# AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions

## AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions and Answers.

### 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 = $$\frac{x}{2}$$                                                                                      y = $$\frac{20-3x}{4}$$   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 = $$\frac{4-x}{2}$$                                                                                  y = $$\frac{6-x}{2}$$
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., $$\frac{a_{1}}{a_{2}}$$ and $$\frac{b_{1}}{b_{2}}$$
⇒ $$\frac{2}{3}$$ ≠ $$\frac{3}{-1}$$
The given lines are intersecting lines.
2x + 3y = 1                                                                     3x – y = 7
3y = 1 – 2x                                                                      y – 3x = 7
y = $$\frac{1-2x}{3}$$  The system of equations has a unique solution (2, – 1). ii) x + 2y = 6
2x + 4y = 12
Answer:
From the given pair of equations,
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{6}{12}$$ = $$\frac{1}{2}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$
∴ 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 = $$\frac{6-x}{2}$$                                                                             y = $$\frac{12-2x}{4}$$ or y = $$\frac{6-x}{2}$$  iii) 3x + 2y = 6
6x + 4y = 18
Answer:
From the given pair of equations,
$$\frac{a_{1}}{a_{2}}$$ = $$\frac{3}{6}$$ = $$\frac{1}{2}$$;
$$\frac{b_{1}}{b_{2}}$$ = $$\frac{2}{4}$$ = $$\frac{1}{2}$$;
$$\frac{c_{1}}{c_{2}}$$ = $$\frac{6}{18}$$ = $$\frac{1}{3}$$
∴ $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$
∴ The lines are parallel and hence no solution.
3x + 2y = 6                                                                                   6x + 4y = 18
2y = 6 – 3x                                                                                    4y = 18 – 6x
y = $$\frac{6-3x}{2}$$                                                                                    y = $$\frac{18-6x}{4}$$   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: $$\frac{a_{1}}{a_{2}}$$ = $$\frac{b_{1}}{b_{2}}$$ = $$\frac{c_{1}}{c_{2}}$$ 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 = $$\frac{-6}{3}$$ = -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 = $$\frac{-14}{7}$$ = -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 = $$\frac{9-3y}{2}$$
Substituting x = $$\frac{9-3y}{2}$$ 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 + $$\frac{6}{y}$$ = 6
3x – $$\frac{8}{y}$$ = 5
Answer:
Given:
x + $$\frac{6}{y}$$ = 6 …….. (1)
3x – $$\frac{8}{y}$$ = 5 …….. (2)
From equation (1) x = 6 – $$\frac{6}{y}$$
Substituting x = 6 – $$\frac{6}{y}$$ in equation (2)
we get Substituting y = 2 in equation (1) we get
x + $$\frac{6}{2}$$ = 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 = $$\frac{13-3y}{2}$$
Substituting x = $$\frac{13-3y}{2}$$ equation (2) we get
$$\frac{13-3y}{2}$$ + 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 = $$\frac{4}{2}$$ = 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 = $$\frac{-16}{8}$$ = -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 = $$\frac{9}{3}$$ = 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 = $$\frac{45}{3}$$ = 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 = $$\frac{48}{4}$$ = 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 = $$\frac{-2ab}{a+b}$$
∴ Solution to given pair of linear equations x = a + b, y = $$\frac{-2ab}{a+b}$$