Systems of Linear Equations Practice Questions — SAT — Page 2

A system of linear equations is shown below.

\begin{aligned} 5x - 2y &= 14 \\ 2x + y &= 11 \end{aligned}

If

(x, y)

is the solution to the system of equations, what is the value of

x

?

3

36

4

2

Show answer & explanation

Answer: 4

Answer

4

The correct answer is

4

. Isolating

y

in the second equation gives

y = 11 - 2x

. Substituting this into the first equation yields

5x - 2(11 - 2x) = 14

. Distributing the

-2

results in

5x - 22 + 4x = 14

. Combining like terms simplifies the equation to

9x - 22 = 14

. Adding

22

to both sides gives

9x = 36

, and dividing by

9

yields

x = 4

.

Step-by-Step Solution

1

Isolate

y

in the second equation.

y = 11 - 2x

This allows for substitution into the first equation.

2

Substitute the expression for

y

into the first equation.

5x - 2(11 - 2x) = 14

This creates an equation with only one variable,

x

.

3

Distribute the

-2

through the parentheses and combine like terms.

9x - 22 = 14

Distributing

-2

to

-2x

yields

+4x

, and combining

5x + 4x

gives

9x

.

4

Add

22

to both sides and divide by

9

.

x = 4

Isolating

x

gives the final value of the variable.

Key Concept

Solving systems of linear equations using the substitution method.

Question 22Question →

In the system of equations below,

c

is a constant.

\begin{aligned} 2x + y &= 15 \\ cx - 2y &= 6 \end{aligned}

If the solution

(x, y)

to the system of equations above lies on the line

y = 3x

in the

xy

-plane, what is the value of the constant

c

?

Show answer & explanation

Answer: 8

Answer

The value of the constant

c

is 8.

Since the solution

(x, y)

lies on the line

y = 3x

, we can substitute

y = 3x

into the first equation,

2x + y = 15

, to get

2x + 3x = 15

, which simplifies to

5x = 15

, yielding

x = 3

. Substituting

x = 3

back into

y = 3x

gives

y = 9

. Thus, the intersection point of the system is

(3, 9)

. Substituting

x = 3

and

y = 9

into the second equation,

cx - 2y = 6

, gives

c(3) - 2(9) = 6

, or

3c - 18 = 6

. Adding 18 to both sides gives

3c = 24

, which simplifies to

c = 8

.

Step-by-Step Solution

1

Substitute the equation of the line

y = 3x

into the first equation of the system,

2x + y = 15

.

2x + 3x = 15

, which simplifies to

5x = 15

. Solving for

x

gives

x = 3

.

Since the solution to the system lies on the line

y = 3x

, the coordinates of the solution must satisfy all three equations.

2

Calculate the corresponding

y

-value using the relation

y = 3x

.

y = 3(3) = 9

.

The

y

-coordinate is three times the

x

-coordinate for any point on the line

y = 3x

.

3

Substitute the point

(3, 9)

into the second equation of the system,

cx - 2y = 6

.

c(3) - 2(9) = 6

, which simplifies to

3c - 18 = 6

. Adding 18 to both sides gives

3c = 24

, so

c = 8

.

Since

(3, 9)

is the solution to the system, it must satisfy the second equation as well.

Key Concept

Solving systems of linear equations using substitution and finding parameter values based on geometric constraints.

Question 23Question →

In the system of equations below,

k

is a positive integer that is a multiple of 9.

\begin{aligned} 3x + 8y &= 213 \\ x - 3y &= k \end{aligned}

If the system has a solution

(x, y)

such that

x

and

y

are both positive integers, what is the value of

k

?

Show answer & explanation

Answer: 54

Answer

54

Substituting

x = 3y + k

into the first equation yields

17y + 3k = 213

. Solving for

k

gives

k = 71 - \frac{17}{3}y

. For

k

to be a positive integer,

y

must be a positive multiple of 3 less than 12.5. Testing the possible values for

y

(3, 6, 9, 12) yields the possible values of

k

as 54, 37, 20, and 3. Since

k

must be a multiple of 9, the correct value is 54.

Step-by-Step Solution

1

Express

x

in terms of

y

and

k

using the second equation.

x = 3y + k

Isolating

x

allows for easy substitution into the first equation to eliminate one of the variables.

2

Substitute the expression for

x

into the first equation.

17y + 3k = 213

Substituting

x = 3y + k

into

3x + 8y = 213

yields

3(3y + k) + 8y = 213

, which simplifies to

9y + 3k + 8y = 213

and then to

17y + 3k = 213

.

3

Solve for

k

in terms of

y

and analyze the divisibility constraint.

k = 71 - \frac{17}{3}y

, where

y

is a multiple of 3

For

k

to be an integer, the fractional term

\frac{17}{3}y

must resolve to an integer, requiring

y

to be divisible by 3.

4

Apply the constraint that

k

must be a positive integer (

k > 0

).

y < 12.5

, meaning

y

can be 3, 6, 9, or 12

Setting

71 - \frac{17}{3}y > 0

yields

17y < 213

, or

y < 12.53

. The positive integer multiples of 3 in this range are 3, 6, 9, and 12.

5

Find the corresponding values of

k

and select the one that is a multiple of 9.

The possible values for

k

are 54, 37, 20, and 3. The only multiple of 9 is 54.

Testing the possible values of

y

gives:

y=3 \rightarrow k=54

,

y=6 \rightarrow k=37

,

y=9 \rightarrow k=20

, and

y=12 \rightarrow k=3

. Among these, 54 is the only multiple of 9.

Key Concept

Systems of Linear Equations with Integer Constraints

Estimated Time:3m 0s

Question 24Question →

A local store sells two types of coffee beans: Arabica and Robusta. On Monday, the store sold $8$ pounds of Arabica coffee and $5$ pounds of Robusta coffee for a total of $\$62.00$ . On Tuesday, the store sold $4$ pounds of Arabica coffee and $7$ pounds of Robusta coffee for a total of $\$58.00$ . What is the cost, in dollars, of one pound of Robusta coffee?

Show answer & explanation

Answer: 6

Answer

The cost of one pound of Robusta coffee is 6 dollars.

The correct answer is 6. By setting up the system of equations

8a + 5r = 62

and

4a + 7r = 58

, we can multiply the second equation by 2 to get

8a + 14r = 116

. Subtracting the first equation from this yields

9r = 54

, which simplifies to

r = 6

.

Step-by-Step Solution

1

Define variables for the price per pound of Arabica coffee (

a

) and Robusta coffee (

r

), then set up a system of linear equations based on the given store sales information.

Equation 1:

8a + 5r = 62

Equation 2:

4a + 7r = 58

This translates the word problem context into a solvable system of mathematical equations.

2

Multiply the second equation by

2

to make the coefficients of

a

equal in both equations.

8a + 14r = 116

This prepares the system for solving by the elimination method.

3

Subtract the first equation (

8a + 5r = 62

) from the new equation (

8a + 14r = 116

) to eliminate

a

and solve for

r

.

9r = 54

, which simplifies to

r = 6

Subtracting the equations eliminates one variable, leaving a single-variable equation that is easily solved.

Key Concept

Solving systems of two linear equations in two variables using the elimination method.

Estimated Time:1m 30s

Question 25Question →

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

\begin{aligned} 3x - 2y &= 3c - 1 \\ x + 3y &= 2c - 8 \end{aligned}

If the solution $(x, y)$ to the system satisfies $x + y = 2$ , what is the value of $c$ ?

4

16

3

-1

Show answer & explanation

Answer: 4

Answer

The correct value of the constant

c

is 4.

To find the value of

c

, the linear constraint

x + y = 2

is rearranged to

x = 2 - y

. Substituting this expression into the first equation yields

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

, which simplifies to

6 - 5y = 3c - 1

, or

y = \frac{7 - 3c}{5}

. Substituting it into the second equation yields

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

, which simplifies to

2 + 2y = 2c - 8

, or

y = c - 5

. Equating these two expressions gives

c - 5 = \frac{7 - 3c}{5}

. Multiplying both sides by

5

results in

5c - 25 = 7 - 3c

. Adding

3c

and

25

to both sides yields

8c = 32

, which gives

c = 4

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the given constraint

x + y = 2

.

x = 2 - y

This allows us to substitute the relation into the system of equations and reduce the variables from two to one.

2

Substitute

x = 2 - y

into the first equation,

3x - 2y = 3c - 1

, and isolate

y

.

3(2 - y) - 2y = 3c - 1 \Rightarrow 6 - 5y = 3c - 1 \Rightarrow 5y = 7 - 3c \Rightarrow y = \frac{7 - 3c}{5}

To represent

y

as a function of the parameter

c

.

3

Substitute

x = 2 - y

into the second equation,

x + 3y = 2c - 8

, and isolate

y

.

(2 - y) + 3y = 2c - 8 \Rightarrow 2 + 2y = 2c - 8 \Rightarrow 2y = 2c - 10 \Rightarrow y = c - 5

To obtain another independent expression for

y

in terms of

c

.

4

Equate the two expressions for

y

and solve for

c

.

c - 5 = \frac{7 - 3c}{5} \Rightarrow 5(c - 5) = 7 - 3c \Rightarrow 5c - 25 = 7 - 3c \Rightarrow 8c = 32 \Rightarrow c = 4

Since both expressions represent the same value

y

, they must be equal, allowing us to determine the constant

c

.

Key Concept

Solving a system of linear equations containing unknown parameters under given linear constraints.

Alternative Method

Alternatively, substitute the relationship

x = 2 - y

into both equations to write them as a system in terms of

y

and

c

:

3(2 - y) - 2y = 3c - 1 \Rightarrow 5y + 3c = 7

and

(2 - y) + 3y = 2c - 8 \Rightarrow 2y - 2c = -10

. This simplified system can be solved for

c

by multiplying the first equation by 2 and the second equation by 5 to eliminate

y

:

2(5y + 3c) - 5(2y - 2c) = 2(7) - 5(-10) \Rightarrow 6c + 10c = 14 + 50 \Rightarrow 16c = 64 \Rightarrow c = 4

.

Estimated Time:2m 0s

Question 26Question →

In the $xy$ -plane, a system of two linear equations has no solutions. One of the equations in the system is $4x - 6y = 15$ . The graph of the second equation is a line that passes through the points $(3, k)$ and $(9, 7)$ , where $k$ is a constant. What is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of

k

is

3

.

For a system of linear equations to have no solutions, the lines representing the equations must be parallel, which means they have equal slopes but different y-intercepts. The first equation,

4x - 6y = 15

, can be rewritten in slope-intercept form as

y = \frac{2}{3}x - 2.5

, showing its slope is

\frac{2}{3}

. The slope of the second line, passing through

(3, k)

and

(9, 7)

, is given by

\frac{7 - k}{9 - 3} = \frac{7 - k}{6}

. Setting the two slopes equal yields

\frac{2}{3} = \frac{7 - k}{6}

. Multiplying both sides by

6

gives

4 = 7 - k

, which solves to

k = 3

. Substituting

k = 3

back into the second line gives

y = \frac{2}{3}x + 1

. Since the slopes are equal and the y-intercepts (

-2.5

and

1

) are different, the lines are parallel and distinct, confirming there are no solutions.

Step-by-Step Solution

1

Find the slope of the first line by converting the equation to slope-intercept form.

Slope is

\frac{2}{3}

and y-intercept is

-2.5

.

Converting

4x - 6y = 15

to

y = \frac{2}{3}x - 2.5

reveals the slope of the first line.

2

Express the slope of the second line using the coordinates of the two points on the line.

Slope expression is

\frac{7 - k}{6}

.

Applying the slope formula

\frac{y_2 - y_1}{x_2 - x_1}

to

(3, k)

and

(9, 7)

defines the slope in terms of

k

.

3

Equate the two slopes and solve for the unknown parameter.

k = 3

Since the system has no solutions, the lines must be parallel and have equal slopes. Setting

\frac{2}{3} = \frac{7 - k}{6}

and solving gives

k = 3

.

Key Concept

For a system of two linear equations to have no solutions, the lines representing the equations must be parallel, which requires them to have the same slope but different y-intercepts.

Estimated Time:2m 0s

Question 27Question →

A system of linear equations is given by

\begin{aligned} kx - 4y &= 12 \\ 3x + ky &= 5 \end{aligned}

where

k

is a constant. If the system has a unique solution

(x, y)

such that

x > 0

and

y < 0

, how many possible integer values of

k

exist?

7

8

9

10

Show answer & explanation

Answer: 9

Answer

9

To find the number of integer values of

k

for which the system has a solution with

x > 0

and

y < 0

, we first express

x

and

y

in terms of

k

. Eliminating

y

by multiplying the first equation by

k

and the second by

4

and adding them yields

(k^2 + 12)x = 12k + 20

, or

x = \frac{12k+20}{k^2+12}

. Similarly, eliminating

x

yields

y = \frac{5k-36}{k^2+12}

. Since the denominator

k^2 + 12

is strictly positive for all real

k

, the sign of

x

and

y

depends solely on their numerators. For

x > 0

, we require

12k + 20 > 0

, which gives

k > -\frac{5}{3} \approx -1.67

. For

y < 0

, we require

5k - 36 < 0

, which gives

k < \frac{36}{5} = 7.2

. Combining these constraints gives the interval

-\frac{5}{3} < k < \frac{36}{5}

. The integers in this interval are

-1, 0, 1, 2, 3, 4, 5, 6,

and

7

, which is a total of 9 integers.

Step-by-Step Solution

1

Solve the system of equations for

x

in terms of

k

by eliminating

y

.

x = \frac{12k + 20}{k^2 + 12}

Multiply the first equation by

k

and the second by 4, then add them:

k(kx - 4y) + 4(3x + ky) = 12k + 20 \implies (k^2 + 12)x = 12k + 20

.

2

Solve the system of equations for

y

in terms of

k

by eliminating

x

.

y = \frac{5k - 36}{k^2 + 12}

Multiply the first equation by 3 and the second by

k

, then subtract the first from the second:

k(3x + ky) - 3(kx - 4y) = 5k - 36 \implies (k^2 + 12)y = 5k - 36

.

3

Apply the condition

x > 0

to find a constraint on

k

.

k > -\frac{5}{3}

Since the denominator

k^2 + 12

is positive for all real values of

k

, the expression for

x

is positive if and only if its numerator is positive:

12k + 20 > 0 \implies k > -\frac{20}{12} = -\frac{5}{3}

.

4

Apply the condition

y < 0

to find another constraint on

k

.

k < \frac{36}{5}

Similarly, since

k^2 + 12 > 0

, the expression for

y

is negative if and only if its numerator is negative:

5k - 36 < 0 \implies k < \frac{36}{5}

.

5

Combine the constraints and count the number of integer values of

k

in the resulting interval.

9 integer values

The combined inequality is

-\frac{5}{3} < k < \frac{36}{5}

, which simplifies to approximately

-1.67 < k < 7.2

. The integers in this range are

-1, 0, 1, 2, 3, 4, 5, 6, 7

, giving a total of 9 integers.

Key Concept

Solving systems of linear equations with parameters under inequality constraints

Question 28Question →

In the system of equations below, $a$ and $b$ are constants.

\begin{aligned} ax + by &= 24 \\ 2x - 5y &= -7 \end{aligned}

If the system has the same unique solution $(x, y)$ for all values of $a$ and $b$ that satisfy the equation $4a + 3b = 24$ , what is the value of $x + y$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x + y

is

7

.

The correct answer is

7

. The solution to the system must satisfy

ax + by = 24

for any constants

a

and

b

that satisfy the constraint

4a + 3b = 24

. By matching the coefficients of

a

and

b

in both equations, we find

x = 4

and

y = 3

. To verify, we substitute these coordinates into the second equation:

2(4) - 5(3) = 8 - 15 = -7

, which is correct. The sum of the coordinates is

x + y = 4 + 3 = 7

.

Step-by-Step Solution

1

Analyze the constraint on the constants

a

and

b

given by

4a + 3b = 24

and relate it to the first equation

ax + by = 24

.

Since the first equation

ax + by = 24

must hold for all pairs of

(a, b)

satisfying

4a + 3b = 24

, the coefficients must correspond directly, meaning

x = 4

and

y = 3

.

This shows that

(4, 3)

is a point that lies on the line

ax + by = 24

regardless of the specific values of

a

and

b

, as long as they satisfy the given constraint.

2

Prove the uniqueness of the point

(4, 3)

by testing specific values for

(a, b)

that satisfy the constraint

4a + 3b = 24

.

If

b = 0

, then

4a = 24 \implies a = 6

. The equation becomes

6x = 24 \implies x = 4

. If

a = 0

, then

3b = 24 \implies b = 8

. The equation becomes

8y = 24 \implies y = 3

. This confirms

(x, y) = (4, 3)

is the unique point.

Choosing convenient test values for

a

and

b

helps isolate the variables

x

and

y

to verify the coordinates of the solution.

3

Substitute the point

(4, 3)

into the second equation of the system,

2x - 5y = -7

, to verify that it is consistent.

2(4) - 5(3) = 8 - 15 = -7

. Since this statement is true,

(4, 3)

is the consistent unique solution to the system of equations.

A solution to a system of equations must satisfy all equations in the system.

4

Find the requested value of

x + y

using the coordinates of the solution.

x + y = 4 + 3 = 7

.

The question asks for the sum of the coordinates of the solution.

Key Concept

Systems of linear equations with parameter constraints

Question 29Question →

A theater sells adult tickets for $x$ dollars each and child tickets for $y$ dollars each. For a morning show, the theater sold 2 adult tickets and 1 child ticket, collecting a total of $14$ dollars. For an afternoon show, the theater sold 3 adult tickets and 2 child tickets, collecting a total of $23$ dollars. What is the cost, in dollars, of 1 adult ticket and 1 child ticket combined?

5

6

9

19

Show answer & explanation

Answer: 9

Answer

The combined cost of 1 adult ticket and 1 child ticket is 9 dollars.

The correct answer is 9. Subtracting the equation representing the morning show (

2x + y = 14

) from the equation representing the afternoon show (

3x + 2y = 23

) directly yields

x + y = 9

. Alternatively, solving the system of equations by substitution or elimination gives

x = 5

(cost of an adult ticket) and

y = 4

(cost of a child ticket), and adding these two values results in

9

.

Step-by-Step Solution

1

Write the system of linear equations based on the problem description.

The system of equations is:

\begin{aligned} 2x + y &= 14 \\ 3x + 2y &= 23 \end{aligned}

To mathematically represent the ticket sales and total revenue for the morning and afternoon shows.

2

Subtract the first equation from the second equation to find the value of

x + y

directly.

(3x + 2y) - (2x + y) = 23 - 14 \implies x + y = 9

Subtracting the equations isolates the expression

x + y

immediately, which represents the combined cost of 1 adult ticket and 1 child ticket.

Key Concept

Solving systems of linear equations by elimination or subtraction to find a linear combination of variables.

Alternative Method

Solve the system using elimination: multiply the first equation by 2 to get

4x + 2y = 28

. Subtract the second equation (

3x + 2y = 23

) from this new equation to find

x = 5

. Substitute

x = 5

into the first equation to find

y = 4

. Finally, add the values to get

x + y = 9

.

Estimated Time:1m 30s

Question 30Question →

In the system of linear equations below,

a

,

b

, and

c

are constants.

\begin{aligned} \frac{3}{2}x - \frac{2}{3}y &= 12 \\ ax + by &= c \end{aligned}

If the system has infinitely many solutions, and the point

(4, b)

lies on the line represented by the second equation, what is the value of

c

?

-162

72

108

162

Show answer & explanation

Answer: 162

Answer

The correct answer is 162.

The correct answer is 162. Since the system of linear equations has infinitely many solutions, the two equations represent the same line. Therefore, any point that lies on the second line must also lie on the first line. Substituting the coordinates of the point (4, b) into the first equation gives 3/2(4) - 2/3(b) = 12, which simplifies to 6 - 2/3(b) = 12. Solving this equation for b yields b = -9. Since the equations represent the same line, the ratio of their corresponding coefficients must be equal: c / 12 = b / (-2/3). Solving this proportion for c gives c = -18b. Substituting the value of b = -9 into this relation yields c = -18(-9) = 162.

Step-by-Step Solution

1

Determine the relationship between the two equations based on the number of solutions.

Since the system has infinitely many solutions, the two equations represent coincident lines in the coordinate plane. Therefore, any point that lies on the second line must also satisfy the first equation.

Infinitely many solutions in a system of two linear equations indicate that they represent the same line.

2

Substitute the point

(4, b)

into the first equation to solve for

b

.

\frac{3}{2}(4) - \frac{2}{3}b = 12 \implies 6 - \frac{2}{3}b = 12 \implies -\frac{2}{3}b = 6 \implies b = -9

Plugging the coordinates of the point into the first equation allows us to find the value of the unknown coordinate.

3

Set up the ratio of corresponding coefficients for the coincident lines.

\frac{a}{\frac{3}{2}} = \frac{b}{-\frac{2}{3}} = \frac{c}{12}

Equivalent equations must have proportional coefficients and constant terms.

4

Solve for

c

using the ratio containing

b

and

c

.

\frac{c}{12} = \frac{b}{-\frac{2}{3}} \implies c = -18b \implies c = -18(-9) = 162

Substituting the solved value of

b = -9

into the proportion yields the value of

c

.

Key Concept

Systems of Linear Equations with Infinitely Many Solutions

Question 31Question →

A local library has two types of study rooms: small rooms, which can accommodate up to $3$ people, and large rooms, which can accommodate up to $8$ people. The library has a total of $15$ study rooms. If the maximum capacity of all the study rooms combined is $80$ people, how many large study rooms does the library have?

Show answer & explanation

Answer: 7

Answer

The correct answer is 7.

By representing the number of small rooms as

s

and large rooms as

l

, we set up the system of linear equations

s + l = 15

and

3s + 8l = 80

. Solving the first equation for

s

gives

s = 15 - l

. Substituting this into the second equation yields

3(15 - l) + 8l = 80

. Distributing and combining like terms results in

45 + 5l = 80

. Subtracting 45 from both sides gives

5l = 35

, which simplifies to

l = 7

. Thus, there are 7 large study rooms.

Step-by-Step Solution

1

Set up the system of linear equations representing the total number of rooms and their total capacity.

s + l = 15

and

3s + 8l = 80

We define

s

as the number of small study rooms and

l

as the number of large study rooms. The total number of rooms is 15, and the total capacity is 80.

2

Express the number of small rooms,

s

, in terms of the number of large rooms,

l

, using the first equation.

s = 15 - l

This allows for substitution into the capacity equation so that we can solve directly for the number of large study rooms.

3

Substitute the expression for

s

into the capacity equation and solve the resulting single-variable equation for

l

.

3(15 - l) + 8l = 80 \implies 45 - 3l + 8l = 80 \implies 45 + 5l = 80 \implies 5l = 35 \implies l = 7

Substituting

s

eliminates one variable, leaving a linear equation in terms of

l

that can be solved using standard algebraic steps.

Key Concept

Solving systems of linear equations in a real-world context using substitution.

Question 32Question →

For the constants $p$ and $q$ , the given system of linear equations in $x$ and $y$ has infinitely many solutions:

\begin{aligned} px - 3y &= q \\ 4x + (p - 7)y &= 12 \end{aligned}

Which of the following is a possible value of $q$ ?

3

6

9

16

Show answer & explanation

Answer: 9

Answer

9

For the system of linear equations to have infinitely many solutions, the equations must be equivalent. This means the ratio of the coefficients of

x

, the coefficients of

y

, and the constant terms must all be equal:

\frac{p}{4} = \frac{-3}{p-7} = \frac{q}{12}

. Cross-multiplying the first two ratios gives

p(p-7) = -12

, which simplifies to

p^2 - 7p + 12 = 0

. Factoring this quadratic equation gives

(p-3)(p-4) = 0

, which yields

p = 3

or

p = 4

. Using the relation between the first and third ratios, we have

\frac{q}{12} = \frac{p}{4}

, which simplifies to

q = 3p

. Substituting the possible values of

p

gives

q = 3(3) = 9

or

q = 3(4) = 12

. Since 9 is one of the options, it is a possible value of

q

.

Step-by-Step Solution

1

Set up the condition for infinitely many solutions by equating the ratios of the coefficients and the constants.

The coefficients and constants must satisfy the proportion:

\frac{p}{4} = \frac{-3}{p - 7} = \frac{q}{12}

For a system of two linear equations to have infinitely many solutions, the two equations must represent the exact same line, meaning all corresponding coefficients and constant terms must be proportional.

2

Solve the equation formed by the first two ratios to find the possible values of the parameter

p

.

\frac{p}{4} = \frac{-3}{p - 7} \Rightarrow p(p - 7) = -12 \Rightarrow p^2 - 7p + 12 = 0

. Factoring the quadratic yields

(p - 3)(p - 4) = 0

, which gives

p = 3

or

p = 4

.

This step determines the values of

p

for which the two lines are parallel (i.e., they have equal slopes).

3

Find the corresponding values of

q

using the relation between the first and third ratios.

\frac{p}{4} = \frac{q}{12} \Rightarrow q = 3p

. Substituting the values of

p

:
- If

p = 3

, then

q = 3(3) = 9

.
- If

p = 4

, then

q = 3(4) = 12

.

This step ensures that the parallel lines are coincident by equating their y-intercepts.

4

Compare the possible values of

q

with the given options.

The value 9 is a possible value of

q

and matches one of the choices.

To identify which of the two mathematically valid solutions for

q

is listed in the multiple-choice options.

Key Concept

Conditions for infinitely many solutions in a system of linear equations

Alternative Method

Instead of using ratios directly, we can write both equations in slope-intercept form:

y = mx + b

. For the first equation,

px - 3y = q \Rightarrow y = \frac{p}{3}x - \frac{q}{3}

. For the second equation,

4x + (p-7)y = 12 \Rightarrow y = -\frac{4}{p-7}x + \frac{12}{p-7}

(assuming

p \neq 7

). For the system to have infinitely many solutions, the two lines must have the same slope and the same y-intercept. Equating the slopes gives

\frac{p}{3} = -\frac{4}{p-7}

, which simplifies to

p^2 - 7p + 12 = 0

, yielding

p = 3

or

p = 4

. Equating the y-intercepts gives

-\frac{q}{3} = \frac{12}{p-7} \Rightarrow q = -\frac{36}{p-7}

. Substituting

p=3

gives

q = -\frac{36}{-4} = 9

, and substituting

p=4

gives

q = -\frac{36}{-3} = 12

.

Estimated Time:3m 0s

Question 33Question →

The system of linear equations shown below contains constants

a

and

b

.

\begin{aligned} \frac{1}{2}(ax - 4y) &= x - \frac{1}{2}y + 6 \\ 2x - y &= b \end{aligned}

If the system has infinitely many solutions, what is the value of

a + b

?

Show answer & explanation

Answer: 12

Answer

The correct answer is 12.

To find the value of

a + b

that yields infinitely many solutions, we rewrite both equations in the standard form

Ax + By = C

. Simplifying the first equation gives

(a - 2)x - 3y = 12

. Multiplying the second equation

2x - y = b

by

3

gives

6x - 3y = 3b

. For the system to have infinitely many solutions, the two equations must be equivalent, meaning

a - 2 = 6

(which gives

a = 8

) and

3b = 12

(which gives

b = 4

). The sum of these values is

8 + 4 = 12

.

Step-by-Step Solution

1

Distribute the fraction

\frac{1}{2}

on the left side of the first equation.

\frac{1}{2}ax - 2y = x - \frac{1}{2}y + 6

To expand the expression and prepare it for simplification.

2

Group the

x

and

y

terms on the left side of the equation and the constants on the right side.

\left(\frac{1}{2}a - 1\right)x - \frac{3}{2}y = 6

To write the equation in standard linear form.

3

Multiply the entire equation by

2

to eliminate the fractional coefficients.

(a - 2)x - 3y = 12

To simplify comparison with the second equation by working with integer coefficients.

4

Multiply the second equation,

2x - y = b

, by

3

to align the

y

-coefficients with the first equation.

6x - 3y = 3b

Two linear equations have infinitely many solutions if they represent the same line, which requires matching coefficients and constants.

5

Equate the corresponding

x

-coefficients and constant terms from

(a - 2)x - 3y = 12

and

6x - 3y = 3b

.

a - 2 = 6 \implies a = 8

and

3b = 12 \implies b = 4

To solve for the values of the constants

a

and

b

.

6

Calculate the sum of

a

and

b

.

8 + 4 = 12

To find the final requested value of

a + b

.

Key Concept

Determining parameters for infinitely many solutions in a system of linear equations

Question 34Question →

\begin{aligned} x - 2y &= 5 \\ 3x - y &= 10 \end{aligned}

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

2

3

6

8

Show answer & explanation

Answer: 2

Answer

2

The correct answer is 2. Isolating

x

in the first equation gives

x = 2y + 5

. Substituting this expression into the second equation yields

3(2y + 5) - y = 10

. Distributing the 3 gives

6y + 15 - y = 10

, which simplifies to

5y + 15 = 10

. Subtracting 15 from both sides gives

5y = -5

, so

y = -1

. Substituting

y = -1

back into the equation for

x

gives

x = 2(-1) + 5 = 3

. The sum of the solutions is

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

.

Step-by-Step Solution

1

Isolate

x

in the first equation.

x = 2y + 5

Isolating one variable makes it straightforward to substitute it into the other equation.

2

Substitute the expression for

x

into the second equation and solve for

y

.

y = -1

Substituting

x = 2y + 5

gives

3(2y + 5) - y = 10

. Distributing the 3 yields

6y + 15 - y = 10

. Combining like terms gives

5y + 15 = 10

. Subtracting 15 from both sides gives

5y = -5

, so

y = -1

.

3

Substitute

y = -1

back into the expression for

x

to find its value.

x = 3

Using the isolated expression

x = 2y + 5

, substituting

y = -1

gives

x = 2(-1) + 5 = -2 + 5 = 3

.

4

Calculate the sum

x + y

.

2

The question asks for the value of

x + y

, which is

3 + (-1) = 2

.

Key Concept

Solving systems of linear equations in two variables using substitution or elimination, and evaluating algebraic combinations of the variables.

Alternative Method

The system can also be solved by elimination. Multiplying the second equation by 2 gives

6x - 2y = 20

. Subtracting the first equation (

x - 2y = 5

) from this new equation eliminates

y

, resulting in

5x = 15

, or

x = 3

. Substituting

x = 3

into either original equation yields

y = -1

. Finally, the sum is

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

.

Estimated Time:1m 30s

Question 35Question →

In the $xy$ -plane, the system of linear equations below has infinitely many solutions, where $a$ , $b$ , and $c$ are constants and $c > 0$ :

\begin{aligned} 3x - 4y &= 10 \\ ax + by &= c \end{aligned}

If the graph of the second equation in the system passes through the point $(a, b)$ , what is the value of $c$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is 4.

Since the system has infinitely many solutions, the two equations are equivalent. This means the coefficients are proportional, so we can write

a = 3k

,

b = -4k

, and

c = 10k

for some constant

k

. Because the line passes through

(a, b)

, substituting these coordinates into the second equation gives

a^2 + b^2 = c

. Substituting the expressions in terms of

k

results in

(3k)^2 + (-4k)^2 = 10k

, which simplifies to

25k^2 = 10k

. Since

c > 0

, we have

k > 0

, and dividing by

k

gives

25k = 10

, so

k = 0.4

. Finally,

c = 10(0.4) = 4

.

Step-by-Step Solution

1

Set up a proportionality constant to relate the coefficients of the two equations.

a = 3k

,

b = -4k

, and

c = 10k

for a constant

k

.

Since the system has infinitely many solutions, the equations represent the same line, meaning their coefficients and constants must be proportional.

2

Substitute the point

(a, b)

into the second equation

ax + by = c

.

a^2 + b^2 = c

The graph of the equation passes through the point

(a, b)

, so the coordinates must satisfy the equation.

3

Substitute the parametric expressions of

a

,

b

, and

c

into the equation

a^2 + b^2 = c

.

25k^2 = 10k

This allows us to solve for the parameter

k

using a single variable quadratic equation.

4

Solve the equation

25k^2 = 10k

for

k

, given that

k > 0

.

k = 0.4

Since

c > 0

and

c = 10k

,

k

must be strictly positive, allowing us to divide both sides by

k

.

5

Compute the final value of

c

using the value of

k

.

c = 4

Substituting

k = 0.4

back into the expression

c = 10k

gives the value of

c

.

Key Concept

Systems of linear equations with infinitely many solutions and coordinate geometry constraints

Question 36Question →

Consider the system of linear equations below, where $k$ is a constant:

\begin{aligned} 3x - 2y &= 7 \\ kx + 3y &= 18 \end{aligned}

If the system has a solution $(x, y)$ such that $x + y = 4$ , what is the value of $k$ ?

3

5

9

1

Show answer & explanation

Answer: 5

Answer

5

The correct answer is

5

. To find the value of

k

, we first express

x

in terms of

y

using the given constraint equation:

x = 4 - y

. Next, we substitute this expression into the first equation of the system:

3(4 - y) - 2y = 7

. Distributing and combining like terms yields

12 - 5y = 7

. Solving for

y

gives

y = 1

. Substituting this back into the constraint equation gives

x = 4 - 1 = 3

. Finally, we substitute the solution point

(3, 1)

into the second equation:

k(3) + 3(1) = 18 \implies 3k + 3 = 18 \implies 3k = 15 \implies k = 5

.

Step-by-Step Solution

1

Express one variable in terms of the other using the given constraint equation

x + y = 4

.

x = 4 - y

This allows us to substitute the expression for

x

into the first equation to reduce it to a single variable.

2

Substitute

x = 4 - y

into the first equation

3x - 2y = 7

and solve for

y

.

3(4-y) - 2y = 7 \implies 12 - 3y - 2y = 7 \implies 12 - 5y = 7 \implies -5y = -5 \implies y = 1

By solving this linear equation, we find the unique y-coordinate of the system's solution.

3

Substitute the value of

y = 1

back into the constraint equation to find

x

.

x = 4 - 1 = 3

This determines the x-coordinate of the solution point, giving us the full solution

(3, 1)

.

4

Substitute the point

(3, 1)

into the second equation

kx + 3y = 18

to solve for the constant

k

.

k(3) + 3(1) = 18 \implies 3k + 3 = 18 \implies 3k = 15 \implies k = 5

Since the solution must satisfy all equations in the system, we can solve for

k

using the coordinates we found.

Key Concept

Solving systems of linear equations under linear constraints using algebraic substitution.

Estimated Time:2m 0s

Question 37Question →

Consider the system of equations below:

\begin{aligned} \frac{1}{3}x + \frac{1}{4}y &= 5 \\ x - y &= 8 \end{aligned}

If

(x, y)

is the solution to the system, what is the value of

y

?

Show answer & explanation

Answer: 4

Answer

The value of

y

is

4

.

Substituting

x = y + 8

from the second equation into the first equation yields

\frac{1}{3}(y + 8) + \frac{1}{4}y = 5

. Multiplying the entire equation by

12

to clear denominators gives

4(y + 8) + 3y = 60

, which simplifies to

7y + 32 = 60

. Solving for

y

gives

7y = 28

, or

y = 4

.

Step-by-Step Solution

1

Express

x

in terms of

y

from the second equation.

x = y + 8

This allows for substitution into the first equation to solve for

y

directly.

2

Substitute

x = y + 8

into the first equation.

_\frac{1}{3}(y + 8) + \frac{1}{4}y = 5

This reduces the system to a single-variable linear equation in terms of

y

.

3

Multiply the entire equation by the least common multiple of the denominators, which is

12

.

4(y + 8) + 3y = 60

This clears the fractional coefficients and simplifies the arithmetic.

4

Distribute the

4

and combine like terms.

7y + 32 = 60

Simplifies the equation to prepare for isolating the variable

y

.

5

Isolate the variable term by subtracting

32

from both sides, then dividing by

7

.

y = 4

This gives the final value of

y

that satisfies the system.

Key Concept

Solving systems of linear equations using substitution and clearing fractional coefficients

Question 380% of students answered this correctlyQuestion →

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

\begin{aligned} 2x + 3y &= 12 \\ cx - y &= 6 \end{aligned}

If the system has a unique solution $(x, y)$ such that $x > 0$ and $y < 0$ , which of the following could be the value of $c$ ?

-2

-1

\frac{1}{2}

2

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The option containing the value

\frac{1}{2}

is correct.

The correct answer is the option containing the value

\frac{1}{2}

. Solving the system of linear equations in terms of the constant

c

gives

x = \frac{30}{2+3c}

and

y = \frac{12c-12}{2+3c}

. For the solution

(x, y)

to lie in the fourth quadrant, we require

x > 0

and

y < 0

. The condition

x > 0

is satisfied when

2+3c > 0

, which simplifies to

c > -\frac{2}{3}

. Using this result, the condition

y < 0

requires the numerator of

y

to be negative, so

12c - 12 < 0

, which simplifies to

c < 1

. Combining these inequalities yields the interval

-\frac{2}{3} < c < 1

. The only value among the given options that falls within this interval is

\frac{1}{2}

.

Step-by-Step Solution

1

Express

y

in terms of

x

and the constant

c

using the second equation.

y = cx - 6

This allows for substitution into the first equation to solve for

x

.

2

Substitute the expression for

y

into the first equation and solve for

x

.

2x + 3(cx - 6) = 12 \implies (2 + 3c)x - 18 = 12 \implies (2 + 3c)x = 30 \implies x = \frac{30}{2 + 3c}

This isolates the variable

x

in terms of the constant

c

.

3

Substitute the expression for

x

back into the equation for

y

to express

y

in terms of

c

.

y = c\left(\frac{30}{2 + 3c}\right) - 6 = \frac{30c - 6(2 + 3c)}{2 + 3c} = \frac{12c - 12}{2 + 3c}

This isolates the variable

y

in terms of the constant

c

.

4

Apply the condition

x > 0

to find the constraint on

c

.

Since

x = \frac{30}{2 + 3c} > 0

and the numerator is positive, the denominator must also be positive:

2 + 3c > 0 \implies c > -\frac{2}{3}

.

This determines the lower bound for the constant

c

.

5

Apply the condition

y < 0

to find the constraint on

c

.

Since

y = \frac{12c - 12}{2 + 3c} < 0

and the denominator

2 + 3c

is positive, the numerator must be negative:

12c - 12 < 0 \implies c < 1

.

This determines the upper bound for the constant

c

.

6

Combine the inequalities to find the complete range for

c

and identify the matching option.

-\frac{2}{3} < c < 1

. The only value among the options that lies in this interval is

\frac{1}{2}

.

This identifies the correct option based on the mathematical constraints.

Key Concept

Solving systems of linear equations with parameters and applying quadrant boundary constraints.

Question 39Question →

A nutritionist is designing a diet plan using two food types, Food A and Food B. The table below shows the amount of protein and carbohydrates, in grams, per gram of each food type:

Food Type	Protein (g)	Carbohydrates (g)
Food A	$0.12$	$0.08$
Food B	$0.18$	$0.05$

The nutritionist wants to combine the two foods to obtain a mixture that has a total mass of $250$ grams and contains exactly $36$ grams of protein. How many grams of carbohydrates will be in this mixture?

Show answer & explanation

Answer: 17

Answer

The total amount of carbohydrates in the mixture is 17 grams.

To find the total amount of carbohydrates in the mixture, we first define

x

as the mass in grams of Food A and

y

as the mass in grams of Food B. Since the total mass of the mixture is

250

grams, we write the equation

x + y = 250

. Since the total protein in the mixture must be

36

grams, and Food A and Food B contain

0.12

grams and

0.18

grams of protein per gram respectively, we write the equation

0.12x + 0.18y = 36

. Multiplying the first equation by

12

yields

12x + 12y = 3000

, and multiplying the second equation by

100

yields

12x + 18y = 3600

. Subtracting the first new equation from the second gives

6y = 600

, which simplifies to

y = 100

. Substituting this back into the first equation gives

x + 100 = 250

, so

x = 150

. Using these values, the total amount of carbohydrates is

0.08(150) + 0.05(100) = 12 + 5 = 17

grams.

Step-by-Step Solution

1

Define variables for the mass of each food type.

Let

x

be the mass, in grams, of Food A, and let

y

be the mass, in grams, of Food B.

To represent the unknown quantities in the system of equations.

2

Set up the system of linear equations.

x + y = 250

(total mass equation) and

0.12x + 0.18y = 36

(total protein equation).

The mixture has a total mass of

250

grams and must contain exactly

36

grams of protein.

3

Solve the system of equations for

x

and

y

.

Multiply the first equation by

12

to get

12x + 12y = 3000

. Multiply the second equation by

100

to get

12x + 18y = 3600

. Subtract the first modified equation from the second to get

6y = 600

, which gives

y = 100

. Substitute

y = 100

back into

x + y = 250

to get

x = 150

.

To find the specific masses of Food A and Food B required for the mixture.

4

Calculate the total mass of carbohydrates in the mixture.

Total carbohydrates

= 0.08x + 0.05y = 0.08(150) + 0.05(100) = 12 + 5 = 17

grams.

The question asks for the total grams of carbohydrates in the final mixture.

Key Concept

Systems of Linear Equations

Question 40Question →

Consider the system of linear equations below, where $k$ is a constant:

\begin{aligned} ky - 3x &= 4 \\ 4y - (k-1)x &= 3 \end{aligned}

If the system of equations has no solution, what is the sum of all possible values of $k$ ?

1

-1

4

7

Show answer & explanation

Answer: 1

Answer

1

For the system of equations to have no solution, the lines must be parallel. In slope-intercept form, the equations are

y = \frac{3}{k}x + \frac{4}{k}

and

y = \frac{k-1}{4}x + \frac{3}{4}

. Setting the slopes equal gives

\frac{3}{k} = \frac{k-1}{4}

, which simplifies to the quadratic equation

k^2 - k - 12 = 0

. Factoring this equation yields

(k-4)(k+3) = 0

, giving

k = 4

and

k = -3

. Since both values yield different y-intercepts for the two lines, they both result in parallel lines with no intersection. The sum of these values is

4 + (-3) = 1

.

Step-by-Step Solution

1

Express both equations in slope-intercept form (

y = mx + b

) to find their slopes.

The first equation becomes

y = \frac{3}{k}x + \frac{4}{k}

(for

k \neq 0

). The second equation becomes

y = \frac{k-1}{4}x + \frac{3}{4}

.

For a system of linear equations to have no solution, the lines must be parallel, meaning they have the same slope but different y-intercepts.

2

Set the slopes of the two lines equal to each other to solve for

k

.

\frac{3}{k} = \frac{k-1}{4} \implies k(k-1) = 12 \implies k^2 - k - 12 = 0

.

Equating the slopes allows us to find the values of

k

that make the lines parallel.

3

Solve the quadratic equation

k^2 - k - 12 = 0

by factoring.

(k-4)(k+3) = 0 \implies k = 4

or

k = -3

.

This determines the specific values of

k

that make the slopes equal.

4

Verify that both values of

k

produce different y-intercepts (so the lines do not coincide) and calculate their sum.

For

k = 4

, the y-intercepts are

\frac{4}{4} = 1

and

\frac{3}{4}

, which are different. For

k = -3

, the y-intercepts are

-\frac{4}{3}

and

\frac{3}{4}

, which are also different. The sum of the values of

k

is

4 + (-3) = 1

.

We must confirm the lines are parallel and not identical, then compute the required sum.

Key Concept

Systems of linear equations with no solution represent parallel lines with equal slopes and unequal y-intercepts.

Systems of Linear Equations

Answer

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