Andhra Pradesh Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables InText Questions Textbooks Solutions PDF
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AP Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables InText Questions Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 10th Maths Chapter 4 Pair of Linear Equations in Two Variables InText Questions Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables InText Questions and Answers
Question 1.
Solve the following systems of equations: i) x – 2y = 0; 3x + 4y = 20 (Page No. 79)
Answer:
i) x – 2y = 0; 3x + 4y = 20
-2y = -x 4y = 20 – 3x
y = 𝑥2 y = 20−3𝑥4
The two lines meet at (4, 2).
The solution set is {(4, 2)}
ii) x + y = 2
2x + 2y = 4
Answer:
x + y = 2
2x + 2y = 4
These two are coincident lines.
∴ There are infinitely many solutions.
iii) 2x – y = 4
4x – 2y = 6
Answer:
2x – y = 4 4x – 2y = 6
⇒ y = 2x – 4 ⇒ 2y = 4x – 6 ⇒ y = 2x – 3
These two are parallel lines.
∴ The pair of linear equations has no solution.
Question 2.
Two rails of a railway track are represented by the equations.
x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation graphically. (Page No. 79)
Answer:
x + 2y – 4 = 0 2x + 4y – 12 = 0
2y = 4 – x 4y = 12 – 2x (or) 4y = 2 (6 – x)
y = 4−𝑥2 y = 6−𝑥2
x + 2y – 4 = 0 2x + 4y – 12 = 0
These lines are parallel and hence no solution.
Question 3.
Check each of the given systems of equations to see if it has a unique solution, infinitely many solutions or no solution. Solve them graphically. (Page No. 83)
i) 2x + 3y = 1
3x – y = 7
Answer:
Let a1x + b1y – c1 = 0 ≃ 2x + 3y – 1 = 0
a2x + b2y + c2 = 0 ≃ 3x – y – 7 = 0
Now comparing their coefficients i.e., 𝑎1𝑎2 and 𝑏1𝑏2
⇒ 23 ≠ 3−1
The given lines are intersecting lines.
2x + 3y = 1 3x – y = 7
3y = 1 – 2x y – 3x = 7
y = 1−2𝑥3
The system of equations has a unique solution (2, – 1).
ii) x + 2y = 6
2x + 4y = 12
Answer:
From the given pair of equations,
𝑎1𝑎2 = 12;
𝑏1𝑏2 = 24 = 12;
𝑐1𝑐2 = 612 = 12
∴ 𝑎1𝑎2 = 𝑏1𝑏2 = 𝑐1𝑐2
∴ The lines are dependent and have infinitely many solutions.
x + 2y = 6 2x + 4y = 12
2y = 6 – x 4y = 12 – 2x (or) 4y = 2(6 – x)
y = 6−𝑥2 y = 12−2𝑥4 or y = 6−𝑥2
iii) 3x + 2y = 6
6x + 4y = 18
Answer:
From the given pair of equations,
𝑎1𝑎2 = 36 = 12;
𝑏1𝑏2 = 24 = 12;
𝑐1𝑐2 = 618 = 13
∴ 𝑎1𝑎2 = 𝑏1𝑏2
∴ The lines are parallel and hence no solution.
3x + 2y = 6 6x + 4y = 18
2y = 6 – 3x 4y = 18 – 6x
y = 6−3𝑥2 y = 18−6𝑥4
Try these
(Page No. 75, 76)
Question 1.
Mark the correct option in the following questions:
Which of the following equations is not a linear equation?
a) 5 + 4x = y + 3
b) x + 2y = y – x
c) 3 – x = y2 + 4
d) x + y = 0
Answer:
[ c ]
Question 2.
Which of the following is a linear equation in one variable?
a) 2x + 1 = y – 3
b) 2t – 1= 2t + 5
c) 2x – 1 = x2
d) x2 – x + 1 =0
Answer:
[ b ]
Question 3.
Which of the following numbers is a solution for the equation 2(x + 3) = 18?
a) 5
b) 6
c) 13
d) 21
Answer:
[b]
Question 4.
The value of x which satisfies the equation 2x – (4 – x) = 5 – x is
a) 4.5
b) 3
c) 2.25
d) 0.5
Answer:
[ c ]
Question 5.
The equation x – 4y = 5 has
a) no solution
b) unique solution
c) two solutions
d) infinitely many solutions
Answer:
[ d ]
Question 6.
In the example given above, can you find the cost of each bat and ball? (Page No. 79)
Answer:
We can’t find the exact values for the costs of bat and ball as there are infinitely many possibilities.
Question 7.
For what value of ‘p’ the following pair of equations has a unique solution. (Page No. 83)
2x + py = – 5 and 3x + 3y = – 6
Answer:
Given: 2x + py = – 5
3x 4- 3y = – 6
To have an unique solution we should have
∴ The pair has unique solution when p ≠ 2.
Question 8.
Find the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines. (Page No. 83)
Answer:
Given: 2x – ky + 3 = 0
4x + 6y – 5 = 0
If the above lines are to be parallel, then
∴ k = – 3 is the required value for which lines are parallel.
Question 9.
For what value of ‘k’, the pair of equation 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines. (Page No. 83)
Answer:
Given: 3x + 4y + 2 = 0
9x + 12y + k = 0
If the lines are to be coincident with each other, then
∴ k = 2 × 3 = 6
Question 10.
For what positive values of ‘p’ the following pair of linear equations have infinitely many solutions? (Page No. 83)
px + 3y – (p – 3) = 0
12x + py – p = 0
Answer:
Given: px + 3y – (p – 3) = 0
12x + py – p = 0
The above equations to have infinitely many solutions
p.p = 12 × 3
⇒ p2 = 36
⇒ p = ±6
Think & Discuss
Question 1.
Two situations are given below:
i) The cost of 1 kg potatoes and 2 kg tomatoes was Rs. 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be Rs. 66.
ii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for Rs. 3900. Later he buys one more bat and 2 balls for Rs. 1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each case. (Page No. 73)
Answer:
i) The unknowns in the first problem are
a) cost of 1 kg tomatoes
b) cost of 1 kg potatoes
ii) In the second problem, the unknowns are
a) cost of each bat
b) cost of each ball
Question 2.
Is a dependent pair of linear equations always consistent? Why or why not? (Page No. 79)
Answer:
Reason: 𝑎1𝑎2 = 𝑏1𝑏2 = 𝑐1𝑐2 always holds. In other words, they have infinitely many solutions.
Do these
Solve each pair of equations by using the substitution method. (Page No. 88)
Question 1.
3x – 5y = -1
x – y = -1
Answer:
Given: 3x – 5y = -1 ……. (1)
x – y = -1 …….. (2)
From equation (2), x – y = – 1
x = y – 1
Substituting x = y – 1 in equation (1)
we get
3 (y – 1) – 5y = – 1
⇒ 3y – 3 – 5y = r 1
⇒ – 2y = – 1 + 3
⇒ 2y = – 2
⇒ y = -1
Substituting y = – 1 in equation (1) we get
3x – 5 (- 1) = -1
3x + 5 = – 1
3x = – 1 – 5
x = −63 = -2
∴ The solution is (-2, -1)
Question 2.
x + 2y = – 1
2x – 3y = 12
Answer:
Given: x + 2y = -1 ……. (1)
2x – 3y = 12 …….. (2)
From equation (1)x + 2y = -l
⇒ x = – 1 – 2y
Substituting x = – 1 – 2y in equation (2), we get
2 (- 1 – 2y) – 3y = 12
– 2 – 4y – 3y = 12
– 2 – 7y = 12
7y = – 2 – 12
∴ y = −147 = -2
Substituting y = – 2 in equation (1), we get
x + 2 (- 2) = – 1
x = – 1 + 4
x = 3
∴ The solution is (3, – 2)
Question 3.
2x + 3y = 9
3x + 4y = 5
Answer:
Given: 2x + 3y – 9 …….. (1)
3x + 4y = 5 ……. (2)
From equation (1);
2x = 9 – 3y
x = 9−3𝑦2
Substituting x = 9−3𝑦2 in equation (2)
we get
Substituting y = + 17 in equation (1) we get
2x + 3 (+ 17) = 9
⇒ 2x = 9 – 51
⇒ 2x = -42
⇒ x = -21
∴ The solution is (-21, 17)
Question 4.
x + 6𝑦 = 6
3x – 8𝑦 = 5
Answer:
Given:
x + 6𝑦 = 6 …….. (1)
3x – 8𝑦 = 5 …….. (2)
From equation (1) x = 6 – 6𝑦
Substituting x = 6 – 6𝑦 in equation (2)
we get
Substituting y = 2 in equation (1) we get
x + 62 = 6 ⇒ x + 3 = 6
∴ x = 3
∴ The solution is (3, 2)
Question 5.
0.2x + 0.3y =1.3
0.4x + 0.5y = 2.3
Answer:
Given:
0.2x + 0.3y = 1.3
⇒ 2x + 3y = 13 …… (1)
0.4x + 0.5y = 2.3
⇒ 4x + 5y = 23 …… (2)
From equation (1)
2x = 13 – 3y
⇒ x = 13−3𝑦2
Substituting x = 13−3𝑦2 equation (2) we get
13−3𝑦2 + 5y = 23
⇒ 26 – 6y + 5y = 23
⇒ -y + 26 = 23
⇒ y = 26 — 23 = 3
Substituting y = 3 in equaion (1) we get
2x + 3(3) = 13
⇒ 2x + 9 = 13
⇒ 2x = 13 – 9
⇒ 2x = 4
⇒ x = 42 = 2
∴ The solution is (2, 3)
Question 6.
√2x + √3y = 0
√3x – √8y = 0
Answer:
Given:
√2x + √3y = 0 ……. (1)
√3x – √8y = 0 ……. (2)
Substitute x = 0 in (1),
√2(0) + √3y = 0
√3y = 0
∴ y = 0
∴ The solution is x = 0, y = 0
Note: a1x + b1y + c1 = 0
a2x + b2y + c2 = 0, if c1 = c2 = 0
then, x = 0, y = 0 is a solution.
Solve each of the following pairs of equations by the elimination method. (Page No. 89)
Question 7.
8x + 5y = 9
3x + 2y = 4
Answer:
Given: 8x + 5y = 9 ……. (1)
3x + 2y = 4 …….. (2)
∴ y = 5
Substituting y = 5 in equation (1) we get
8x + 5 × 5 = 9
⇒ 8x = 9 – 25
x = −168 = -2
∴ The solution is (- 2, 5)
Question 8.
2x + 3y = 8
4x + 6y = 7
Answer:
Given: 2x + 3y = 8 ……. (1)
4x + 6y = 7 …….. (2)
The lines are parallel.
∴ The pair of lines has no solution.
Question 9.
3x + 4y = 25
5x – 6y = -9
Answer:
Given: 3x + 4y = 25 ……. (1)
5x – 6y = -9 …….. (2)
Substituting y = 4 in equation (1) we get
3x + 4 × 4 = 25
3x = 25 – 16
⇒ x = 93 = 3
∴ (3,4) is the solution for given pair of lines.
Question 10.
In a competitive exam, 3 marks are awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test? (Madhu attempted all the questions) .
Now use the elimination method to solve the above example – 9.
Answer:
The equations formed are
3x – y = 40 ……. (1)
4x – 2y = 50 …….. (2)
Substituting y = 5 in equation (1) we get
3x – 5 = 40
⇒ 3x = 40 + 5
⇒ x = 453 = 15
Total number of questions = Number of correct questions + Number of wrong answers
= x + y
= 15 + 5 = 20
Question 11.
Mary told her daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Find the present age of Mary and her daughter. Solve example – 10 by the substitution method.
Answer:
The equations formed are
The equations formed are
x – 7y + 42 = 0 ……. (1)
x – 3y – 6 = 0 …….. (2)
From (1), x = – 42 + 7y
Substituting x = – 42 + 7y in equation (2) we get
-42 + 7y – 3y – 6 = 0
⇒ 4y – 48 = 0
⇒ y = 484 = 12
Substituting y = 12 in equation (2) we get
x – 3 × 12 – 6 = 0
x – 36 – 6 = 0
x = 42
Try this
Question 1.
Solve the given pair of linear equations, (a – b)x + (a + b)y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2 (Page No. 89)
Answer:
x(-2b) = – 2b(a + b)
⇒ x = (a + b)
Put this value of ‘x’ in eq (1) we get
(a – b) (a + b) + (a + b)y = a2 – 2ab – b2
a2 – b2 + (a + b)y = a2 – 2ab – b2
⇒ y = −2𝑎𝑏𝑎+𝑏
∴ Solution to given pair of linear equations x = a + b, y = −2𝑎𝑏𝑎+𝑏
AP Board Textbook Solutions PDF for Class 10th Maths
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