NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 3 PAIR OF LINEAR EQUATIONS IN TWO VARIABLES EX 3.3
Question 1. Solve the following pairs of linear equations by the substitution method:
Solution:
(i) Given,
x + y = 14 and x – y = 4 are the two linear equations in two variables.
From 1st equation x + y = 14 , we get ,
x = 14 – y …………………….Eq. (1)
Now, substitute the value of x from Eq. (1) in second equation x – y = 4 to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
y = 5
By the value of y, we can now find the exact value of x;
Put the value y = 5 in Eq. (1)
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.
(ii) Given,
s – t = 3 and (s / 3) + (t / 2) = 6 are the two linear equations in two variables.
From 1st equation s – t = 3 , we get,
s = 3 + t …………………….Eq. (1)
Now, substitute the value of s from Eq. (1) in second equation (s / 3) + (t / 2) = 6 to get,
{(3 + t) / 3)} + (t / 2) = 6
Taking LCM of the denominators 3 and 2
⇒ [ {2 ( 3 + t ) } + { 3t } ] / 6 = 6
⇒ {( 6 + 2t + 3t ) / 6} = 6
⇒ (6 + 5t) = 36
⇒ 5t = 30
⇒ t = 6
Now, substitute the value of t in Eq. (1)
s = 3 + 6 = 9
Therefore, s = 9 and t = 6.
(iii) Given,
3x – y = 3 and 9x – 3y = 9 are the two linear equations in two variables.
From 1st equation 3x – y = 3, we get,
x = (3 + y) / 3…………………….Eq. (1)
Now, substitute the value of x from Eq. (1) in the given second equation 9x – 3y = 9 to get,
9 {(3 + y) / 3} – 3y = 9
⇒9 +3y – 3y = 9
⇒ 9 = 9 .
Therefore, y has infinite values and since, x = (3+y) /3,
so x also has infinite values.
(iv) Given,
0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3 are the two linear equations in two variables.
From 1st equation 0.2x + 0.3y = 1.3 , we get, x = (1.3 – 0.3y) / 0.2 …………Eq. (1)
Now, substitute the value of x from Eq. (1) in the given second equation 0.4x + 0.5y = 2.3 to get,
[0.4{(1.3 – 0.3y) / 0.2}] + 0.5y = 2.3
⇒2(1.3 – 0.3y) + 0.5y = 2.3
⇒ 2.6 – 0.6y + 0.5y = 2.3
⇒ 2.6 – 0.1 y = 2.3
⇒ 0.1 y = 0.3
⇒ y = 3
Now, substitute the value of y in Eq. (1) , we get,
x = {1.3 – 0.3(3)} / 0.2 = (1.3 – 0.9) / 0.2 = 0.4 / 0.2 = 2
Therefore, x = 2 and y = 3.
(v) Given,
√2 x + √3 y = 0 and √3 x – √8 y = 0 are the two linear equations in two variables.
From 1st equation √2 x + √3 y = 0 , we get, x = – (√3 /√2) y …………………….Eq. (1)
Putting the value of x from Eq. (1) in the given second equation √3 x – √8 y = 0 to get,
√3 {–(√3 /√2)y} – √8y = 0
⇒ (–3/√2)y – √8 y = 0.
Taking y common, we get y{(–3 /√2 ) – 8 }= 0.
As {(–3/√2) – √8 } is Non-Zero.
... y should be zero to make y{(–3/√2) – √8 } as Zero.
⇒ y = 0
Now, substitute the value of y in Eq. (1), we get,
⇒ x = 0
Therefore, x = 0 and y = 0.
(vi) Given,
(3x / 2) – (5y / 3) = –2 and (x / 3) + (y / 2) = 13 / 6 are the two linear equations in two variables.
From 1st equation (3x / 2) – (5y / 3) = –2 , we get, (3 / 2) x = –2 + (5y / 3) ,
On further simplification, we get
⇒ x = 2(–6 + 5y) / 9 .
... x = (–12 + 10y) / 9 …………….Eq. (1)
Putting the value of x from Eq. (1) in the given second equation (x / 3) + (y / 2) = 13 / 6 to get,
{((–12 + 10y) / 9 ) / 3} + (y / 2) = 13 / 6
⇒{(–12 + 10y) / 27 } + (y / 2) =13 / 6…………Eq. (2)
Taking LCM of the denominators 27 & 2 , and further solving Eq. (2) ,
we get {(–24 + 20y + 27y) / 54 } = 13 / 6
y = 3
Now, substitute the value of y = 3 in Eq. (1), we get,
x = (–12 + 10 x 3) / 9
⇒ x = 18 / 9
⇒ x = 2
Therefore, x = 2 and y = 3
Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of’m’ for which y = mx +3.
Solution:
Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method:
(i) The difference between the two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charges per km? How much does a person have to pay for traveling a distance of 25 km?
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
(i) The difference between the two numbers is 26 and one number is three times the other. Find them.
Solution:
Let the two numbers be x and y respectively, such that y > x.
According to the question,
y = 3x ………………..Eq.(1)
y – x = 26 ……………Eq.(2)
Substituting the value of Eq.(1) into Eq.(2), we get
3x – x = 26
x = 13 …………………Eq.(3)
Substituting Eq.(3) in Eq.(1), we get y = 39
Hence, the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
Let the larger angle be x° and the smaller angle be y°.
We know that the sum of two supplementary pairs of angles is always 180°.
According to the question,
x + y = 180° …………….Eq. (1)
x – y = 18° ………………Eq.(2)
From (1), we get x = 180° – y …………. Eq.(3)
Substituting x = 180° – y in Eq. (2), we get
180° – y – y =18°
162° = 2y
y = 81° ………….. Eq.(4)
Using the value of y in Eq.(3), we get
x = 180° – 81°
x = 99°
Hence, the angles are 99° and 81° .
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.
Solution:
Let the cost a bat be x and cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. Eq.(1)
3x + 5y = 1750 ………………. Eq.(2)
From Eq.(1), we get
y = (3800 –7x)/6………………..Eq.(3)
Substituting Eq.(3) in Eq.(2) we get,
3x+5(3800 – 7x)/6 =1750
⇒ 3x+ 9500/3 – 35x/6 = 1750
⇒3x – 35x/6 = 1750 – 9500/3
⇒(18x – 35x) / 6 = (5250 – 9500)/3
⇒ –17x/6 = – 4250 / 3
⇒ –17x = – 8500
x = 500 ………………………..Eq.(4)
Substituting the value of x in Eq.(3), we get
y = (3800 –7 × 500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500, and the cost of a ball is Rs 50.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?
Solution:
Let the fixed charge be Rs x and per km charge be Rs y.
According to the question,
x + 10y = 105 …………….. Eq.(1)
x + 15y = 155 …………….. Eq.(2)
From (1), we get x = 105 – 10y ………………. Eq.(3)
Substituting the value of x in Eq.(2), we get
105 – 10y + 15y = 155
5y = 50
y = 10 …………….. Eq.(4)
Putting the value of y in Eq.(3), we get
x = 105 – 10 × 10 = 5
Hence, fixed charge is Rs 5 and per km charge = Rs 10
Charge for 25 km = x + 25y = 5 + 250 = Rs 255
(v) A fraction becomes 9 / 11 , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5 / 6. Find the fraction.
Solution:
Let the fraction be x / y.
According to the question,
(x + 2) / (y + 2) = 9 / 11
11x + 22 = 9y + 18
11x – 9y = – 4 …………….. Eq.(1)
(x+3) /(y+3) = 5/6
6x + 18 = 5y +15
6x – 5y = – 3 ………………. Eq.(2)
From (1), we get x = (– 4 + 9y )/11 …………….. Eq.(3)
Substituting the value of x in Eq.(2), we get
6 (–4 + 9y)/ 11 –5y = –3
–24 + 54y – 55y = –33
–y = –9
y = 9 ………………… Eq.(4)
Substituting the value of y in Eq.(3), we get
x = (–4 + 9 × 9 )/11 = 7
Hence the fraction is 7/9.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solutions:
Let the age of Jacob and his son be x and y respectively.
According to the question,
(x + 5) = 3(y + 5)
x – 3y = 10 ……………………………………..Eq.(1)
(x – 5) = 7(y – 5)
x – 7y = -30 ……………………………………….Eq.(2)
From (1), we get x = 3y + 10 ……………………. Eq.(3)
Substituting the value of x in Eq.(2), we get
3y + 10 – 7y = –30
– 4y = – 40
y = 10 ………………… Eq.(4)
Substituting the value of y in Eq.(3), we get
x = 3 x 10 + 10 = 40
Hence, the present age of Jacob’s and his son is 40 years and 10 years respectively.