Systems of Linear Equations Practice Questions — GRE General Test

Question 1Question →

If $x + y = 8$ and $2x - y = 7$ , what is the value of $x$ ?

5

3

1

15

11

Show answer & explanation

Answer: 5

Answer

5

Adding the two equations yields

(x + y) + (2x - y) = 8 + 7

, which simplifies to

3x = 15

. Dividing both sides by 3 gives the correct value,

x = 5

.

Step-by-Step Solution

1

Add the two equations together to eliminate the variable

y

.

(x + y) + (2x - y) = 8 + 7 \implies 3x = 15

Since the coefficients of

y

are

+1

and

-1

, adding the equations eliminates

y

directly.

2

Solve for

x

by dividing both sides of the equation by 3.

x = \frac{15}{3} = 5

Isolating

x

gives the required solution.

Key Concept

Elimination Method in Systems of Linear Equations

Question 2Question →

Consider the system of linear equations in two variables

x

and

y

:

\begin{aligned} 2x - 3y &= a \\ 4x - 6y &= b \end{aligned}

where

a

and

b

are real constants. Which of the following statements MUST be true? Select all that apply.

Select all that apply

If

b = 2a

, the system has infinitely many solutions.

If

b \neq 2a

, the system has no solution.

If

a = 0

and

b = 0

, the system has a unique solution at

(0, 0)

.

There exist real values of

a

and

b

for which the system has exactly one solution.

If

b = 2a + 1

, the lines represented by the equations in the

xy

-plane are parallel and distinct.

Show answer & explanation

Answer: If $b = 2a$ , the system has infinitely many solutions.; If $b \neq 2a$ , the system has no solution.; If $b = 2a + 1$ , the lines represented by the equations in the $xy$ -plane are parallel and distinct.

Answer

The correct statements are: 'If

b = 2a

, the system has infinitely many solutions.', 'If

b \neq 2a

, the system has no solution.', and 'If

b = 2a + 1

, the lines represented by the equations in the

xy

-plane are parallel and distinct.'

Statements asserting that the system has infinitely many solutions when

b = 2a

, no solution when

b \neq 2a

, and parallel distinct lines when

b = 2a + 1

are all mathematically sound. Multiplying the first equation by 2 reveals that the left sides are identical (

4x - 6y

). Equality of the right sides (

b = 2a

) makes the lines identical, whereas inequality (

b \neq 2a

) makes them parallel and distinct.

Step-by-Step Solution

1

Analyze the coefficients of the system of equations.

The coefficients of

x

and

y

in the second equation (

4

and

-6

) are exactly twice the coefficients of

x

and

y

in the first equation (

2

and

-3

).

Comparing coefficient ratios determines line relationships (slopes).

2

Multiply the first equation by 2.

2(2x - 3y) = 2(a) \implies 4x - 6y = 2a

.

This puts the left side of the first equation in exact alignment with the second equation (

4x - 6y = b

).

3

Evaluate the condition for infinitely many solutions.

If

b = 2a

, the two equations become

4x - 6y = 2a

and

4x - 6y = 2a

, which describe the exact same line, giving infinitely many solutions.

Coincident lines intersect at every point along the line.

4

Evaluate the condition for no solution.

If

b \neq 2a

, subtracting the equations gives

0 = b - 2a \neq 0

, a contradiction. Hence, the lines are parallel and distinct, meaning no solution exists.

Distinct parallel lines never intersect.

5

Examine specific cases such as

b = 2a + 1

and

a=0, b=0

.

For

b = 2a + 1

, since

2a + 1 \neq 2a

,

b \neq 2a

, confirming parallel distinct lines. For

a=0, b=0

,

b = 2(0) = 0

, which yields infinitely many solutions rather than a unique solution.

Verifies specific option claims.

Key Concept

Systems of Linear Equations (Solvability and Geometric Interpretation)

Question 3Question →

If the pair $(x, y)$ satisfies the system of linear equations below, what is the value of $x - y$ ?

\begin{aligned} 3x - 5y &= 14 \\ 2x + 7y &= -1 \end{aligned}

-1

2

3

4

5

Show answer & explanation

Answer: 4

Answer

The value of

x - y

is

4

.

Solving the system of linear equations by elimination gives

x = 3

and

y = -1

. Substituting these values into the target expression

x - y

yields

3 - (-1) = 3 + 1 = 4

.

Step-by-Step Solution

1

Eliminate one variable using the elimination method.

Multiply the first equation by

2

and the second equation by

3

:

6x - 10y = 28

6x + 21y = -3

Creating matching coefficients for

x

allows elimination of

x

by subtraction.

2

Subtract the first modified equation from the second modified equation to solve for

y

.

(6x + 21y) - (6x - 10y) = -3 - 28

31y = -31 \implies y = -1

Subtracting eliminates

x

, isolating

y

.

3

Substitute

y = -1

back into one of the original equations to solve for

x

.

3x - 5(-1) = 14 \implies 3x + 5 = 14 \implies 3x = 9 \implies x = 3

Plugging in

y

allows direct solution for

x

.

4

Evaluate the requested expression

x - y

.

x - y = 3 - (-1) = 3 + 1 = 4

Subtracting a negative value is equivalent to adding its positive value.

Key Concept

Solving systems of linear equations using elimination and evaluating algebraic expressions.

Estimated Time:1m 30s

Question 4Question →

A chemistry laboratory prepares a $100\text{ mL}$ mixture using three solutions: Solution $X$ ( $10\%$ acid by volume), Solution $Y$ ( $30\%$ acid by volume), and Solution $Z$ ( $50\%$ acid by volume). The resulting mixture is $37\%$ acid by volume. If the volume of Solution $Y$ used is $5\text{ mL}$ more than twice the volume of Solution $X$ , what is the volume, in $\text{mL}$ , of Solution $Z$ used in the mixture?

15

35

40

50

55

Show answer & explanation

Answer: $50$

Answer

50\text{ mL}

Setting up the system of equations gives

x + y + z = 100

,

x + 3y + 5z = 370

, and

y = 2x + 5

. Substituting

y = 2x + 5

into the first two equations yields

3x + z = 95

and

7x + 5z = 355

. Solving for

x

gives

x = 15

, which leads to

z = 95 - 3(15) = 50

. Therefore,

50\text{ mL}

of Solution Z was used.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

x

,

y

, and

z

be the volumes in

\text{mL}

of Solutions

X

,

Y

, and

Z

, respectively.
1) Total volume:

x + y + z = 100

2) Total acid volume:

0.10x + 0.30y + 0.50z = 0.37(100) \implies x + 3y + 5z = 370

3) Relationship between

Y

and

X

:

y = 2x + 5

Translate the word problem statements into mathematical equations.

2

Substitute

y = 2x + 5

into equations (1) and (2) to reduce to a two-variable system

From equation (1):

x + (2x + 5) + z = 100 \implies 3x + z = 95 \implies z = 95 - 3x

From equation (2):

x + 3(2x + 5) + 5z = 370 \implies 7x + 15 + 5z = 370 \implies 7x + 5z = 355

Eliminating

y

simplifies the system to two equations in

x

and

z

.

3

Substitute

z = 95 - 3x

into

7x + 5z = 355

and solve for

x

7x + 5(95 - 3x) = 355 \implies 7x + 475 - 15x = 355 \implies -8x = -120 \implies x = 15

Solves for the unknown volume of Solution X.

4

Calculate

z

using

z = 95 - 3x

z = 95 - 3(15) = 95 - 45 = 50

Finds the requested volume of Solution Z.

Key Concept

Setting up and solving 3x3 systems of linear equations using substitution or elimination

Estimated Time:2m 0s

Question 5Question →

Consider the following system of linear equations in variables

x

and

y

, where

k

is a real constant:

\begin{aligned} kx + 4y &= 8 \\ x + ky &= k + 2 \end{aligned}

Which of the following statements are true? Select all that apply.

Select all that apply

If

k = 2

, the system has infinitely many solutions.

If

k = -2

, the system has no solutions.

If

k = 0

, the unique solution to the system is

(x, y) = (2, 2)

.

If

k = 3

, the system has no solutions.

If

k = 1

, the solution to the system satisfies

x + y = 4

.

Show answer & explanation

Answer: If $k = 2$ , the system has infinitely many solutions.; If $k = -2$ , the system has no solutions.; If $k = 0$ , the unique solution to the system is $(x, y) = (2, 2)$ .

Answer

The correct statements are that setting

k = 2

results in infinitely many solutions, setting

k = -2

results in no solutions, and setting

k = 0

yields the unique solution

(2, 2)

.

The system has a coefficient matrix determinant of

k^2 - 4

. Setting

k = 2

produces identical equations (

x + 2y = 4

), giving infinitely many solutions. Setting

k = -2

produces parallel equations with different constants (

-x + 2y = 4

vs

-x + 2y = 0

), giving no solutions. Setting

k = 0

reduces the system directly to

y = 2

and

x = 2

, confirming the unique point

(2, 2)

.

Step-by-Step Solution

1

Analyze the determinant of the coefficient matrix to identify conditions for unique vs. non-unique solutions.

The coefficient matrix determinant is

\Delta = k\cdot k - 4\cdot 1 = k^2 - 4 = (k - 2)(k + 2)

.

If

\Delta \neq 0

(i.e.,

k \neq \pm 2

), the system has a unique solution. If

\Delta = 0

(i.e.,

k = 2

or

k = -2

), the lines are either identical or parallel.

2

Test

k = 2

in the system.

Equation 1 becomes

2x + 4y = 8 \implies x + 2y = 4

. Equation 2 becomes

x + 2y = 4

.

Since the equations are identical, the lines coincide, resulting in infinitely many solutions.

3

Test

k = -2

in the system.

Equation 1 becomes

-2x + 4y = 8 \implies -x + 2y = 4

. Equation 2 becomes

x - 2y = 0 \implies -x + 2y = 0

.

The slopes are equal (

1/2

) but the y-intercepts differ (

2

vs

0

), meaning the lines are parallel and distinct, yielding zero solutions.

4

Test

k = 0

and

k = 1

to evaluate remaining choices.

For

k = 0

,

4y = 8 \implies y = 2

and

x + 0 = 2 \implies x = 2

, giving

(2, 2)

. For

k = 1

, Equation 2 is directly

x + y = 3

.

This verifies that the statement for

k = 0

is correct, while the statement for

k = 1

claiming

x + y = 4

is false.

Key Concept

Parametric Linear Systems and Conditions for Solvability

Question 6Question →

A logistics warehouse uses two sizes of shipping crates: small crates and large crates. A shipment containing $3$ small crates and $5$ large crates has a total weight of $47$ pounds. A second shipment containing $6$ small crates and $2$ large crates has a total weight of $38$ pounds. What is the total weight, in pounds, of a shipment containing $2$ small crates and $3$ large crates?

11

22

26

29

33

Show answer & explanation

Answer: 29

Answer

29 pounds

By setting up the linear system

3s + 5l = 47

and

6s + 2l = 38

, using elimination gives

s = 4

pounds for a small crate and

l = 7

pounds for a large crate. Substituting these values into the target expression

2s + 3l

yields

2(4) + 3(7) = 29

pounds.

Step-by-Step Solution

1

Set up a system of linear equations from the given context.

Let

s

be the weight of a small crate and

l

be the weight of a large crate. The equations are: (1)

3s + 5l = 47

and (2)

6s + 2l = 38

.

Translate the physical constraints of each shipment into algebraic equations.

2

Solve for the variable

l

by eliminating

s

.

Multiply Equation (1) by

2

to get

6s + 10l = 94

. Subtract Equation (2) (

6s + 2l = 38

) from this new equation:

(6s + 10l) - (6s + 2l) = 94 - 38 \implies 8l = 56 \implies l = 7

.

Align the coefficients of

s

so that elimination via subtraction yields a single-variable linear equation for

l

.

3

Substitute

l = 7

back into Equation (2) to solve for

s

.

6s + 2(7) = 38 \implies 6s + 14 = 38 \implies 6s = 24 \implies s = 4

.

Determine the individual weight of a small crate.

4

Calculate the requested total weight for

2

small crates and

3

large crates.

2s + 3l = 2(4) + 3(7) = 8 + 21 = 29

.

Substitute the individual values of

s

and

l

into the target expression.

Key Concept

Solving 2x2 Systems of Linear Equations by Elimination and Linear Combination Evaluation

Estimated Time:1m 30s

Question 7Question →

A chemist mixes $x$ ounces of a 40% acid solution with $y$ ounces of a 70% acid solution to produce a 20-ounce mixture that is 52% acid. What is the value of $x - y$ ?

-4

4

8

12

20

Show answer & explanation

Answer: 4

Answer

4

Setting up the linear system

x + y = 20

and

0.40x + 0.70y = 10.4

leads to

x = 12

and

y = 8

. Subtracting

y

from

x

yields

12 - 8 = 4

.

Step-by-Step Solution

1

Set up the system of linear equations based on total solution volume and pure acid content.

Total volume equation:

x + y = 20

. Pure acid equation:

0.40x + 0.70y = 0.52(20) = 10.4

.

The sum of the component volumes equals the total volume, and the sum of the pure acid contents equals the total acid content.

2

Multiply the acid equation by 10 to eliminate decimals.

4x + 7y = 104

.

Clearing decimals simplifies the subsequent elimination calculation.

3

Solve for

y

using the elimination method.

Multiply

x + y = 20

by 4 to get

4x + 4y = 80

. Subtract this from

4x + 7y = 104

:

(4x + 7y) - (4x + 4y) = 104 - 80 \implies 3y = 24 \implies y = 8

.

Eliminating the variable

x

isolates

y

.

4

Solve for

x

and compute

x - y

.

x = 20 - 8 = 12

. Therefore,

x - y = 12 - 8 = 4

.

Substitute

y = 8

back into the first equation and calculate the requested expression.

Key Concept

Systems of Linear Equations in Mixture Problems

Question 8Question →

Consider the following system of linear equations:

\begin{aligned} 2x + 3y &= 12 \\ 4x + 6y &= 24 \end{aligned}

Which of the following statements about this system must be true? Select all that apply.

Select all that apply

The system has infinitely many solutions.

The graphs of the two equations represent the exact same line in the

xy

-plane.

The ordered pair

(3, 2)

is a solution to the system.

The system is inconsistent and has no solution.

The

y

-intercept of the line is

(6, 0)

.

Show answer & explanation

Answer: The system has infinitely many solutions.; The graphs of the two equations represent the exact same line in the $xy$ -plane.; The ordered pair $(3, 2)$ is a solution to the system.

Answer

The correct statements are that the system has infinitely many solutions, the graphs represent the exact same line, and

(3, 2)

is a solution to the system.

Dividing

4x + 6y = 24

by

2

produces

2x + 3y = 12

, showing that both equations represent the exact same line. Therefore, the system has infinitely many solutions. Substituting

x = 3

and

y = 2

yields

2(3) + 3(2) = 12

, confirming that

(3, 2)

is one of the infinitely many valid solutions.

Step-by-Step Solution

1

Analyze the relationship between the two linear equations

Dividing the second equation

4x + 6y = 24

by

2

gives

2x + 3y = 12

, which is identical to the first equation.

Comparing coefficients and constants determines whether equations in a system are dependent, independent, or inconsistent.

2

Determine the number of solutions and geometric structure

Because the equations are mathematically equivalent, they describe the same line in the coordinate plane and have infinitely many intersection points.

Identical linear equations form a dependent system with infinitely many solutions.

3

Test the ordered pair

(3, 2)

Evaluating

2(3) + 3(2) = 6 + 6 = 12

confirms that

(3, 2)

lies on the line.

Any point satisfying one equation satisfies the entire system of equivalent equations.

4

Determine the axis intercepts

Setting

y=0

gives

2x=12 \Rightarrow x=6

(the

x

-intercept is

(6,0)

); setting

x=0

gives

3y=12 \Rightarrow y=4

(the

y

-intercept is

(0,4)

).

Checking coordinates of axis intersections prevents mislabeling

x

- and

y

-intercepts.

Key Concept

Dependent Systems of Linear Equations

Question 9Question →

Consider the following system of linear equations:

\begin{aligned} x + 2y &= 10 \\ 3x - y &= 9 \end{aligned}

Which of the following statements are true? Select all that apply.

Select all that apply

The value of

x

is

4

.

The value of

x + y

is

7

.

The value of

y

is

4

.

The value of

2x + y

is

11

.

The value of

x - y

is

5

.

Show answer & explanation

Answer: The value of $x$ is $4$ .; The value of $x + y$ is $7$ .; The value of $2x + y$ is $11$ .

Answer

The true statements are those asserting that

x = 4

, that

x + y = 7

, and that

2x + y = 11

.

Solving the system of linear equations yields the unique solution

x = 4

and

y = 3

. Substituting these values into the given choices demonstrates that the statements claiming

x = 4

,

x + y = 7

, and

2x + y = 11

are all mathematically correct.

Step-by-Step Solution

1

Express

y

in terms of

x

using the second equation.

y = 3x - 9

Isolating

y

allows for simple substitution into the first equation.

2

Substitute

y = 3x - 9

into the first equation

x + 2y = 10

.

x + 2(3x - 9) = 10 \implies x + 6x - 18 = 10 \implies 7x = 28 \implies x = 4

This yields a single linear equation in terms of

x

.

3

Calculate

y

using

x = 4

.

y = 3(4) - 9 = 3

Substituting

x = 4

back gives the unique solution for

y

.

4

Evaluate the given statements with

(x, y) = (4, 3)

.

x = 4

is true;

x + y = 4 + 3 = 7

is true;

y = 4

is false (

y = 3

);

2x + y = 2(4) + 3 = 11

is true;

x - y = 4 - 3 = 1

is false.

Testing each condition determines which options are valid.

Key Concept

Solving a 2x2 system of linear equations using substitution to evaluate linear expressions.

Estimated Time:1m 0s

Question 10Question →

An investor divides a total of $\$15,000$ between two accounts, Account P and Account Q. Account P earns an annual simple interest rate of $6\%$ , while Account Q earns an annual simple interest rate of $8\%$ . If the total interest earned from both accounts combined after one year is $\$1,020$ , how much more money was invested in Account P than in Account Q?

\$1,500

\$3,000

\$4,500

\$6,000

\$9,000

Show answer & explanation

Answer: $\$3,000$

Answer

\$3,000

Setting up the system of linear equations

p + q = 15,000

and

0.06p + 0.08q = 1,020

yields

p = 9,000

and

q = 6,000

. Subtracting the amount in Account Q from Account P gives

9,000 - 6,000 = 3,000

. Thus,

\$3,000

is the correct answer.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

p

be the amount invested in Account P and

q

be the amount invested in Account Q. The equations are

p + q = 15,000

and

0.06p + 0.08q = 1,020

.

The first equation represents the total investment amount, and the second equation represents the total annual interest earned.

2

Simplify the interest equation and solve for one variable using elimination

Multiply

0.06p + 0.08q = 1,020

by 100 to obtain

6p + 8q = 102,000

, which simplifies to

3p + 4q = 51,000

. Multiplying the first equation by 3 yields

3p + 3q = 45,000

. Subtracting this from

3p + 4q = 51,000

gives

q = 6,000

.

Eliminating

p

allows for direct calculation of the amount invested in Account Q.

3

Calculate the amount invested in Account P

p = 15,000 - 6,000 = 9,000

.

Substitute the value of

q

back into the total investment equation.

4

Determine the requested difference

p - q = 9,000 - 6,000 = 3,000

.

The question specifically asks how much more money was invested in Account P than in Account Q.

Key Concept

Solving Systems of 2x2 Linear Equations for Word Problems

Question 11Question →

A logistics company packages cargo using three types of containers: small, medium, and large.

- A shipment of $3$ small, $2$ medium, and $1$ large container has a total weight of $130$ kilograms.
- A shipment of $1$ small, $4$ medium, and $2$ large containers has a total weight of $185$ kilograms.
- A shipment of $2$ small, $1$ medium, and $3$ large containers has a total weight of $160$ kilograms.

What is the weight, in kilograms, of one large container?

Show answer & explanation

Answer: 35

Answer

The weight of one large container is 35 kilograms.

Representing the weights of small, medium, and large containers as variables

S

,

M

, and

L

yields the 3x3 system of linear equations:
1)

3S + 2M + L = 130

2)

S + 4M + 2L = 185

3)

2S + M + 3L = 160

Solving for

S

in equation (2) gives

S = 185 - 4M - 2L

. Substituting this into equations (1) and (3) reduces the system to:
-

2M + L = 85

-

7M + L = 210

Subtracting the first equation from the second yields

5M = 125

, so

M = 25

. Substituting

M = 25

into

2M + L = 85

gives

50 + L = 85

, which simplifies to

L = 35

.

Step-by-Step Solution

1

Set up a system of three linear equations based on the shipment descriptions

3S + 2M + L = 130

,

S + 4M + 2L = 185

, and

2S + M + 3L = 160

Translate the physical constraints of the three shipments into mathematical relationships

2

Isolate variable

S

in the second equation and substitute into the first and third equations

Two equations in two variables:

2M + L = 85

and

7M + L = 210

Reduce the 3x3 system to a 2x2 system to eliminate variable

S

3

Subtract the two simplified equations to solve for

M

5M = 125 \implies M = 25

Eliminate variable

L

to obtain the value of

M

4

Substitute

M = 25

back into

2M + L = 85

to solve for

L

L = 35

Find the requested value for the weight of one large container

Key Concept

Solving 3x3 Systems of Linear Equations using Substitution and Elimination

Question 12Question →

Consider the following system of linear equations in

x

,

y

, and

z

, where

c

is a real constant:

\frac{3}{2}x - y + 2z = 5

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

6x - y + z = c

If the system has at least one solution $(x, y, z)$ , what is the value of $7x - \frac{2}{3}y$ ?

18

16

24

7

2

Show answer & explanation

Answer: 18

Answer

18

To find the value of

7x - \frac{2}{3}y

without individual values for

x, y,

and

z

, we express

7x - \frac{2}{3}y

as a linear combination

m(\text{Eq. 1}) + n(\text{Eq. 2})

. Matching the

z

-coefficients requires

2m - n = 0 \implies n = 2m

. Matching the

y

-coefficients yields

-m + \frac{1}{3}(2m) = -\frac{2}{3} \implies m = 2

, which gives

n = 4

. Verifying the

x

-coefficient gives

2\left(\frac{3}{2}\right) + 4(1) = 7

. Applying these multipliers to the right-hand sides gives

2(5) + 4(2) = 10 + 8 = 18

.

Step-by-Step Solution

1

Identify the target expression

7x - \frac{2}{3}y

as a linear combination of the first two equations

Express

m\left(\frac{3}{2}x - y + 2z\right) + n\left(x + \frac{1}{3}y - z\right) = 7x - \frac{2}{3}y + 0z

Because the system is dependent when consistent, individual variable values cannot be uniquely determined, but specific linear combinations independent of

z

can be evaluated.

2

Set up a system of equations for the scalar multipliers

m

and

n

Equating coefficients of

z

:

2m - n = 0 \implies n = 2m

. Equating coefficients of

y

:

-m + \frac{1}{3}n = -\frac{2}{3}

.

Eliminating

z

requires the net coefficient of

z

to equal 0.

3

Solve for

m

and

n

Substitute

n = 2m

into the

y

-coefficient equation:

-m + \frac{2}{3}m = -\frac{1}{3}m = -\frac{2}{3} \implies m = 2

, which gives

n = 4

.

Determining the exact linear multipliers needed to match the target expression.

4

Verify

x

-coefficient consistency and compute the target value

x

-coefficient:

2\left(\frac{3}{2}\right) + 4(1) = 3 + 4 = 7

. Value:

2(5) + 4(2) = 10 + 8 = 18

.

Applying the scalars

m = 2

and

n = 4

to the right-hand side constants gives the exact numerical value of

7x - \frac{2}{3}y

.

Key Concept

Linear combinations of dependent systems of equations

Alternative Method

Multiply the first equation by 2 to clear fractions:

3x - 2y + 4z = 10

. Multiply the second equation by 3 to clear fractions:

3x + y - 3z = 6

. Eliminate

z

by forming

3(3x - 2y + 4z) + 4(3x + y - 3z) = 3(10) + 4(6) \implies 21x - 2y = 54

. Dividing both sides of

21x - 2y = 54

by 3 directly gives

7x - \frac{2}{3}y = 18

.

Estimated Time:2m 0s

Question 13Question →

An electronics manufacturer produces three types of circuit boards: Alpha, Beta, and Gamma. Production requires processing across three specialized workstations: Solder, Component Placement, and Inspection.

- Each Alpha board requires 2 hours of Solder, 3 hours of Component Placement, and 1 hour of Inspection.
- Each Beta board requires 1 hour of Solder, 4 hours of Component Placement, and 2 hours of Inspection.
- Each Gamma board requires 3 hours of Solder, 2 hours of Component Placement, and 4 hours of Inspection.

During a given production cycle, the Solder station was operated for 55 hours, the Component Placement station for 85 hours, and the Inspection station for 65 hours. If all three workstations were operated at full capacity with no downtime, what was the total number of circuit boards produced?

Show answer & explanation

Answer: 28

Answer

The total number of circuit boards produced is 28.

Setting up equations for total machine hours yields

2x + y + 3z = 55

,

3x + 4y + 2z = 85

, and

x + 2y + 4z = 65

, where

x

,

y

, and

z

represent the quantities of Alpha, Beta, and Gamma boards produced, respectively. Subtracting the third equation from the first equation gives

(2x + y + 3z) - (x + 2y + 4z) = 55 - 65

, which simplifies to

x - y - z = -10

, or

x = y + z - 10

. Substituting

x = y + z - 10

into the second and third equations produces the 2x2 system

7y + 5z = 115

and

3y + 5z = 75

. Subtracting these two equations eliminates

z

, giving

4y = 40

, so

y = 10

. Substituting

y = 10

into

3y + 5z = 75

gives

30 + 5z = 75

, so

z = 9

. Substituting

y = 10

and

z = 9

into

x = y + z - 10

gives

x = 9

. The total number of circuit boards produced is

x + y + z = 9 + 10 + 9 = 28

.

Step-by-Step Solution

1

Set up the 3x3 system of linear equations based on workstation hours.

2x + y + 3z = 55

,

3x + 4y + 2z = 85

, and

x + 2y + 4z = 65

Each equation models the total operational hours used across the three product types.

2

Subtract the third equation from the first equation to isolate

x

in terms of

y

and

z

.

x = y + z - 10

Eliminating terms directly reduces coefficient complexity.

3

Substitute

x = y + z - 10

into the second and third equations to construct a 2x2 system.

3y + 5z = 75

and

7y + 5z = 115

Reducing to a two-variable system allows direct elimination.

4

Subtract the two reduced equations to solve for

y

and

z

.

y = 10

and

z = 9

The

5z

terms cancel out upon subtraction.

5

Determine

x

and compute the total sum

x + y + z

.

x = 9

, total

= 9 + 10 + 9 = 28

The question asks for the total quantity of circuit boards produced.

Key Concept

Systems of Linear Equations

Question 14Question →

Consider the following system of three linear equations in variables

x

,

y

, and

z

, where

k

is a real constant:

\begin{aligned} 2x + 3y - z &= 11 \\ x - 2y + 4z &= -3 \\ 4x - y + 7z &= k \end{aligned}

For what value of

k

does the system have at least one solution

(x, y, z)

?

-5

5

8

17

23

Show answer & explanation

Answer: $5$

Answer

5

The left-hand side of the third equation is a linear combination of the first two equations:

1 \cdot (2x + 3y - z) + 2 \cdot (x - 2y + 4z) = 4x - y + 7z

. For the linear system to be consistent and possess at least one solution, the same linear combination must hold for the constant terms on the right-hand side:

1(11) + 2(-3) = 11 - 6 = 5

. Therefore, the value of

k

must be

5

.

Step-by-Step Solution

1

Analyze the variable coefficients across the three equations for linear dependence.

Notice that the coefficients of the third equation can be expressed as a linear combination of the first two equations.

If the left-hand side of the third equation is a linear combination of the first two equations, the system will only be consistent if the right-hand side constants satisfy the exact same linear combination.

2

Determine the multiplier needed to produce the third equation's left-hand side.

Multiply the second equation by

2

and add it to the first equation:

(2x + 3y - z) + 2(x - 2y + 4z) = 4x - y + 7z

.

This yields the exact expression

4x - y + 7z

present on the left-hand side of the third equation.

3

Apply the identical combination to the right-hand side constant terms.

The combined constant value is

11 + 2(-3) = 11 - 6 = 5

.

For the system to have at least one solution (i.e., to avoid contradiction and be consistent),

k

must equal this computed value of

5

.

Key Concept

Linear Dependence and Consistency in 3x3 Linear Systems

Estimated Time:2m 0s

Question 15Question →

If

x

and

y

are real numbers such that

x > y > 0

and they satisfy the following system of equations:

\frac{3}{x+y} + \frac{4}{x-y} = \frac{11}{4}

\frac{5}{x+y} - \frac{2}{x-y} = \frac{1}{4}

what is the value of

x

?

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Substituting

u = \frac{1}{x+y}

and

v = \frac{1}{x-y}

transforms the non-linear looking equations into the linear system

3u + 4v = \frac{11}{4}

and

5u - 2v = \frac{1}{4}

. Solving this system yields

u = \frac{1}{4}

and

v = \frac{1}{2}

. Consequently,

x + y = 4

and

x - y = 2

. Adding these two equations gives

2x = 6

, so

x = 3

.

Step-by-Step Solution

1

Introduce auxiliary variables to linearize the system.

Let

u = \frac{1}{x+y}

and

v = \frac{1}{x-y}

. The system becomes

3u + 4v = \frac{11}{4}

and

5u - 2v = \frac{1}{4}

.

Replacing non-linear reciprocal terms with simple variables allows elimination or substitution methods for linear systems.

2

Solve the system of linear equations for

u

and

v

using elimination.

Multiply

5u - 2v = \frac{1}{4}

by 2 to get

10u - 4v = \frac{1}{2}

. Add this to

3u + 4v = \frac{11}{4}

:

13u = \frac{11}{4} + \frac{2}{4} = \frac{13}{4} \implies u = \frac{1}{4}

. Then

4v = \frac{11}{4} - 3\left(\frac{1}{4}\right) = 2 \implies v = \frac{1}{2}

.

Eliminating

v

yields a single equation in

u

, which provides the values of both auxiliary variables.

3

Convert auxiliary values back to equations in

x

and

y

.

Since

u = \frac{1}{x+y} = \frac{1}{4}

, we get

x + y = 4

. Since

v = \frac{1}{x-y} = \frac{1}{2}

, we get

x - y = 2

.

Inverting the fractions restores the original variables in a standard 2x2 linear system.

4

Solve for

x

by adding the two linear equations.

(x + y) + (x - y) = 4 + 2 \implies 2x = 6 \implies x = 3

.

Adding the equations eliminates

y

directly, isolating

x

.

Key Concept

Solving systems of linear equations using substitution variables for algebraic simplification

Question 16Question →

Consider the following system of linear equations in variables

x

,

y

, and

z

, where

a

is a real constant:

\begin{aligned} x + y + z &= 6 \\ x + 2y + 3z &= 10 \\ x + 2y + (a^2 - 1)z &= a + 8 \end{aligned}

Which of the following statements must be true? Select all that apply.

Select all that apply

If

a = 2

, the system has infinitely many solutions.

If

a = -2

, the system has no solution.

If

a = 2

, every solution to the system satisfies

2x + y = 8

.

If

a = 3

, the unique solution has

y = 3

.

The system has a unique solution for all real values of

a

except

a = 2

.

Show answer & explanation

Answer: If $a = 2$ , the system has infinitely many solutions.; If $a = -2$ , the system has no solution.; If $a = 2$ , every solution to the system satisfies $2x + y = 8$ .

Answer

The correct statements are that

a = 2

yields infinitely many solutions,

a = -2

results in no solution, and for

a = 2

every solution satisfies

2x + y = 8

.

The reduced equation

(a^2 - 4)z = a - 2

determines the behavior of the system. Setting

a = 2

gives

0 = 0

, leading to infinitely many solutions where

x = z + 2

and

y = 4 - 2z

, which identically satisfies

2x + y = 8

. Setting

a = -2

gives

0 = -4

, an inconsistency yielding no solutions.

Step-by-Step Solution

1

Eliminate

x

and

y

using elimination between the second and third equations.

(x + 2y + (a^2 - 1)z) - (x + 2y + 3z) = (a + 8) - 10 \implies (a^2 - 4)z = a - 2

Isolating the parameter dependence onto a single variable

z

reveals existence and uniqueness conditions.

2

Analyze the equation

(a - 2)(a + 2)z = a - 2

for key parameter values.

If

a = 2

,

0 \cdot z = 0

(infinitely many solutions). If

a = -2

,

0 \cdot z = -4

(no solution). If

a \neq \pm 2

,

z = \frac{1}{a + 2}

(unique solution).

Determining system consistency depends on whether the leading coefficient and right-hand side evaluate to zero.

3

Express

x

and

y

in terms of

z

for the consistent case

a = 2

.

Subtracting the first equation from the second gives

y + 2z = 4 \implies y = 4 - 2z

. Substituting into the first gives

x = z + 2

.

Parameterizing the solution set allows verification of linear combinations.

4

Evaluate the linear combination

2x + y

when

a = 2

.

2x + y = 2(z + 2) + (4 - 2z) = 2z + 4 + 4 - 2z = 8

.

Verifies that

2x + y = 8

is an invariant across all parametric solutions.

Key Concept

Parametric Analysis of 3x3 Systems of Linear Equations

Question 17Question →

A financial analyst models a company's weekly metrics using three variables—revenue

R

, operating cost

C

, and advertising expenditure

A

, measured in thousands of dollars. The metrics satisfy the following system of linear equations, where

k

is a real constant:

\begin{aligned} 4R - 3C + 2A &= 18 \\ R + 5C - 6A &= 14 \\ 5R + 2C - 4A &= k \end{aligned}

If this system of linear equations is consistent (has at least one solution), what is the value of

k

?

4

16

32

-32

252

Show answer & explanation

Answer: 32

Answer

32

Adding the left-hand sides of the first two equations yields

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

, which is identical to the left-hand side of the third equation. For the linear system to have at least one solution (to be consistent), the right-hand side constant must satisfy the exact same linear combination:

k = 18 + 14 = 32

.

Step-by-Step Solution

1

Examine the linear combination of the left-hand sides of the first two equations

(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A

Observing that the sum of the coefficients of the first two equations matches the left-hand side of the third equation.

2

Apply the condition for system consistency

Right-hand side of Equation 3 must equal Right-hand side of Equation 1 + Right-hand side of Equation 2

For a system with linearly dependent left-hand sides to be consistent, the same linear combination must hold for the right-hand constants.

3

Calculate the value of

k

k = 18 + 14 = 32

Adding the constants from the right-hand side of the first two equations gives the consistent value for

k

.

Key Concept

Consistency and Linear Dependence in Systems of Linear Equations

Question 18Question →

For how many real values of the constant

a

does the following system of linear equations in

x

,

y

, and

z

have no solution?

\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}

None

Exactly one

Exactly two

Exactly three

Infinitely many

Show answer & explanation

Answer: Exactly one

Answer

Exactly one

To find when the system has no solution, we first eliminate

x

by subtracting the first equation from the second and third equations. This produces a two-variable system in

y

and

z

:

(a-2)y + 4z = 2

and

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

. The determinant of this system's coefficients is

(a-2)(a+2) - 12 = a^2 - 16

. Setting the determinant to zero yields two critical values:

a = 4

and

a = -4

. Testing

a = 4

simplifies both reduced equations to

y + 2z = 1

, which means the system is consistent with infinitely many solutions. Testing

a = -4

yields

-3y + 2z = 1

and

-3y + 2z = 5

, which is impossible (

1 = 5

), making the system inconsistent. Thus, there is exactly one real value of

a

(

a = -4

) for which the system has no solution.

Step-by-Step Solution

1

Eliminate the variable

x

from the second and third equations using the first equation.

Subtracting the first equation

x + y - z = 3

from the second equation yields:

(a-2)y + 4z = 2

Subtracting the first equation from the third equation yields:

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

Reducing the

3 \times 3

system to a

2 \times 2

system in

y

and

z

simplifies the analysis of linear dependence and consistency.

2

Determine the values of

a

for which the reduced

2 \times 2

system lacks a unique solution by setting its coefficient determinant to zero.

The determinant of the coefficient matrix is:

D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16

Setting

D = 0

yields

a^2 = 16

, which gives

a = 4

or

a = -4

.

A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.

3

Test

a = 4

in the reduced system.

Substituting

a = 4

into the reduced equations gives:

2y + 4z = 2 \implies y + 2z = 1

3y + 6z = 3 \implies y + 2z = 1

Since both equations are identical, the system is consistent and has infinitely many solutions.

When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.

4

Test

a = -4

in the reduced system.

Substituting

a = -4

into the reduced equations gives:

-6y + 4z = 2 \implies -3y + 2z = 1

3y - 2z = -5 \implies -3y + 2z = 5

Comparing these gives

1 = 5

, which is a contradiction. Thus, for

a = -4

, the system has no solution.

When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Key Concept

Parametric Systems of Linear Equations and Consistency Conditions

Question 19Question →

Two water pumps, Pump A and Pump B, working simultaneously at their respective constant rates, can fill an empty storage tank in $4$ hours. If Pump A operates alone at its constant rate for $2$ hours and then Pump B operates alone at its constant rate for $7$ hours, the tank is also filled completely. How many hours would it take Pump A, working alone at its constant rate, to fill the entire storage tank?

\frac{10}{3}

hours

\frac{20}{3}

hours

10

hours

12

hours

15

hours

Show answer & explanation

Answer: $\frac{20}{3}$ hours

Answer

\frac{20}{3}

hours (or

6\frac{2}{3}

hours)

The correct answer is derived by setting up two linear equations representing the total work accomplished:

4r_A + 4r_B = 1

and

2r_A + 7r_B = 1

. Solving this system gives Pump A's rate

r_A = \frac{3}{20}

tanks per hour. The time required for Pump A working alone is the reciprocal of its rate, which equals

\frac{20}{3}

hours.

Step-by-Step Solution

1

Define variables and formulate the system of linear equations

Let

r_A

be the rate of Pump A (tanks/hour) and

r_B

be the rate of Pump B (tanks/hour).
Combined work equation:

4(r_A + r_B) = 1 \implies 4r_A + 4r_B = 1

Sequential work equation:

2r_A + 7r_B = 1

Work done equals rate multiplied by time, and completing one full tank corresponds to total work

= 1

.

2

Express

r_B

in terms of

r_A

using the first equation

r_A + r_B = \frac{1}{4} \implies r_B = \frac{1}{4} - r_A

Simplifying the combined rate equation allows substitution into the second linear equation.

3

Substitute

r_B

into the second equation and solve for

r_A

2r_A + 7\left(\frac{1}{4} - r_A\right) = 1 \implies 2r_A + \frac{7}{4} - 7r_A = 1 \implies -5r_A = 1 - \frac{7}{4} = -\frac{3}{4} \implies r_A = \frac{3}{20}

Eliminating

r_B

isolates

r_A

as a single-variable linear equation.

4

Calculate the time required for Pump A to fill the tank alone

\text{Time}_A = \frac{1}{r_A} = \frac{1}{\frac{3}{20}} = \frac{20}{3}\text{ hours}

The total time to complete 1 unit of work is the reciprocal of the unit work rate.

Key Concept

Formulating and solving systems of two linear equations in two variables derived from work-rate relationships.

Question 20Question →

Consider the following system of linear equations in variables

x

,

y

, and

z

, where

a

,

b

, and

c

are real constants:

\begin{aligned} 2x - y + 3z &= a \\ x + 2y - z &= b \\ 7x + 4y + 3z &= c \end{aligned}

Which of the following statements must be true? Select all such statements.

Select all that apply

If

c = 2a + 3b

, the system has infinitely many solutions.

There exist no real values of

a

,

b

, and

c

for which the system has a unique solution.

If

a = 0

,

b = 0

, and

c = 0

, the system has

(0, 0, 0)

as its only solution.

If

c \neq 2a + 3b

, the system has infinitely many solutions.

If

a = 1

,

b = 2

, and

c = 7

, the system has at least one solution.

Show answer & explanation

Answer: If $c = 2a + 3b$ , the system has infinitely many solutions.; There exist no real values of $a$ , $b$ , and $c$ for which the system has a unique solution.

Answer

The correct statements are: 'If

c = 2a + 3b

, the system has infinitely many solutions' and 'There exist no real values of

a

,

b

, and

c

for which the system has a unique solution.'

The correct options accurately reflect the structural properties of the system. First, scaling the first equation by 2 and the second by 3 yields

7x + 4y + 3z = 2a + 3b

. Comparing this with the third equation,

7x + 4y + 3z = c

, shows that when

c = 2a + 3b

, the third equation provides no new constraints, leaving 2 independent equations in 3 variables and thus producing infinitely many solutions. Second, because the coefficient matrix has linearly dependent rows, its rank is 2 (less than the 3 variables), making a unique solution impossible regardless of the constants

a

,

b

, and

c

.

Step-by-Step Solution

1

Analyze the linear dependence of the left-hand sides of the equations.

Observe that

2(2x - y + 3z) + 3(x + 2y - z) = (4x + 3x) + (-2y + 6y) + (6z - 3z) = 7x + 4y + 3z

.

Finding a linear combination of the first two equations that produces the left-hand side of the third equation allows us to analyze system consistency.

2

Determine the condition for consistency.

The system is consistent if and only if

2a + 3b = c

.

If

c = 2a + 3b

, the third equation is a linear combination of the first two, resulting in a system of 2 independent equations in 3 variables, which yields infinitely many solutions.

3

Evaluate the possibility of a unique solution.

The rank of the coefficient matrix is 2, which is strictly less than the number of variables (3).

A system of linear equations has a unique solution if and only if the rank of the coefficient matrix equals the number of variables. Thus, no choice of

a, b, c

can produce a unique solution.

4

Verify specific numerical options.

For

a=0, b=0, c=0

,

c = 2(0)+3(0)=0

, giving infinitely many solutions. For

a=1, b=2, c=7

,

2(1)+3(2)=8 \neq 7

, giving zero solutions.

Testing specific constant values confirms consistency or inconsistency based on whether

c = 2a + 3b

is satisfied.

Key Concept

Consistency and Number of Solutions in 3x3 Systems of Linear Equations

Systems of Linear Equations

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

Answer

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Common Mistakes

Answer

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Key Concept

Common Mistakes

Alternative Method

Answer

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Key Concept

Common Mistakes

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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