Solve the following pair of equations by substitution method

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..