Practice your knowledge of linear systems, an important math topic that involves the study of simultaneous equations. With many practical applications, they are used to solve problems involving different variables.
All questions are resolved step by step, where we will use different methods, such as: substitution, addition, elimination, scaling and Cramer's rule.
Question 1 (substitution method)
Determine the ordered pair that solves the following system of linear equations.
Response:
Isolating x in the first equation:
Substituting x into the second equation:
Substituting the value of y into the first equation.
So, the ordered pair that solves the system is:
Question 2 (scaling method)
The solution to the following system of linear equations is:
Answer: x = 5, y = 1, z = 2
The system is already in echelon form. The third equation has two zero coefficients (y = 0 and x = 0), the second equation has a zero coefficient (x = 0), and the third equation has no zero coefficients.
In an echelon system, we solve "from bottom to top", that is, we start with the third equation.
Moving to the top equation, we substitute z = 2.
Finally, we substitute z = 2 and y = 1 in the first equation, in order to obtain x.
Solution
x = 5, y = 1, z = 2
Question 3 (Cramer's rule or method)
Solve the following system of linear equations:
Answer: x = 4, y = 0.
Using Cramer's rule.
Step 1: determine the determinants D, Dx and Dy.
The matrix of coefficients is:
Its determinant:
D = 1. 1 - 2. (-1)
D = 1 - (-2) = 1 + 2 = 3
For the calculation of Dx, we replace the column of terms of x with the column of independent terms.
Dx = 4. 1 - 8. (-1)
Dx = 4 + 8 = 12
For the calculation of Dy, we replace the terms of y with the independent terms.
Dy = 1. 8 - 2. 4
Dy = 8 - 8
Dy = 0
step 2: determine x and y.
To determine x, we do:
To determine y, we do:
question 4
A t-shirt and cap seller at a sporting event sold 3 t-shirts and 2 caps, raising a total of R$220.00. The next day, he sold 2 shirts and 3 caps, raising R$190.00. What would be the price of a T-shirt and the price of a hat?
a) T-shirt: BRL 60.00 | Cap: BRL 40.00
b) T-shirt: BRL 40.00 | Cap: BRL 60.00
c) T-shirt: BRL 56.00 | Cap: BRL 26.00
d) T-shirt: BRL 50.00 | Cap: BRL 70.00
e) T-shirt: BRL 80.00 | Cap: BRL 30.00
Let's label the price of T-shirts c and the price of hats b.
For the first day we have:
3c + 2b = 220
For the second day we have:
2c + 3b = 190
We form two equations with two unknowns each, c and b. So we have a system of 2x2 linear equations.
Resolution
Using Cramer's Rule:
1st step: determinant of the matrix of coefficients.
2nd step: determinant Dc.
We replace the column of c with the matrix of independent terms.
3rd step: determinant Db.
4th step: determine the value of c and b.
Response:
The price of the T-shirt is R$56.00 and the cap R$26.00.
question 5
A movie theater charges R$10.00 per ticket for adults and R$6.00 per ticket for children. In one day, 80 tickets were sold and the total collection was R$ 700.00. How many tickets of each type were sold?
a) Adults: 75 | Children: 25
b) Adults: 40 | Children: 40
c) Adults: 65 | Children: 25
d) Adults: 30 | Children: 50
e) Adults: 25 | Children: 75
We will name it as The the ticket price for adults and w for kids.
In relation to the total number of tickets we have:
a + c = 80
Regarding the obtained value we have:
10a + 6c = 700
We form a system of linear equations with two equations and two unknowns, that is, a 2x2 system.
Resolution
We will use the substitution method.
Isolating a in the first equation:
a = 80 - c
Substituting a into the second equation:
10.(80 - c) + 6c = 700
800 -10c + 6c = 700
800 - 700 = 10c - 6c
100 = 4c
c = 100/4
c = 25
Substituting c in the second equation:
6a + 10c = 700
6a+10. 25 = 700
6y + 250 = 700
6a = 700 - 250
6a = 450
a = 450/6
a = 75
question 6
A store sells T-shirts, shorts and shoes. On the first day, 2 T-shirts, 3 shorts and 4 pairs of shoes were sold, totaling R$ 350.00. On the second day, 3 T-shirts, 2 shorts and 1 pair of shoes were sold, totaling R$ 200.00. On the third day, 1 T-shirt, 4 shorts and 2 pairs of shoes were sold, totaling R$ 320.00. How much would a T-shirt, shorts and a pair of shoes cost?
a) T-shirt: BRL 56.00 | Bermuda: R$ 24.00 | Shoes: BRL 74.00
b) T-shirt: BRL 40.00 | Bermuda: R$ 50.00 | Shoes: BRL 70.00
c) T-shirt: BRL 16.00 | Bermuda: R$ 58.00 | Shoes: BRL 36.00
d) T-shirt: BRL 80.00 | Bermuda: R$ 50.00 | Shoes: BRL 40.00
e) T-shirt: BRL 12.00 | Bermuda: R$ 26.00 | Shoes: BRL 56.00
- c is the price of shirts;
- b is the price of the shorts;
- s is the price of the shoes.
For the first day:
2c + 3b + 4s = 350
For the second day:
3c + 2b + s = 200
For the third day:
c + 4b + 2s = 320
We have three equations and three unknowns, forming a 3x3 system of linear equations.
Using Cramer's rule.
The matrix of coefficients is
Its determinant is D = 25.
The column matrix of responses is:
To calculate Dc, we replace the column matrix of responses with the first column in the matrix of coefficients.
dc = 400
For the calculation of Db:
Db = 1450
For the calculation of Ds:
Ds = 900
To determine c, b, and s, we divide the determinants Dc, Db, and Ds by the principal determinant D.
question 7
A restaurant offers three dish options: meat, salad and pizza. On the first day, 40 meat dishes, 30 salad dishes and 10 pizzas were sold, totaling R$ 700.00 in sales. On the second day, 20 meat dishes, 40 salad dishes and 30 pizzas were sold, totaling R$ 600.00 in sales. On the third day, 10 meat dishes, 20 salad dishes and 40 pizzas were sold, totaling R$ 500.00 in sales. How much would each dish cost?
a) meat: BRL 200.00 | salad: R$ 15.00 | pizza: BRL 10.00
b) meat: R$ 150.00 | salad: R$ 10.00 | pizza: BRL 60.00
c) meat: BRL 100.00 | salad: R$ 15.00 | pizza: BRL 70.00
d) meat: BRL 200.00 | salad: R$ 10.00 | pizza: BRL 15.00
e) meat: BRL 140.00 | salad: R$ 20.00 | pizza: BRL 80.00
Using:
- c for meat;
- s for salad;
- p for pizza.
On the first day:
In the second day:
On the third day:
The price of each dish can be obtained by solving the system:
Resolution
Using the elimination method.
Multiply 20c + 40s + 30p = 6000 by 2.
Subtract the second matrix equation obtained from the first.
In the matrix above, we replace this equation with the second one.
We multiply the third equation above by 4.
Subtracting the third from the first equation, we get:
Substituting the equation obtained by the third one.
Subtracting equations two and three, we have:
From the third equation, we get p = 80.
Substituting p in the second equation:
50s + 50.80 = 5000
50s + 4000 = 5000
50s = 1000
s = 1000/50 = 20
Substituting the values of s and p in the first equation:
40c + 30.20 + 10.80 = 7000
40c + 600 + 800 = 7000
40c = 7000 - 600 - 800
40c = 5600
c = 5600 / 40 = 140
Solution
p=80, s=20 and c=140
question 8
(UEMG) In the plan, the system represents a pair of lines
a) coincident.
b) distinct and parallel.
c) concurrent lines at the point ( 1, -4/3 )
d) concurrent lines at the point ( 5/3, -16/9 )
Multiplying the first equation by two and adding the two equations:
Substituting x in equation A:
question 9
(PUC-MINAS) A certain laboratory sent 108 orders to pharmacies A, B and C. It is known that the number of orders sent to pharmacy B was twice the total number of orders sent to the two other pharmacies. In addition, three orders more than half the amount shipped to pharmacy A were dispatched to pharmacy C.
Based on this information, it is CORRECT to state that the total number of orders sent to pharmacies B and C was
a) 36
b) 54
c) 86
d) 94
According to the statement we have:
A + B + C = 108.
Also, that the amount of B was twice that of A + C.
B = 2(A + C)
Three orders were dispatched to pharmacy C, more than half of the quantity dispatched to pharmacy A.
C = A/2 + 3
We have equations and three unknowns.
Using the substitution method.
Step 1: replace the third with the second.
Step 2: Substitute the result obtained and the third equation in the first.
Step 3: Substitute the value of A to determine the values of B and C.
B = 3A + 6 = 3.22 + 6 = 72
For C:
Step 4: add the values of B and C.
72 + 14 = 86
question 10
(UFRGS 2019) So that the system of linear equations possible and determinate, it is necessary and sufficient that
a) a ∈ R.
b) a = 2.
c) a = 1.
d) a ≠ 1.
c) a ≠ 2.
One of the ways to classify a system as possible and determinate is through Cramer's method.
The condition for this is that the determinants are different from zero.
Making the determinant D of the main matrix equal to zero:
To learn more about linear systems:
- Linear Systems: what they are, types and how to solve
- Systems of Equations
- Scaling of Linear Systems
- Cramer's Rule
For more exercises:
- Systems of Equations of the 1st Degree
ASTH, Rafael. Exercises on solved linear systems.All Matter, [n.d.]. Available in: https://www.todamateria.com.br/exercicios-de-sistemas-lineares-resolvidos/. Access at:
See too
- Linear systems
- Scaling of Linear Systems
- Systems of Equations
- 11 exercises on matrix multiplication
- Second degree equation
- Inequality Exercises
- 27 Basic Mathematics exercises
- Cramer's Rule