3. Pair of Linear Equation in two variables

Class X
Exercise 3.3

Q.No. 1. Solve the following pair of linear equation by substitution method.
(i) 
x + y = 14
x - y = 4
(ii)
s - t = 3
s/3 + t/2 = 6
(iii)
3x - y = 3
9x - 3y = 9
(iv)
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)
√2x + √3y = 0
√3x - √8y = 0
(vi)
3x/2 + 5y/3 = -2
x/3 + y/2 = 13/6

Solution:

(i)
From equation no. (i)
x = 14 - y ------------- (iii)
Now, substitute the value of x in equation no. (ii)
14 - y - y = 4
or, 14 - 2y = 4
or, 2y = 14 - 4 = 10
∴ y  = 10/2 = 5
Now, putting the value of y in equation no. (iii)
 x = 14 - 5 = 9
Hence, the value of x = 5 and y = 9 Ans.

(ii)
From equation no. (i)
s = 3 + t ---------- (iii)
Now, from equation no. (ii)
(3 + t)/3 + t/2 = 6
or, 2(3 + t) + 3t = 6x6
or, 6 + 2t + 3t = 36
or, 6 + 5t = 36
or, 5t = 36 - 6 
or, t = 30/5 = 6
Now putting the value of t in equation no. (iii)
s = 3 + 6 = 9
Hence, the value of s = 9 ant t = 6 Ans.

(iii)
From equation no. (i)
x = (3 + y)/3 ---------------- (iii)
Now, substitute the value of x in equation no. (ii)
9(3 + y)/3 - 3y = 9
9 + 3y - 3y = 9
9 = 9
This is independent of y.
Hence, y has infinite many solution.
Therefore x has infinite many solution.

(iv)

From equation no. (i)
x = (1.3 - 0.3y)/0.2 ----------- (iii)
Now, substitute the value of x in equation no. (ii)
0.4(1.3 - 0.3y)/0.2 + 0.5y = 2.3
or, 2.6 - 0.6y + 0.5y = 2.3
or, 0.1y = 0.3
or, y = 3
Now, putting the value of y in equation no. (iii)
x = (1.3 - 0.9)/0.2
or, x = 2
Hence, x = 2 and y = 4.

(v)
From equation no. (i)
x = -√3y/√2 ------------ (iii)
Now substitute the value of x in equation no. (ii)
-√3x√3y/√2 - √8y = 0
or, 3y - 4y = 0
or, y = 0
Now, putting the value of y in equation no. (iii)
x = 0
Hence, x = 0 and y = 0. Ans.

(vi)
From equation no. (i)
x = (-12 - 10y)/9 -------------- (iii)
Now, substitute the value of x in equation no.(ii)
(-12 - 10y)/27 + y/3 = 13/6
or, -12 - 10y + 9y = 13x27/6
or, -12 - y = 39/2
y = (-24 - 39)/2
y = -63/2
Now, putting the value of y in equation no. (iii)
x = (-12 - 165)/9
x = - 177/9
x = -59/3
x = -59/3 and y = -63/2 Ans.

Exercise 3.4

Q.No.1. Solve the following pair of linear equations by the elimination methods and substitution method:

(i) x + y = 5 and 2x - 3y = 4
(ii) 3x + 4y = 10 and 2x - 3y = 4
(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y = -1 and x - y/3 = 3

(i) Solution:
Elimination Method
Multiply equation no. (i) by 3 and (ii) by 1
3x + 3y = 15 ------------ (iii)
2x - 3y = 4 --------------- (iv)
Now, add equation no. (iii) and (iv)
5x = 19
x = 19/5
Now, putting the value of x in equation no. (iii)
3x19/5 + 3y = 15
3y = 15 - 57/5
3y = (75 - 57)/5
y = 18/15 = 6/5
Required answer 
x = 19/5 and y = 6/5

(ii)
Solution:
Elimination Method
Multiply the equation no. (i) by 3 and (ii) by 4
9x + 12y = 30 ---------- (iii)
8x - 12y = 16 ----------- (iv)
Now, add the equations
17x = 46
x = 46/17
Now, putting the value of x in equation no.(ii)
92/17 - 3y = 4
3y = 92/17 - 4
3y = (92 - 68)/17
y = 24/51
Required answer 
x = 46/17 and y = 24/51 Ans.

(iii)
Solution:
Elimination Method
3x - 5y = 4 ------------ (i)
9x - 2y = 7 ----------- (ii)
Multiply equation no. (i) by 2 and (ii) by 5
6x - 10y = 8 ----------- (iii)
45x - 10y = 35 -------- (iv)
Now, subtract equation no.(iii) from (iv)
39x = 27
x = 27/39
x = 9/13
Now, substitute the value of x in equation no.(i)
27/13 - 5y = 4
5y = 27/13 - 4
5y = (27 - 52)/13
y = -25/65
y = -5/13
Required Answer
x = 9/13 and y = -5/13 Ans.

(iv)
Solution:
x/2 + 2y = -1 ------- (i)
x - y/3 = 3
Multiply equation no.(i) by 1/3 and (ii) by 2
x/6 + 2y/3 = -1/3 ---------- (iii)
2x - 2y/3 = 6 ---------- (iv)
Now, add equation no. (iii) and (iv)
x/6 + 2x = -1/3 + 6
13x/6 = 17/3
x = 34/13
Now, substitute the value of x in equation no.(ii)
34/13 - y/3 = 3
y/3 = 34/13 - 3
y/3 = (34 - 39)/13
y = -5/39
Required Answer
x = 34/13 and y = -5/39 Ans.
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