Nonlinear Systems of Equations Practice Questions — SAT — Page 3

A circle and a line intersect at two points in the $xy$ -plane. The equations of the circle and the line are:

(x - 3)^2 + y^2 = 25

y = x - 2

If $(x_1, y_1)$ is the intersection point in Quadrant I and $(x_2, y_2)$ is the intersection point in Quadrant III, what is the value of $x_1 + y_2$ ?

3

5

1

10

Show answer & explanation

Answer: 3

Answer

The value of

x_1 + y_2

is 3.

To solve the system, we substitute the linear equation

y = x - 2

into the circle equation, obtaining

(x - 3)^2 + (x - 2)^2 = 25

. Expanding this gives

x^2 - 6x + 9 + x^2 - 4x + 4 = 25

, which simplifies to

2x^2 - 10x - 12 = 0

. Dividing by 2, we get

x^2 - 5x - 6 = 0

, which factors as

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

. The solutions are

x = 6

and

x = -1

. Substituting these back into the linear equation gives the points of intersection:

(6, 4)

in Quadrant I (since both coordinates are positive) and

(-1, -3)

in Quadrant III (since both coordinates are negative). Therefore,

x_1 = 6

and

y_2 = -3

, and their sum is

6 + (-3) = 3

. This matches the correct option.

Step-by-Step Solution

1

Substitute the linear equation into the circle equation to obtain an equation in terms of

x

only.

(x - 3)^2 + (x - 2)^2 = 25

This reduces the system to a single quadratic equation in one variable, allowing us to find the x-coordinates of the intersection points.

2

Expand and simplify the quadratic equation into standard form.

x^2 - 6x + 9 + x^2 - 4x + 4 = 25 \implies 2x^2 - 10x - 12 = 0 \implies x^2 - 5x - 6 = 0

Simplifying the equation makes it solvable by standard quadratic factoring methods.

3

Solve the simplified quadratic equation for

x

by factoring.

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

or

x = -1

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

4

Find the corresponding

y

-coordinates by substituting the

x

-values back into the linear equation

y = x - 2

.

For

x = 6

,

y = 6 - 2 = 4

, giving the point

(6, 4)

in Quadrant I. For

x = -1

,

y = -1 - 2 = -3

, giving the point

(-1, -3)

in Quadrant III.

This yields the coordinates of the two points of intersection and identifies which point lies in which quadrant.

5

Identify

x_1

and

y_2

and calculate the sum

x_1 + y_2

.

x_1 = 6

and

y_2 = -3

, so

x_1 + y_2 = 6 + (-3) = 3

.

This answers the specific question using the coordinate components of the two points.

Key Concept

Nonlinear Systems of Equations

Question 42Question →

\begin{aligned} y &= 2x^2 - 5x - 1 \\ y &= 3x - 7 \end{aligned}

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

-11

5

-3

1

Show answer & explanation

Answer: 5

Answer

5

Substituting

y = 3x - 7

into

y = 2x^2 - 5x - 1

yields the quadratic equation

3x - 7 = 2x^2 - 5x - 1

. Moving all terms to one side gives

2x^2 - 8x + 6 = 0

. Dividing by 2 simplifies this to

x^2 - 4x + 3 = 0

, which factors as

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

. This gives the solutions

x = 1

and

x = 3

. Substituting these into the linear equation yields the coordinate points

(1, -4)

and

(3, 2)

. Since the problem states that

y > 0

, we choose the solution

(3, 2)

. The value of

x + y

is

3 + 2 = 5

.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the quadratic equation.

3x - 7 = 2x^2 - 5x - 1

Since both equations are solved for

y

, they can be set equal to each other to find the

x

-coordinates of the intersection points.

2

Rearrange the terms to set the quadratic equation equal to zero, and then divide by the common factor.

2x^2 - 8x + 6 = 0

, which simplifies to

x^2 - 4x + 3 = 0

Putting the equation in standard form

ax^2 + bx + c = 0

allows us to factor and solve for

x

.

3

Factor the quadratic equation to find the values of

x

.

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

, so

x = 1

or

x = 3

Factoring allows us to find the roots of the equation directly.

4

Substitute the

x

values back into the linear equation to solve for the corresponding

y

values.

For

x = 1

,

y = 3(1) - 7 = -4

, giving the solution

(1, -4)

. For

x = 3

,

y = 3(3) - 7 = 2

, giving the solution

(3, 2)

.

Finding the

y

-coordinates completes the solutions

(x, y)

to the system.

5

Apply the given constraint

y > 0

and calculate

x + y

.

The solution

(3, 2)

satisfies

y > 0

since

2 > 0

. Thus,

x + y = 3 + 2 = 5

.

The problem asks for the sum of the coordinates of the solution that has a positive

y

-value.

Key Concept

Solving a system consisting of a linear equation and a quadratic equation by substitution and applying coordinate constraints.

Estimated Time:1m 30s

Question 43Question →

\begin{aligned} y + 11 &= x^2 \\ y - 3x &= 7 \end{aligned}

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

Show answer & explanation

Answer: 31

Answer

The correct answer is 31.

Solving the system of equations by expressing

y

in terms of

x

from the linear equation gives

y = 3x + 7

. Substituting this expression for

y

in the first equation yields

(3x + 7) + 11 = x^2

, which simplifies to

x^2 - 3x - 18 = 0

. Factoring the quadratic expression gives

(x - 6)(x + 3) = 0

, so

x = 6

or

x = -3

. Since the problem specifies that

x > 0

, the

x

-value of the solution must be

6

. Substituting

x = 6

back into the linear equation gives

y = 3(6) + 7 = 25

. The sum of

x

and

y

is

6 + 25 = 31

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation.

y = 3x + 7

This isolates the variable

y

to facilitate substitution.

2

Substitute the expression for

y

into the quadratic equation.

(3x + 7) + 11 = x^2

This eliminates the variable

y

, yielding a single equation in terms of

x

.

3

Rearrange the equation into standard quadratic form.

x^2 - 3x - 18 = 0

Standard form (

ax^2 + bx + c = 0

) is required to solve by factoring.

4

Factor the quadratic equation.

(x - 6)(x + 3) = 0

Finding factors helps determine the possible values of

x

.

5

Find the values of

x

and apply the positive constraint.

x = 6

The equation yields

x = 6

and

x = -3

. The condition

x > 0

restricts the solution to

x = 6

.

6

Calculate the corresponding value of

y

.

y = 25

Substituting

x = 6

into the linear equation gives

y = 3(6) + 7 = 25

.

7

Calculate the sum of

x

and

y

.

31

The problem asks for the value of

x + y

.

Key Concept

Solving systems of nonlinear equations algebraically using substitution and quadratic factoring.

Alternative Method

Alternatively, solve the linear equation for

x

to get

x = \frac{y - 7}{3}

and substitute this into the quadratic equation to solve for

y

first. This approach is more complex because it introduces fractional terms.

Estimated Time:1m 30s

Question 44Question →

In the $xy$ -plane, the circle with equation $x^2 + y^2 = 13$ intersects the line with equation $y - x = 1$ at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the value of $y_1 + y_2$ ?

-1

1

3

-5

Show answer & explanation

Answer: 1

Answer

The sum of the

y

-coordinates of the intersection points is

1

.

The correct answer is

1

. Substituting

x = y - 1

from the linear equation into the circle equation yields

(y-1)^2 + y^2 = 13

. Expanding and simplifying gives the quadratic equation

2y^2 - 2y - 12 = 0

. Dividing by 2 yields

y^2 - y - 6 = 0

, which factors into

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

. Therefore, the

y

-coordinates of the intersection points are

3

and

-2

. The sum of these coordinates is

3 + (-2) = 1

.

Step-by-Step Solution

1

Rearrange the linear equation to solve for

x

in terms of

y

.

x = y - 1

Expressing

x

in terms of

y

allows us to substitute it into the circle equation to find the

y

-coordinates directly.

2

Substitute

x = y - 1

into the circle equation

x^2 + y^2 = 13

.

(y - 1)^2 + y^2 = 13

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

y

.

3

Expand and simplify the quadratic equation into standard form

ay^2 + by + c = 0

.

y^2 - 2y + 1 + y^2 = 13 \implies 2y^2 - 2y - 12 = 0

Expanding the squared term and combining like terms puts the equation in a form that can be solved.

4

Divide the equation by 2 and factor the simplified quadratic equation.

y^2 - y - 6 = 0 \implies (y - 3)(y + 2) = 0

Simplifying and factoring helps find the individual

y

-coordinates of the two intersection points.

5

Solve for

y

and calculate the sum of the coordinates.

y = 3

or

y = -2

, and

3 + (-2) = 1

Solving the factored equation gives the

y

-coordinates of the two intersection points, which are then added to find the final value.

Key Concept

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

Alternative Method

Instead of solving for

x

first, we can substitute

y = x + 1

into the circle equation to find the

x

-coordinates:

x^2 + (x + 1)^2 = 13 \implies 2x^2 + 2x - 12 = 0 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0

. This gives

x = -3

or

x = 2

. We then find the corresponding

y

-values using

y = x + 1

: for

x = -3

,

y = -2

; for

x = 2

,

y = 3

. The sum of the

y

-coordinates is

-2 + 3 = 1

.

Estimated Time:1m 30s

Question 45Question →

If $(x, y)$ is a solution to the system of equations below and $x > 1$ , what is the value of $x$ ?

\begin{aligned} y - 2x &= 3 \\ y^2 - 5x^2 &= 20 \end{aligned}

Show answer & explanation

Answer: 11

Answer

The value of

x

is 11.

The system can be solved by substituting the first equation into the second. Rearranging the first equation gives

y = 2x + 3

. Substituting this expression for

y

in the second equation gives

(2x + 3)^2 - 5x^2 = 20

. Expanding the binomial yields

4x^2 + 12x + 9 - 5x^2 = 20

, which simplifies to

-x^2 + 12x + 9 = 20

. Rearranging into standard quadratic form gives

x^2 - 12x + 11 = 0

. Factoring the quadratic equation results in

(x - 11)(x - 1) = 0

. Thus,

x = 11

or

x = 1

. Since the problem specifies

x > 1

, the value of

x

must be 11.

Step-by-Step Solution

1

Rearrange the first equation to express

y

in terms of

x

.

y = 2x + 3

This allows for substitution into the second equation.

2

Substitute

y = 2x + 3

into the second equation.

(2x + 3)^2 - 5x^2 = 20

To eliminate

y

and obtain an equation in terms of

x

only.

3

Expand and simplify the equation.

x^2 - 12x + 11 = 0

Expanding

(2x + 3)^2

gives

4x^2 + 12x + 9

. Subtracting

5x^2

yields

-x^2 + 12x + 9 = 20

. Subtracting 20 from both sides gives

-x^2 + 12x - 11 = 0

. Multiplying the entire equation by

-1

results in standard form.

4

Factor the quadratic equation.

(x - 11)(x - 1) = 0

To find the possible values of

x

.

5

Apply the given constraint

x > 1

.

x = 11

The solutions are

x = 11

and

x = 1

. The constraint

x > 1

excludes

x = 1

, leaving

x = 11

as the only valid solution.

Key Concept

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

Question 46Question →

In the $xy$ -plane, the system of equations below has exactly one real solution.

y = -x^2 + 6x + 2

y + 2x = k

If $k$ is a constant, what is the value of $k$ ?

Show answer & explanation

Answer: 18

Answer

18

To find the value of the constant

k

for which the system of equations has exactly one real solution, we can solve the system by substitution. Substituting

y = k - 2x

from the second equation into the first equation gives

k - 2x = -x^2 + 6x + 2

. Rearranging the terms to write this quadratic equation in standard form,

ax^2 + bx + c = 0

, yields

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

. A quadratic equation has exactly one real solution when its discriminant,

b^2 - 4ac

, is equal to zero. Substituting

a = 1

,

b = -8

, and

c = k - 2

into the discriminant formula gives

(-8)^2 - 4(1)(k - 2) = 0

. Simplifying this expression results in

64 - 4k + 8 = 0

, which simplifies further to

72 - 4k = 0

. Solving for

k

yields

k = 18

.

Step-by-Step Solution

1

Rewrite the linear equation to express

y

in terms of

x

and

k

.

y = k - 2x

This allows substituting the expression for

y

into the quadratic equation to eliminate

y

.

2

Substitute the expression for

y

into the quadratic equation.

k - 2x = -x^2 + 6x + 2

To construct a single quadratic equation in terms of

x

whose solutions correspond to the system's intersection points.

3

Rearrange the equation into standard quadratic form

ax^2 + bx + c = 0

.

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

To identify the coefficients

a = 1

,

b = -8

, and

c = k - 2

.

4

Set the discriminant of the quadratic equation to zero.

(-8)^2 - 4(1)(k - 2) = 0

A quadratic equation has exactly one real solution if and only if its discriminant is zero.

5

Solve the linear equation for

k

.

k = 18

Simplify the equation:

64 - 4k + 8 = 0 \implies 72 - 4k = 0 \implies 4k = 72 \implies k = 18

.

Key Concept

Nonlinear Systems of Equations

Question 47Question →

The system of equations consists of a quadratic equation and a linear equation, where

c

is a constant:

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

If this system has exactly one real solution, what is the value of

c

?

-3

-1

1

3

Show answer & explanation

Answer: $-1$

Answer

The value of the constant

c

is

-1

.

To find the value of

c

for which the system has exactly one real solution, we equate the two equations to get

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

. Subtracting

x

and

c

from both sides gives the quadratic equation

2x^2 - 4x + (1 - c) = 0

. A quadratic equation has exactly one real solution when its discriminant,

b^2 - 4ac

, is equal to zero. Substituting

a = 2

,

b = -4

, and the constant term

(1 - c)

into the discriminant formula gives

(-4)^2 - 4(2)(1 - c) = 0

, which simplifies to

16 - 8(1 - c) = 0

. Solving for

c

yields

8 + 8c = 0

, or

c = -1

.

Step-by-Step Solution

1

Equate the two expressions for

y

to set up a single equation in terms of

x

.

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

Since both equations are equal to

y

, their right-hand sides must be equal at any point of intersection.

2

Rearrange the equation into standard quadratic form,

ax^2 + bx + c_0 = 0

.

2x^2 - 4x + (1 - c) = 0

Subtracting

x

and

c

from both sides groups the terms by their powers of

x

.

3

Write the expression for the discriminant of this quadratic equation.

D = (-4)^2 - 4(2)(1 - c)

For a quadratic equation

ax^2 + bx + c_0 = 0

, the discriminant is given by

b^2 - 4ac_0

where

a = 2

,

b = -4

, and the constant term is

1 - c

.

4

Set the discriminant to zero and solve for

c

to find the condition for exactly one real solution.

16 - 8(1 - c) = 0 \implies 16 - 8 + 8c = 0 \implies 8 + 8c = 0 \implies c = -1

A quadratic equation has exactly one real solution if and only if its discriminant is equal to zero.

Key Concept

Determining the number of solutions of a nonlinear system by setting the discriminant of the combined quadratic equation to zero.

Estimated Time:1m 30s

Question 48Question →

Consider the system of equations below:

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

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

Show answer & explanation

Answer: 8

Answer

The value of

y

is 8.

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

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

. Simplifying this equation by moving all terms to one side yields

2x^2 - 3x - 9 = 0

. Factoring this quadratic equation gives

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

, which has solutions

x = -1.5

and

x = 3

. Since the problem specifies that

x > 0

, we must use

x = 3

. Substituting

x = 3

into the second equation, we find

y = (3)^2 - 2(3) + 5 = 8

. Substituting into the first equation also yields

y = 3(3)^2 - 5(3) - 4 = 8

. Therefore, the value of

y

is 8.

Step-by-Step Solution

1

Set the quadratic expressions equal to each other.

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

Since both equations are solved for

y

, their right-hand sides must be equal at any point of intersection.

2

Rearrange the terms to set the quadratic equation to zero.

2x^2 - 3x - 9 = 0

Putting the equation in standard form

ax^2 + bx + c = 0

allows us to solve it by factoring.

3

Factor the quadratic expression to find the roots.

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

, which yields

x = -1.5

or

x = 3

.

Factoring shows the values of

x

that satisfy the system.

4

Select the positive root and substitute it back to find

y

.

y = 8

The problem specifies

x > 0

, so we use

x = 3

. Substituting

x = 3

into

y = x^2 - 2x + 5

gives the corresponding

y

-value.

Key Concept

Solving a system of nonlinear equations by setting the equations equal to each other and solving the resulting quadratic equation.

Nonlinear Systems of Equations

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Step-by-Step Solution

Key Concept

Common Mistakes

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Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes