Systems of Linear Equations Practice Questions — GMAT

Question 1Question →

A manufacturing plant uses three assembly lines, $L_1$ , $L_2$ , and $L_3$ , to produce three custom components, $X$ , $Y$ , and $Z$ .

- Producing one unit of component $X$ requires 2 hours on $L_1$ , 1 hour on $L_2$ , and 3 hours on $L_3$ .
- Producing one unit of component $Y$ requires 3 hours on $L_1$ , 4 hours on $L_2$ , and 2 hours on $L_3$ .
- Producing one unit of component $Z$ requires 1 hour on $L_1$ , 2 hours on $L_2$ , and 4 hours on $L_3$ .

During a given week, assembly lines $L_1$ , $L_2$ , and $L_3$ were operated for a total of 140 hours, 165 hours, and 235 hours, respectively, with zero idle time. Assuming full capacity utilization, how many units of component $Z$ were produced during that week?

Show answer & explanation

Answer: 30

Answer

30 units of component Z were produced.

Translating the assembly line operational hours into a 3x3 system of linear equations yields

2X + 3Y + Z = 140

,

X + 4Y + 2Z = 165

, and

3X + 2Y + 4Z = 235

. Eliminating

X

results in two equations in

Y

and

Z

:

5Y + 3Z = 190

and

5Y + Z = 130

. Subtracting these equations gives

2Z = 60

, so

Z = 30

.

Step-by-Step Solution

1

Formulate a system of 3 linear equations representing total hours logged on each assembly line.

Line 1:

2X + 3Y + Z = 140

; Line 2:

X + 4Y + 2Z = 165

; Line 3:

3X + 2Y + 4Z = 235

.

Each component requires specific line processing time, and total time per line equals total available capacity.

2

Eliminate variable

X

by substituting

X = 165 - 4Y - 2Z

into the other two equations.

Equation A:

5Y + 3Z = 190

and Equation B:

5Y + Z = 130

.

Reducing a 3-variable system to a 2-variable system simplifies linear elimination.

3

Subtract Equation B from Equation A.

2Z = 60, so Z = 30.

Since the coefficients of

Y

in both reduced equations are identical (

5Y

), subtraction directly isolates

Z

.

Key Concept

Solving Systems of Three Linear Equations via Variable Substitution and Elimination

Question 2Question →

A software analytics company offers three annual subscription plans: Standard, Professional, and Enterprise.

• A client purchasing 3 Standard, 2 Professional, and 1 Enterprise plan pays a total of $1,110. �� A client purchasing 1 Standard, 4 Professional, and 2 Enterprise plans pays a total of$ 1,620.
• A client purchasing 4 Standard, 1 Professional, and 3 Enterprise plans pays a total of $1,730.

What is the cost, in dollars, of 1 Enterprise plan?

Show answer & explanation

Answer: 350

Answer

The cost of 1 Enterprise plan is 350 dollars.

Setting up equations for each purchase tier gives

3x + 2y + z = 1110

,

x + 4y + 2z = 1620

, and

4x + y + 3z = 1730

. Isolating

x

in the second equation gives

x = 1620 - 4y - 2z

. Substituting

x

into the first and third equations yields

2y + z = 750

and

3y + z = 950

, respectively. Subtracting these reduced equations gives

y = 200

, which upon back-substitution into

2y + z = 750

reveals

z = 350

. Thus, 1 Enterprise plan costs 350 dollars.

Step-by-Step Solution

1

Formulate the linear system of equations from the given conditions.

Let

x

be the price of a Standard plan,

y

the price of a Professional plan, and

z

the price of an Enterprise plan:
(1)

3x + 2y + z = 1110

(2)

x + 4y + 2z = 1620

(3)

4x + y + 3z = 1730

Translating the verbal conditions into algebraic equations creates a solvable system.

2

Isolate variable

x

in equation (2) and substitute it into equations (1) and (3).

From (2),

x = 1620 - 4y - 2z

.
Substituting into (1):

3(1620 - 4y - 2z) + 2y + z = 1110 \implies 4860 - 12y - 6z + 2y + z = 1110 \implies 10y + 5z = 3750 \implies 2y + z = 750

(Equation 4)

Substituting into (3):

4(1620 - 4y - 2z) + y + 3z = 1730 \implies 6480 - 16y - 8z + y + 3z = 1730 \implies 15y + 5z = 4750 \implies 3y + z = 950

(Equation 5)

Eliminating

x

reduces the system to two linear equations with two variables.

3

Solve the 2x2 system of equations for

y

and

z

.

Subtracting Equation (4) from Equation (5):

(3y + z) - (2y + z) = 950 - 750 \implies y = 200

.

Substitute

y = 200

back into Equation (4):

2(200) + z = 750 \implies 400 + z = 750 \implies z = 350

.

Solving the reduced system yields the exact values of

y

and

z

.

Key Concept

Solving Systems of Three Linear Equations via Gaussian Elimination / Variable Substitution

Question 3Question →

A boutique bookstore sells hardcover books for $\$25$ each and paperback books for $\$12$ each. On a certain day, the store sold a total of $85$ books and generated a total revenue of $\$1,579$ . How many hardcover books were sold on that day?

Show answer & explanation

Answer: 43

Answer

The total number of hardcover books sold on that day is 43.

By setting up the two linear equations

h + p = 85

(quantity) and

25h + 12p = 1579

(revenue), eliminating

p

yields

13h = 559

, which gives

h = 43

.

Step-by-Step Solution

1

Define variables and construct the system of linear equations

Let

h

be the number of hardcover books and

p

be the number of paperback books. The equations are

h + p = 85

and

25h + 12p = 1579

.

The total quantity of items sold gives a sum equation, and the individual prices multiplied by their quantities yield the total revenue equation.

2

Use substitution or elimination to solve for

h

Multiplying the total books equation by 12 yields

12h + 12p = 1020

. Subtracting this from

25h + 12p = 1579

gives

13h = 559

.

Eliminating

p

directly isolates the variable

h

representing hardcover books.

3

Calculate the value of

h

h = \frac{559}{13} = 43

.

Dividing the remaining total revenue by the coefficient of

h

yields the precise quantity.

Key Concept

Solving two-variable systems of linear equations using elimination or substitution

Question 4Question →

A research laboratory placed two orders for high-precision sensors. The first order, consisting of $5$ acoustic sensors and $3$ thermal sensors, cost a total of $\$1{}160$ . The second order, consisting of $3$ acoustic sensors and $5$ thermal sensors, cost a total of $\$1{}080$ . If each acoustic sensor costs the same amount and each thermal sensor costs the same amount, what is the combined cost of $1$ acoustic sensor and $1$ thermal sensor?

\$140

\$160

\$280

\$320

\$560

Show answer & explanation

Answer: $\$280$

Answer

The combined cost of 1 acoustic sensor and 1 thermal sensor is

\$280

.

By representing the acoustic sensor price as

a

and thermal sensor price as

t

, we form two equations:

5a + 3t = 1160

and

3a + 5t = 1080

. Adding both equations gives

8a + 8t = 2240

. Dividing the entire equation by

8

isolates

a + t = 280

.

Step-by-Step Solution

1

Define variables and write the system of equations based on the problem statement.

Let

a

be the price of one acoustic sensor and

t

be the price of one thermal sensor.
Equation 1:

5a + 3t = 1160

Equation 2:

3a + 5t = 1080

Translate word problem conditions into algebraic equations.

2

Add the two equations together to find a direct linear combination.

(5a + 3t) + (3a + 5t) = 1160 + 1080 \implies 8a + 8t = 2240

Since the target is

a + t

, adding symmetric coefficients yields a common factor of

8

, avoiding the need to solve for

a

and

t

individually.

3

Factor out

8

and solve for

(a + t)

.

8(a + t) = 2240 \implies a + t = \frac{2240}{8} = 280

Dividing the combined equation by

8

directly isolates the required quantity

(a + t)

.

Key Concept

Solving Linear Systems via Algebraic Combinations

Question 5Question →

A bakery sells blueberry muffins for $3 each and chocolate chip muffins for$ 4 each. On Saturday, the bakery sold a total of 50 muffins for $170. How many chocolate chip muffins were sold on Saturday?

Show answer & explanation

Answer: 20

Answer

The bakery sold 20 chocolate chip muffins on Saturday.

Let

b

represent the number of blueberry muffins and

c

represent the number of chocolate chip muffins sold. Based on the problem text, we construct the system of equations:

b + c = 50

and

3b + 4c = 170

. From the first equation,

b = 50 - c

. Substituting this into the second equation yields

3(50 - c) + 4c = 170

, which expands to

150 - 3c + 4c = 170

. Simplifying gives

150 + c = 170

, so

c = 20

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations.

b + c = 50

and

3b + 4c = 170

The total number of muffins sold is 50, and the total revenue from selling blueberry muffins at

3 each and chocolate chip muffins at

4 each is $170.

2

Substitute

b = 50 - c

into the cost equation.

3(50 - c) + 4c = 170 \implies 150 + c = 170

Substituting

b

in terms of

c

creates a single-variable linear equation for

c

.

3

Solve for

c

.

c = 20

Subtracting 150 from both sides gives the exact number of chocolate chip muffins sold.

Key Concept

Solving Systems of Linear Equations by Substitution or Elimination

Question 6Question →

An event planner ordered a total of $30$ gift baskets for a corporate conference. Standard gift baskets cost $\$25$ each, and Deluxe gift baskets cost $\$40$ each. If the total cost of all $30$ gift baskets was $\$900$ , how many Deluxe gift baskets were ordered?

10

12

15

20

25

Show answer & explanation

Answer: $10$

Answer

The correct number of Deluxe gift baskets ordered is

10

.

By setting up the linear system

x + y = 30

and

25x + 40y = 900

, substituting

x = 30 - y

into the cost equation yields

750 + 15y = 900

, which solves directly to

y = 10

. Therefore,

10

Deluxe baskets were ordered.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

x

be the number of Standard baskets and

y

be the number of Deluxe baskets. Equation 1:

x + y = 30

. Equation 2:

25x + 40y = 900

.

The total number of baskets establishes a quantity relation, while the individual prices establish a cost relation.

2

Express

x

in terms of

y

using Equation 1

x = 30 - y

Isolating

x

allows direct substitution into the total cost equation to solve for

y

.

3

Substitute

x = 30 - y

into Equation 2 and solve for

y

25(30 - y) + 40y = 900 \implies 750 - 25y + 40y = 900 \implies 15y = 150 \implies y = 10

.

Simplifying the single-variable equation yields the exact number of Deluxe gift baskets.

Key Concept

Solving Systems of Linear Equations using Substitution or Elimination

Alternative Method

Use the elimination method: multiply the total quantity equation by

25

(

25x + 25y = 750

) and subtract it from the total cost equation (

25x + 40y = 900

) to get

15y = 150

, leading directly to

y = 10

.

Estimated Time:45s

Question 7Question →

If $3x + 2y = 18$ and $x + 4y = 16$ , what is the value of $x + y$ ?

3

4

7

10

17

Show answer & explanation

Answer: 7

Answer

7

Solving the system of linear equations

3x + 2y = 18

and

x + 4y = 16

yields

x = 4

and

y = 3

. Adding these values together produces

4 + 3 = 7

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the second equation

x = 16 - 4y

Isolating one variable allows straightforward substitution into the other equation.

2

Substitute

x = 16 - 4y

into the first equation

3(16 - 4y) + 2y = 18 \implies 48 - 12y + 2y = 18 \implies 48 - 10y = 18

This creates an equation containing only the variable

y

.

3

Solve for

y

-10y = 18 - 48 = -30 \implies y = 3

Isolating the numerical term isolates the value of

y

.

4

Substitute

y = 3

back to find

x

x = 16 - 4(3) = 16 - 12 = 4

Determines the specific value of

x

.

5

Calculate the requested sum

x + y

x + y = 4 + 3 = 7

Provides the final requested expression value.

Key Concept

Solving a system of linear equations by substitution or elimination to evaluate a combined linear expression.

Estimated Time:45s

Question 8Question →

A logistics company ships three types of packages: Small, Medium, and Large. A shipment containing $3$ Small, $5$ Medium, and $2$ Large packages has a total weight of $170$ kilograms. A second shipment containing $1$ Small, $2$ Medium, and $1$ Large package has a total weight of $65$ kilograms. What is the total weight, in kilograms, of a shipment containing $5$ Small, $9$ Medium, and $4$ Large packages?

235

275

300

335

405

Show answer & explanation

Answer: $300$

Answer

The total weight of the shipment is

300

kilograms.

The target quantity

5S + 9M + 4L

can be expressed directly as

1 \times (3S + 5M + 2L) + 2 \times (1S + 2M + 1L)

. Substituting the given total weights yields

1(170) + 2(65) = 300

kilograms.

Step-by-Step Solution

1

Formulate linear equations representing the package weights.

Let

S

,

M

, and

L

represent the weight of a Small, Medium, and Large package respectively. Equation 1:

3S + 5M + 2L = 170

. Equation 2:

1S + 2M + 1L = 65

.

Setting up algebraic representations for the system based on the problem statement.

2

Determine if the requested expression

5S + 9M + 4L

can be formed as a linear combination

c_1(3S + 5M + 2L) + c_2(1S + 2M + 1L)

.

Match coefficients:

3c_1 + c_2 = 5

,

5c_1 + 2c_2 = 9

, and

2c_1 + c_2 = 4

.

Since individual values of

S

,

M

, and

L

cannot be uniquely determined from two equations with three variables, we seek scalar constants

c_1

and

c_2

.

3

Solve for the multipliers

c_1

and

c_2

.

Subtracting

2c_1 + c_2 = 4

from

3c_1 + c_2 = 5

yields

c_1 = 1

. Substituting

c_1 = 1

into

2c_1 + c_2 = 4

gives

c_2 = 2

. Checking the middle equation:

5(1) + 2(2) = 9

, which holds true.

Finding the scalar weights that recreate the exact combination requested.

4

Compute the total weight using the linear combination of the known values.

Total Weight

= 1 \times 170 + 2 \times 65 = 170 + 130 = 300

kilograms.

Applying the solved linear combination to the total weights of the shipments.

Key Concept

Solving for a linear combination of variables in an underdetermined system without finding individual variable values.

Question 9Question →

An investor allocates a sum of money across three accounts— $A$ , $B$ , and $C$ —which earn simple annual interest rates of $5\%$ , $4\%$ , and $6\%$ , respectively. The total interest earned from all three accounts in one year is $\$1,060$ . The amount invested in Account $C$ equals the total amount invested in Accounts $A$ and $B$ combined. If the amount in Account $A$ were increased by $50\%$ and the amount in Account $B$ were decreased by $25\%$ while Account $C$ remained unchanged, the total annual interest earned would increase by $\$110$ . What is the total amount, in dollars, invested across all three accounts?

\$10,000

\$15,000

\$18,000

\$20,000

\$24,000

Show answer & explanation

Answer: $\$20,000$

Answer

The total amount invested across all three accounts is

\$20,000

.

The system of linear equations derived from the problem constraints uniquely solves to

a = \$6,000

,

b = \$4,000

, and

c = \$10,000

. Summing these three amounts gives a total investment of

\$20,000

.

Step-by-Step Solution

1

Define variables and translate the problem statements into algebraic equations.

Let

a

,

b

, and

c

be the amounts invested in Accounts

A

,

B

, and

C

.
Equation 1 (Total Interest):

0.05a + 0.04b + 0.06c = 1,060 \implies 5a + 4b + 6c = 106,000

.
Equation 2 (Relationship among principal amounts):

c = a + b \implies a + b - c = 0

.
Equation 3 (Change in interest): Increasing

a

by

50\%

adds

0.05(0.50a) = 0.025a

interest. Decreasing

b

by

25\%

reduces interest by

0.04(0.25b) = 0.01b

. The net change is

0.025a - 0.01b = 110 \implies 25a - 10b = 110,000 \implies 5a - 2b = 22,000

.

Establishing a complete 3-variable system of linear equations is necessary to determine the unknowns.

2

Substitute

c = a + b

into Equation 1 to reduce the system to two variables.

5a + 4b + 6(a + b) = 106,000 \implies 11a + 10b = 106,000

.

Using substitution eliminates variable

c

, leaving a linear system in

a

and

b

.

3

Solve the two-variable system using elimination.

We have:
(1)

11a + 10b = 106,000

(2)

5a - 2b = 22,000

Multiply (2) by 5:

25a - 10b = 110,000

.
Add this to (1):

(11a + 10b) + (25a - 10b) = 106,000 + 110,000 \implies 36a = 216,000 \implies a = 6,000

.
Substitute

a = 6,000

back into (2):

5(6,000) - 2b = 22,000 \implies 30,000 - 2b = 22,000 \implies 2b = 8,000 \implies b = 4,000

.

Elimination allows straightforward calculation of individual values for

a

and

b

.

4

Calculate

c

and find the total sum invested across all three accounts.

c = a + b = 6,000 + 4,000 = 10,000

.
Total investment

= a + b + c = 6,000 + 4,000 + 10,000 = 20,000

.

The question asks for the total amount invested in all three accounts combined.

Key Concept

Systems of Linear Equations in Three Variables

Estimated Time:2m 0s

Question 10Question →

A municipal water treatment facility operates three types of filtration units: Model X, Model Y, and Model Z.

\bullet \text{ 2 Model X units, 3 Model Y units, and 1 Model Z unit together process 134,000 gallons per hour.}

\bullet \text{ 1 Model X unit, 4 Model Y units, and 2 Model Z units together process 156,000 gallons per hour.}

\bullet \text{ 3 Model X units, 1 Model Y unit, and 4 Model Z units together process 196,000 gallons per hour.}

What is the processing capacity, in thousands of gallons per hour, of a single Model Y filtration unit?

Show answer & explanation

Answer: 20

Answer

The processing capacity of a single Model Y filtration unit is 20 thousand gallons per hour.

By setting up the system of three linear equations

2x + 3y + z = 134

,

x + 4y + 2z = 156

, and

3x + y + 4z = 196

, and eliminating variables systematically through substitution and combination, we find

y = 20

. Thus, a single Model Y filtration unit processes 20 thousand gallons per hour.

Step-by-Step Solution

1

Set up a system of linear equations in three variables.

Let

x

,

y

, and

z

represent the hourly capacities (in thousands of gallons) of Model X, Model Y, and Model Z respectively.

\begin{aligned} (1) \quad 2x + 3y + z &= 134 \\ (2) \quad x + 4y + 2z &= 156 \\ (3) \quad 3x + y + 4z &= 196 \end{aligned}

Expressing the given conditions algebraically translates the word problem into a solvable linear system.

2

Express

x

in terms of

y

and

z

using Equation (2).

x = 156 - 4y - 2z

Equation (2) has a coefficient of 1 for

x

, making it ideal for algebraic substitution.

3

Substitute

x

into Equations (1) and (3) to eliminate

x

.

Substituting into Equation (1):

2(156 - 4y - 2z) + 3y + z = 134 \implies 312 - 8y - 4z + 3y + z = 134

312 - 5y - 3z = 134 \implies (4) \quad 5y + 3z = 178

Substituting into Equation (3):

3(156 - 4y - 2z) + y + 4z = 196 \implies 468 - 12y - 6z + y + 4z = 196

468 - 11y - 2z = 196 \implies (5) \quad 11y + 2z = 272

Reducing a 3-variable system to a 2-variable system simplifies the calculation.

4

Eliminate variable

z

from Equations (4) and (5) to solve for

y

.

Multiply Equation (4) by 2:

10y + 6z = 356

Multiply Equation (5) by 3:

33y + 6z = 816

Subtract the first result from the second:

(33y + 6z) - (10y + 6z) = 816 - 356

23y = 460 \implies y = 20

Eliminating

z

directly yields the required value of

y

, which is the capacity of Model Y.

Key Concept

Solving systems of three linear equations in three variables via substitution and elimination.

Question 11Question →

An electronics manufacturing plant produces two types of circuit modules: Module X and Module Y. Producing one Module X requires $3$ minutes on Machine A and $5$ minutes on Machine B. Producing one Module Y requires $4$ minutes on Machine A and $2$ minutes on Machine B. During a certain shift, Machine A was operated for a total of $230$ minutes and Machine B was operated for a total of $220$ minutes. If both machines operated at full capacity with no idle time, how many Module X units were produced?

Show answer & explanation

Answer: 30

Answer

30

Formulating equations from the machine times gives

3x + 4y = 230

for Machine A and

5x + 2y = 220

for Machine B. Multiplying the second equation by

2

yields

10x + 4y = 440

. Subtracting

3x + 4y = 230

from

10x + 4y = 440

eliminates

y

and yields

7x = 210

, which solves to

x = 30

.

Step-by-Step Solution

1

Define variables for the unknown quantities.

Let

x

be the number of Module X units produced, and let

y

be the number of Module Y units produced.

Representing unknown quantities with algebraic variables allows for model formulation.

2

Formulate a system of linear equations from the machine time constraints.

Machine A equation:

3x + 4y = 230

Machine B equation:

5x + 2y = 220

The total operational time on each machine equals the sum of times spent producing each module type.

3

Multiply the Machine B equation by

2

to enable elimination of

y

.

2 \times (5x + 2y) = 2 \times 220 \implies 10x + 4y = 440

Matching the coefficient of

y

with the first equation (

4y

) allows elimination via subtraction.

4

Subtract the Machine A equation from the modified Machine B equation and solve for

x

.

(10x + 4y) - (3x + 4y) = 440 - 230 \implies 7x = 210 \implies x = 30

Subtracting eliminates

y

directly, leaving a single linear equation in terms of

x

.

Key Concept

Solving a system of two linear equations in two variables using elimination

Estimated Time:1m 30s

Question 12Question →

A café sells small cold brew coffees for $\$4$ each and large cold brew coffees for $\$6$ each. On Monday morning, the café sold a total of $40$ cold brew coffees for $\$196$ . How many large cold brew coffees were sold?

Show answer & explanation

Answer: 18

Answer

The café sold 18 large cold brew coffees.

Setting up the system

s + l = 40

and

4s + 6l = 196

, substituting

s = 40 - l

yields

4(40 - l) + 6l = 196

. Simplifying gives

160 + 2l = 196

, so

2l = 36

and

l = 18

.

Step-by-Step Solution

1

Set up equations based on total count and total cost.

Let

s

= number of small coffees and

l

= number of large coffees. The system is

s + l = 40

and

4s + 6l = 196

.

The problem gives two distinct relationships between the quantity sold and the total dollar value.

2

Substitute

s = 40 - l

into the second equation.

4(40 - l) + 6l = 196 \implies 160 + 2l = 196

.

Substituting reduces the system from two variables to a single equation in

l

.

3

Solve the linear equation for

l

.

2l = 36 \implies l = 18

.

Subtracting 160 from both sides and dividing by 2 yields the number of large cold brews.

Key Concept

Solving systems of linear equations in two variables using substitution.

Question 13Question →

For what value of the constant $k$ does the system of linear equations below have no solution?

\begin{aligned} (k - 2)x + 3y &= 6 \\ 4x + (k + 2)y &= 12 \end{aligned}

-2

4

-4

Both

-4

and

4

16

Show answer & explanation

Answer: $-4$

Answer

The constant value

k = -4

results in a system with no solution.

For a

2 \times 2

linear system

a_1 x + b_1 y = c_1

and

a_2 x + b_2 y = c_2

to have no solution, the equations must have proportional variable coefficients but non-proportional constant terms:

\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}

. Solving

\frac{k-2}{4} = \frac{3}{k+2}

gives

k^2 - 4 = 12

, so

k = \pm 4

. Substituting

k = -4

yields

-6x + 3y = 6

and

4x - 2y = 12

, which simplify to

-2x + y = 2

and

-2x + y = -6

. Since the constants

2

and

-6

differ, the lines are parallel and distinct, meaning the system has no solution.

Step-by-Step Solution

1

Set up the condition for parallel lines (equal slopes) by equating the ratio of the coefficients of

x

and

y

.

\frac{k - 2}{4} = \frac{3}{k + 2}

A system of two linear equations has either zero solutions (parallel, non-intersecting lines) or infinitely many solutions (coincident lines) when the slopes are equal.

2

Cross-multiply and solve the quadratic equation for

k

.

(k - 2)(k + 2) = 12 \implies k^2 - 4 = 12 \implies k^2 = 16 \implies k = 4 \text{ or } k = -4

Finding all values of

k

where the coefficient matrix determinant is zero.

3

Test

k = 4

in the original system.

(4 - 2)x + 3y = 6 \implies 2x + 3y = 6 \quad \text{and} \quad 4x + (4 + 2)y = 12 \implies 4x + 6y = 12

Dividing

4x + 6y = 12

by

2

yields

2x + 3y = 6

, which is identical to the first equation. Thus,

k = 4

yields infinitely many solutions.

4

Test

k = -4

in the original system.

(-4 - 2)x + 3y = 6 \implies -6x + 3y = 6 \quad \text{and} \quad 4x + (-4 + 2)y = 12 \implies 4x - 2y = 12

Simplifying both equations gives

-2x + y = 2

and

-2x + y = -6

. The lines have identical slopes but different constants, so they are parallel and distinct, producing no solution.

Key Concept

System Solvability and Linear Consistency

Estimated Time:2m 0s

Question 14Question →

A biotechnology laboratory formulates three custom reagent mixtures—Solution $X$ , Solution $Y$ , and Solution $Z$ —using three chemical compounds: Alpha, Beta, and Gamma.

- Solution $X$ contains $2\text{ mL}$ of Alpha, $3\text{ mL}$ of Beta, and $1\text{ mL}$ of Gamma, and costs $\$13.00$ .
- Solution $Y$ contains $1\text{ mL}$ of Alpha, $2\text{ mL}$ of Beta, and $4\text{ mL}$ of Gamma, and costs $\$11.00$ .
- Solution $Z$ contains $3\text{ mL}$ of Alpha, $1\text{ mL}$ of Beta, and $2\text{ mL}$ of Gamma, and costs $\$13.00$ .

What is the cost of a mixture containing $5\text{ mL}$ of Alpha, $4\text{ mL}$ of Beta, and $3\text{ mL}$ of Gamma?

\$22.00

\$24.00

\$25.00

\$26.00

\$30.00

Show answer & explanation

Answer: $\$26.00$

Answer

\$26.00

The target quantity consists of

5\text{ mL}

of Alpha,

4\text{ mL}

of Beta, and

3\text{ mL}

of Gamma. By inspecting the given system, adding Solution

X

(

2\text{ mL}

Alpha,

3\text{ mL}

Beta,

1\text{ mL}

Gamma) and Solution

Z

(

3\text{ mL}

Alpha,

1\text{ mL}

Beta,

2\text{ mL}

Gamma) directly yields

5\text{ mL}

Alpha,

4\text{ mL}

Beta, and

3\text{ mL}

Gamma. Therefore, the required cost is simply the sum of the costs of Solution

X

and Solution

Z

:

\$13.00 + \$13.00 = \$26.00

.

Step-by-Step Solution

1

Set up the linear system of equations representing the cost of each solution.

Let

a

,

b

, and

g

be the cost per mL of Alpha, Beta, and Gamma, respectively.
Equation 1:

2a + 3b + g = 13

Equation 2:

a + 2b + 4g = 11

Equation 3:

3a + b + 2g = 13

Translate the given word problem into algebraic equations representing system relationships.

2

Identify the requested quantity and evaluate whether it can be formed as a linear combination of the given equations.

Target expression:

5a + 4b + 3g

Recognizing linear combinations avoids solving for individual variable values when not required.

3

Add Equation 1 and Equation 3.

(2a + 3b + g) + (3a + b + 2g) = 13 + 13 \implies 5a + 4b + 3g = 26

The sum of coefficients for Alpha (

2+3=5

), Beta (

3+1=4

), and Gamma (

1+2=3

) exactly matches the target mixture.

Key Concept

Solving systems of linear equations using linear combinations without full variable elimination

Estimated Time:2m 0s

Question 15Question →

A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Roast Gamma—using three varieties of single-origin beans: Grade A, Grade B, and Grade C.

- One batch of Roast Alpha requires 3 kg of Grade A, 1 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of $64. - One batch of Roast Beta requires 1 kg of Grade A, 4 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of$ 64.
- One batch of Roast Gamma requires 2 kg of Grade A, 2 kg of Grade B, and 5 kg of Grade C beans, and has a total raw material cost of $90.

What is the cost, in dollars, of 1 kg of Grade A beans?

Show answer & explanation

Answer: 12

Answer

The cost of 1 kg of Grade A beans is 12 dollars.

Setting up the 3-variable linear system

3x + y + 2z = 64

,

x + 4y + 2z = 64

, and

2x + 2y + 5z = 90

allows us to eliminate

z

by subtracting the second equation from the first, yielding

2x - 3y = 0

, or

x = 1.5y

. Substituting this relationship back into the system leads to

y = 8

and

x = 12

. Thus, 1 kg of Grade A beans costs 12 dollars.

Step-by-Step Solution

1

Formulate linear equations representing the total cost of each coffee blend batch.

Let

x

be the cost per kg of Grade A beans,

y

be the cost per kg of Grade B beans, and

z

be the cost per kg of Grade C beans:

\begin{aligned} (1)\quad 3x + y + 2z &= 64 \\ (2)\quad x + 4y + 2z &= 64 \\ (3)\quad 2x + 2y + 5z &= 90 \end{aligned}

Translating the word problem into a system of 3 linear equations with 3 variables.

2

Eliminate variable

z

by subtracting Equation (2) from Equation (1).

(3x + y + 2z) - (x + 4y + 2z) = 64 - 64 \implies 2x - 3y = 0 \implies x = 1.5y

Since both equations (1) and (2) contain the term

+2z

, subtracting them removes

z

directly and provides a simple relation between

x

and

y

.

3

Substitute

x = 1.5y

into Equation (1) and Equation (3) to obtain a system in terms of

y

and

z

.

From Equation (1):

3(1.5y) + y + 2z = 64 \implies 5.5y + 2z = 64 \implies 11y + 4z = 128 \quad (4)

From Equation (3):

2(1.5y) + 2y + 5z = 90 \implies 5y + 5z = 90 \implies y + z = 18 \implies z = 18 - y

Reducing the system from 3 variables down to 2 variables.

4

Substitute

z = 18 - y

into Equation (4) to solve for

y

, and subsequently calculate

x

.

11y + 4(18 - y) = 128 \implies 7y + 72 = 128 \implies 7y = 56 \implies y = 8

Using

x = 1.5y

:

x = 1.5 \times 8 = 12

Solving the single-variable linear equation for

y

, then substituting back to find the required cost

x

for Grade A beans.

Key Concept

Solving a 3-Variable System of Linear Equations via Variable Elimination

Question 16Question →

An event design company offers three distinct decorative bundles—Bundle X, Bundle Y, and Bundle Z—for corporate gala setups.

• Bundle X contains 3 floral arrangements, 2 LED uplights, and 1 table runner.
• Bundle Y contains 1 floral arrangement, 3 LED uplights, and 2 table runners.
• Bundle Z contains 2 floral arrangements, 1 LED uplight, and 3 table runners.

To decorate a venue, a coordinator orders a combination of these bundles containing a total of 26 floral arrangements, 23 LED uplights, and 23 table runners. If every bundle ordered is used in its entirety, what is the total number of bundles ordered by the coordinator?

10

12

15

18

24

Show answer & explanation

Answer: 12

Answer

The total number of bundles ordered by the coordinator is 12.

Summing all three equations gives

6x + 6y + 6z = 72

. Factoring out

6

yields

6(x + y + z) = 72

, so dividing by

6

gives

x + y + z = 12

. Alternatively, solving the system explicitly gives

x = 5

,

y = 4

, and

z = 3

, whose sum is

5 + 4 + 3 = 12

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations based on item counts.

Let

x

,

y

, and

z

represent the number of Bundle X, Bundle Y, and Bundle Z ordered, respectively.

\begin{aligned} 3x + y + 2z &= 26 \quad \text{(Floral arrangements)} \\ 2x + 3y + z &= 23 \quad \text{(LED uplights)} \\ x + 2y + 3z &= 23 \quad \text{(Table runners)} \end{aligned}

Translate the word problem into a standard system of 3 linear equations in 3 variables.

2

Sum the three equations to find a direct linear combination for

(x + y + z)

.

(3x + 2x + x) + (y + 3y + 2y) + (2z + z + 3z) = 26 + 23 + 23

6x + 6y + 6z = 72

Notice that the sum of coefficients for each variable across all three equations is identical (

3 + 2 + 1 = 6

).

3

Factor out 6 and solve for the total number of bundles

(x + y + z)

.

6(x + y + z) = 72 \implies x + y + z = 12

Dividing both sides by 6 directly gives the required total quantity without needing to solve for individual variables

x

,

y

, and

z

separately.

Key Concept

Linear combinations in systems of equations

Estimated Time:2m 0s

Question 17Question →

A corporate catering service offers two lunch options: a Standard Meal and a Premium Meal. On Monday, an order of $40$ Standard Meals and $20$ Premium Meals cost a total of $\$1,300$ . On Tuesday, an order of $20$ Standard Meals and $30$ Premium Meals cost a total of $\$1,250$ . What is the total cost of an order consisting of $15$ Standard Meals and $15$ Premium Meals?

\$650.00

\$712.50

\$812.50

\$975.00

\$1,162.50

Show answer & explanation

Answer: $\$712.50$

Answer

The total cost of 15 Standard Meals and 15 Premium Meals is $712.50.

By setting up the system

40x + 20y = 1300

and

20x + 30y = 1250

, we simplify to

2x + y = 65

and

2x + 3y = 125

. Subtracting the first from the second gives

2y = 60

, so

y = 30

. Substituting

y = 30

into

2x + y = 65

yields

2x = 35

, so

x = 17.50

. The sum of one Standard Meal and one Premium Meal is

x + y = 47.50

. Therefore,

15

of each costs

15 \times 47.50 = \$712.50

. Alternatively, adding the two simplified equations directly yields

4x + 4y = 190 \implies x + y = 47.50

, so

15(x + y) = 15 \times 47.50 = \$712.50

.

Step-by-Step Solution

1

Set up a system of linear equations using variables

x

for the price of a Standard Meal and

y

for the price of a Premium Meal.

40x + 20y = 1300

and

20x + 30y = 1250

Translate the scenario information into algebraic equations.

2

Simplify both equations by dividing by their greatest common factors.

Equation 1:

2x + y = 65

; Equation 2:

2x + 3y = 125

Simplifying equations reduces computation complexity.

3

Subtract Equation 1 from Equation 2 to eliminate

x

and solve for

y

.

(2x + 3y) - (2x + y) = 125 - 65 \implies 2y = 60 \implies y = 30

Elimination isolates variable

y

.

4

Substitute

y = 30

back into Equation 1 to solve for

x

.

2x + 30 = 65 \implies 2x = 35 \implies x = 17.50

Finding the value of

x

gives the cost of one Standard Meal.

5

Calculate the target linear combination

15x + 15y = 15(x + y)

.

15(17.50 + 30.00) = 15(47.50) = 712.50

Multiply the combined price of one of each meal by 15.

Key Concept

Solving systems of two linear equations in two variables using elimination and finding linear combinations.

Estimated Time:1m 45s

Question 18Question →

A logistics company operates two types of delivery trucks: Model A and Model B. Model A consumes $0.25$ liters of fuel per mile and incurs $\$1.50$ in maintenance costs per mile. Model B consumes $0.40$ liters of fuel per mile and incurs $\$1.00$ in maintenance costs per mile. On a given day, the fleet consumed a total of $300$ liters of fuel and incurred a total of $\$1,100$ in maintenance costs. What was the total number of miles traveled by the Model A trucks on that day?

Show answer & explanation

Answer: 400

Answer

The Model A trucks traveled a total of 400 miles on that day.

Let

x

be the number of miles driven by Model A trucks and

y

be the number of miles driven by Model B trucks. Set up two equations:

0.25x + 0.40y = 300

(fuel) and

1.50x + 1.00y = 1100

(maintenance). Substituting

y = 1100 - 1.5x

into the fuel equation yields

0.25x + 0.40(1100 - 1.5x) = 300

, which simplifies to

-0.35x + 440 = 300

, leading directly to

x = 400

.

Step-by-Step Solution

1

Formulate the linear system

0.25x + 0.40y = 300

and

1.50x + 1.00y = 1100

, where

x

is miles driven by Model A and

y

is miles driven by Model B.

Translates the rates of fuel consumption and maintenance per mile into total daily figures.

2

Express

y

in terms of

x

y = 1100 - 1.5x

Isolating

y

makes substitution into the fuel equation straightforward.

3

Substitute

y

into the first equation and solve for

x

0.25x + 0.40(1100 - 1.5x) = 300 \implies -0.35x + 440 = 300 \implies -0.35x = -140 \implies x = 400

Reduces the system to a single linear equation in

x

to find the distance for Model A.

Key Concept

Setting up and solving a two-variable system of linear equations via substitution or elimination.

Question 19Question →

A financial advisory firm manages two types of investment accounts: Growth portfolios and Conservative portfolios. Each Growth portfolio generates $\$5,000$ in annual dividends and $\$12,000$ in annual capital gains. Each Conservative portfolio generates $\$8,000$ in annual dividends and $\$4,000$ in annual capital gains. Last year, the firm generated a total of $\$186,000$ in dividends and $\$188,000$ in capital gains from these portfolios. What was the total number of portfolios (Growth and Conservative combined) managed by the firm?

10

17

27

31

35

Show answer & explanation

Answer: $27$

Answer

The total number of portfolios managed by the firm is

27

.

Setting up the system of equations with

x

as Growth portfolios and

y

as Conservative portfolios gives

5x + 8y = 186

and

12x + 4y = 188

. Multiplying the second equation by

2

yields

24x + 8y = 376

. Subtracting the first equation gives

19x = 190

, so

x = 10

. Substituting

x = 10

gives

y = 17

. The total number of portfolios is

10 + 17 = 27

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations in thousands of dollars.

Let

x

be the number of Growth portfolios and

y

be the number of Conservative portfolios.
Dividends equation:

5x + 8y = 186

Capital gains equation:

12x + 4y = 188

Dividing all dollar amounts by

1,000

simplifies the coefficients.

2

Solve for

x

using the elimination method.

Multiply the capital gains equation by

2

:

2(12x + 4y) = 2(188) \implies 24x + 8y = 376

Subtract the dividends equation (

5x + 8y = 186

) from this new equation:

(24x + 8y) - (5x + 8y) = 376 - 186

19x = 190 \implies x = 10

Eliminating

y

isolates

x

to determine the number of Growth portfolios.

3

Substitute

x = 10

into the capital gains equation to solve for

y

.

Divide

12(10) + 4y = 188

simplify to

120 + 4y = 188 \implies 4y = 68 \implies y = 17

Finds the number of Conservative portfolios.

4

Calculate the total number of portfolios.

Total portfolios =

x + y = 10 + 17 = 27

The question asks for the sum of both types of portfolios.

Key Concept

Solving Systems of Two Linear Equations with Word Problem Constraints

Estimated Time:1m 30s

Question 20Question →

A boutique hotel offers two types of rooms: Executive Suites and Deluxe Suites. During a convention weekend, booking $12$ Executive Suites and $18$ Deluxe Suites generated total revenue of $\$7{}800$ . The following weekend, booking $15$ Executive Suites and $10$ Deluxe Suites generated total revenue of $\$7{}250$ . If the nightly rate for each type of suite remained constant, what was the nightly rental rate, in dollars, of an Executive Suite?

Show answer & explanation

Answer: 350

Answer

The nightly rental rate of an Executive Suite was 350 dollars.

Formulating the revenue statements as linear equations gives

12x + 18y = 7,800

and

15x + 10y = 7,250

. Simplifying these by dividing by

6

and

5

yields

2x + 3y = 1,300

and

3x + 2y = 1,450

. Multiplying the first simplified equation by

2

(

4x + 6y = 2,600

) and the second by

3

(

9x + 6y = 4,350

) aligns the

y

-coefficients. Subtracting the equations leads to

5x = 1,750

, giving

x = 350

.

Step-by-Step Solution

1

Define variables and construct the system of equations.

12x + 18y = 7,800

and

15x + 10y = 7,250

, where

x

represents the Executive Suite rate and

y

represents the Deluxe Suite rate.

Translating word problem information into algebraic equations.

2

Simplify the system equations.

Dividing the first equation by

6

gives

2x + 3y = 1,300

. Dividing the second equation by

5

gives

3x + 2y = 1,450

.

Reducing the coefficients minimizes computation steps and errors.

3

Eliminate variable

y

to solve for

x

.

Multiply

2x + 3y = 1,300

by

2

to get

4x + 6y = 2,600

. Multiply

3x + 2y = 1,450

by

3

to get

9x + 6y = 4,350

. Subtracting gives

5x = 1,750

, so

x = 350

.

Eliminating

y

yields the requested value of

x

directly.

Key Concept

Solving Systems of Two Linear Equations via Elimination

Estimated Time:2m 0s

Systems of Linear Equations

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