

(k + 4)^2 - 144 = 0

. Solving this equation gives

k + 4 = 12 \implies k = 8

, or

k + 4 = -12 \implies k = -16

. The product of these possible values is

8 \times (-16) = -128

.

Step-by-Step Solution

1

Equate the quadratic and linear equations to find the condition for intersection.

2x^2 - kx + 8 = 4x - 10

To find the points where the two graphs intersect, their

y

-values must be set equal to each other.

2

Rearrange the terms to form a standard quadratic equation of the form

Ax^2 + Bx + C = 0

.

2x^2 - (k + 4)x + 18 = 0

Grouping terms allows us to determine the coefficients

A = 2

,

B = -(k + 4)

, and

C = 18

to use in the discriminant.

3

Set the discriminant

\Delta = B^2 - 4AC

of the quadratic equation to

0

.

[-(k + 4)]^2 - 4(2)(18) = 0 \implies (k + 4)^2 - 144 = 0

A system of a line and a parabola has exactly one real solution (meaning the line is tangent to the parabola) when the resulting quadratic equation has exactly one real root, which corresponds to a discriminant of zero.

4

Solve for the possible values of

k

and calculate their product.

k + 4 = \pm 12 \implies k = 8

or

k = -16

. The product is

8 \times (-16) = -128

.

Extracting the square root yields two possible values for

k

, and multiplying them gives the requested product.

Key Concept

Using the discriminant of a quadratic equation to find parameter values that make a linear-quadratic system have exactly one real solution.

Alternative Method

Once the quadratic equation

(k + 4)^2 - 144 = 0

is simplified to

k^2 + 8k - 128 = 0

, Vieta's formulas can be applied directly to find the product of the possible values of

k

. According to Vieta's formulas, the product of the roots of a quadratic equation

ak^2 + bk + c = 0

is given by

c/a

. Here,

a = 1

and

c = -128

, so the product of the roots is

-128 / 1 = -128

.

Estimated Time:2m 0s

Question 39Question →

Consider the system of equations below:

\begin{aligned} y &= 2x^2 + kx + 9 \\ y &= -2x + 1 \end{aligned}

In the system of equations above, $k$ is a positive constant. If the system has exactly one real solution, what is the value of $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

k

is 6.

By substituting the linear equation into the quadratic equation, we obtain the equation

2x^2 + (k+2)x + 8 = 0

. For the system to have exactly one real solution, the discriminant of this quadratic equation,

\Delta = b^2 - 4ac

, must be equal to zero. Substituting

a = 2

,

b = k+2

, and

c = 8

into the discriminant formula gives

(k+2)^2 - 4(2)(8) = 0

, which simplifies to

(k+2)^2 = 64

. Solving for

k

gives

k+2 = 8

or

k+2 = -8

, resulting in

k = 6

or

k = -10

. Since

k

is specified to be a positive constant, the value of

k

must be 6.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the quadratic equation.

2x^2 + kx + 9 = -2x + 1

Setting the two expressions for

y

equal to each other allows us to solve for the

x

-coordinates of the intersection points.

2

Rearrange the equation into the standard quadratic form

ax^2 + bx + c = 0

.

2x^2 + (k+2)x + 8 = 0

Grouping the terms by power of

x

lets us identify the coefficients

a = 2

,

b = k+2

, and

c = 8

.

3

Set the discriminant of the quadratic equation to zero.

(k+2)^2 - 4(2)(8) = 0

, which simplifies to

(k+2)^2 - 64 = 0

A quadratic equation has exactly one real solution if and only if its discriminant,

b^2 - 4ac

, is equal to zero.

4

Solve for the positive constant

k

.

k = 6

Solving

(k+2)^2 = 64

yields

k+2 = 8

or

k+2 = -8

, which means

k = 6

or

k = -10

. Since

k

must be a positive constant, we select 6.

Key Concept

Determining the number of solutions of a quadratic-linear system using the discriminant.

Alternative Method

Alternatively, the quadratic equation

2x^2 + (k+2)x + 8 = 0

has exactly one real solution when it is a perfect square trinomial. Factoring out the leading coefficient gives

2(x^2 + \frac{k+2}{2}x + 4) = 0

. For the term inside the parentheses to be a perfect square trinomial of the form

x^2 + 2dx + d^2

, we must have

d^2 = 4

, which means

d = \pm 2

. Thus, the middle coefficient

\frac{k+2}{2}

must equal

2d = \pm 4

. Since

k

is positive, we set

\frac{k+2}{2} = 4

, which solves to

k = 6

.

Estimated Time:1m 30s

Question 40Question →

In the $xy$ -plane, a system of equations consists of the circle with equation $(x - 7)^2 + (y - 4)^2 = 1$ and the line $y = kx$ , where $k$ is a positive constant. If the system has exactly one real solution and $k > 0.5$ , what is the value of $k$ ?

Show answer & explanation

Answer: 0.75

Answer

The correct answer is 3/4 (or 0.75).

The correct answer is

3/4

(or

0.75

). Substituting the line

y = kx

into the circle equation and setting the discriminant of the resulting quadratic equation to zero yields a quadratic in

k

:

48k^2 - 56k + 15 = 0

. Solving this equation gives two positive tangent slopes:

k = 5/12

and

k = 3/4

. Since the problem specifies that

k > 0.5

, we choose

k = 3/4

, which equals

0.75

. Alternatively, using geometry, the distance from the center

(7,4)

to the line

kx - y = 0

must equal the radius

1

. This gives

\frac{|7k-4|}{\sqrt{k^2+1}} = 1

. Squaring both sides yields

(7k-4)^2 = k^2+1

, which simplifies to

48k^2 - 56k + 15 = 0

, giving the same values of

k

.

Step-by-Step Solution

1

Substitute the linear equation

y = kx

into the circle's quadratic equation

(x - 7)^2 + (y - 4)^2 = 1

.

(x - 7)^2 + (kx - 4)^2 = 1

To find the points of intersection between the line and the circle, substitute the expression for

y

from the linear equation into the circle's equation.

2

Expand the terms and write the equation in the standard quadratic form

Ax^2 + Bx + C = 0

.

(1 + k^2)x^2 - (14 + 8k)x + 64 = 0

Expanding

(x - 7)^2

yields

x^2 - 14x + 49

and expanding

(kx - 4)^2

yields

k^2x^2 - 8kx + 16

. Grouping the terms by powers of

x

and subtracting

1

from both sides gives the standard form.

3

Set the discriminant of the quadratic equation to

0

to satisfy the condition of exactly one real solution.

\Delta = [-(14 + 8k)]^2 - 4(1 + k^2)(64) = 0

A quadratic equation has exactly one real solution if and only if its discriminant,

B^2 - 4AC

, is equal to zero. Geometrically, this means the line is tangent to the circle.

4

Simplify the discriminant equation and solve for

k

.

48k^2 - 56k + 15 = 0

, which factors as

(12k - 5)(4k - 3) = 0

. Thus,

k = \frac{5}{12}

or

k = \frac{3}{4}

.

Simplifying

(14 + 8k)^2 - 256(1 + k^2) = 0

gives

196 + 224k + 64k^2 - 256 - 256k^2 = 0

, which simplifies to

-192k^2 + 224k - 60 = 0

. Dividing the entire equation by

-4

yields the quadratic equation in

k

.

5

Select the value of

k

that satisfies the given condition

k > 0.5

.

k = \frac{3}{4}

(or

0.75

)

Since

\frac{5}{12} \approx 0.417

is less than

0.5

and

\frac{3}{4} = 0.75

is greater than

0.5

, the correct value is

3/4

.

Key Concept

Solving a nonlinear system of equations involving a circle and a line by setting the discriminant of the substituted quadratic equation to zero to find the slope of the tangent lines.

Estimated Time:2m 30s

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