Question 1 Solve the following pair of linear equations by the substitution method.
(i) x+y=14 (ii) s-t=3
x-y=4 s/3 + t/2 = 6
(iii) 3x-y=3 (iv) 0.2x+0.3y=1.3
9x-3y=9 0.4x+0.5y=2.3
(v) 2^0.5 x + 3^0.5 y = 0 (vi) 3x/2 - 5y/3 = -2
3^0.5 x - 8^0.5 y = 0 x/3 + y/2 = 13/6
Class 10 - Math - Pair of Linear Equations in Two Variables Page 53
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Firstly , find the value of one variable in terms of other variable from one of the given equation and then put this value in other equation to obtain a linear equation in one variable and solve it.
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Solution:
i) x + y = 14 …....................... (i)
x – y = 4 ….......................... (ii)
From equation (i), we get
x = 14 – y …......................... (iii)
Putting this value in equation (ii), we get
(14 – y) – y = 4
14 – 2y = 4
10 = 2y
y = 5
Putting this in equation (iii), we get
x =14 - y
x= 14 - 5
x = 9
∴ x = 9 and y = 5
(ii)
s – t = 3 …........................ (i)
s/3 + t/2 = 6 …................. (ii)
From equation (i),
s= t + 3........................ ......(iii)
Putting this value in equation (ii)
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5
t= 6
Putting t=6 in equation (iii)
s= t+3
s= 6+3
s = 9
∴ s = 9, t = 6
(iii)
3x – y = 3 …............................ (i)
9x – 3y = 9 …......................... (ii)
From equation (i)
y = 3x – 3 …........................... (iii)
Putting this value in equation (ii)
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This statement has no variable but I it is a true statement for all values of x,
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x – 3
Therefore, one of its possible solutions is x = 1, y = 0.
(iv)
0.2x + 0.3y = 1.3 …................. (i)
0.4x + 0.5y = 2.3 …................. (ii)
On multiplying both equation by 10 we get
2x + 3y = 13........................(iii)
4x+5y=23...........................(iv)
From eq iii
3y = 13- 2x
y=( 13-2x) /3................(v)
On substituting the value of y in equation iv
4x + 5/3(13 -2x) =23
(4x×3 + 5(13-2x))/3=23
12x+ 65 -10x = 23×3
12x-10x= 69-65
2x= 4
x= 4/2= 2
x= 2
On substituting x = 2 in eq v,
y= (13- 2×2)/3=( 13-4)/3= 9/3= 3
y=3
∴ x = 2 and y = 3
PART V & VI ARE IN THE ATTACHMENT..
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Solution:
i) x + y = 14 …....................... (i)
x – y = 4 ….......................... (ii)
From equation (i), we get
x = 14 – y …......................... (iii)
Putting this value in equation (ii), we get
(14 – y) – y = 4
14 – 2y = 4
10 = 2y
y = 5
Putting this in equation (iii), we get
x =14 - y
x= 14 - 5
x = 9
∴ x = 9 and y = 5
(ii)
s – t = 3 …........................ (i)
s/3 + t/2 = 6 …................. (ii)
From equation (i),
s= t + 3........................ ......(iii)
Putting this value in equation (ii)
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5
t= 6
Putting t=6 in equation (iii)
s= t+3
s= 6+3
s = 9
∴ s = 9, t = 6
(iii)
3x – y = 3 …............................ (i)
9x – 3y = 9 …......................... (ii)
From equation (i)
y = 3x – 3 …........................... (iii)
Putting this value in equation (ii)
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This statement has no variable but I it is a true statement for all values of x,
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x – 3
Therefore, one of its possible solutions is x = 1, y = 0.
(iv)
0.2x + 0.3y = 1.3 …................. (i)
0.4x + 0.5y = 2.3 …................. (ii)
On multiplying both equation by 10 we get
2x + 3y = 13........................(iii)
4x+5y=23...........................(iv)
From eq iii
3y = 13- 2x
y=( 13-2x) /3................(v)
On substituting the value of y in equation iv
4x + 5/3(13 -2x) =23
(4x×3 + 5(13-2x))/3=23
12x+ 65 -10x = 23×3
12x-10x= 69-65
2x= 4
x= 4/2= 2
x= 2
On substituting x = 2 in eq v,
y= (13- 2×2)/3=( 13-4)/3= 9/3= 3
y=3
∴ x = 2 and y = 3
PART V & VI ARE IN THE ATTACHMENT..
Attachments:
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