Question

Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4

(ii) s – t = 3
$\frac{s}{3} + \frac{t}{2}$ =6

(iii) 3x – y = 3
9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

(v)$\sqrt2x$ + $\sqrt3y$=0
$\sqrt3x$ - $\sqrt8y$ = 0

(vi) $\frac{3x}{2} - \frac{5y}{3}$ =-2,
$\frac{ x}{3} + \frac{y}{2}$ = $\frac{ 13}{6}$

Answer 1

(i) x + y = 14 ……………………..(1)
x − y = 4 ..…………………...(2)

From (1), we obtain
x = 14 − y .……………..........(3)

Substituting this value in equation (2), we obtain
(14-y) -y =4
14 -2y = 4
10=2y
y=5

Substituting this in equation (3), we obtain
x=9
∴ x=9, y=5

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(ii) s-t =3 ..……………..(1)
$\frac{s}{3} + \frac{t}{2}$ =6 ……………….(2)

From (1), we obtain
s= t+3 ………..(3)

Substituting this value in equation (2), we obtain
$\frac{t+3}{3}$ + $\frac{t}{2}$ =6
2t +6 +3t = 36
5t =30
t = 6
Substituting in equation (3), we obtain
s=9
∴s=9 , t=6

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(iii) 3x − y = 3 ......................(1)
9x − 3y = 9 .....................(2)

From (1), we obtain
y = 3x − 3 ......................(3)

Substituting this value in equation (2), we obtain
9x -3 (3x-3) =9
9x -9x +9 =9
9 = 9

This is always true. 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.

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(iv) 0.2x + 0.3y = 1.3 ........................(1)
0.4x + 0.5y = 2.3 ........................(2)

From equation (1), we obtain
x=$\frac{1.3-0.3y}{0.2}$ ........................(3)

Substituting this value in equation (2), we obtain
0.4($\frac{1.3-0.3y}{0.2}$) + 0.5 y = 2.3

2.6 -0.6y + 0.5 y =2.3
2.6 -2.3= 0.1y
03=0.1y
y=3

Substituting this value in equation (3), we obtain
x= $\frac{1.3 -0.3 \times 3 }{ 0.2 }$

= $\frac{1.3 -0.9}{0.2}$ =$\frac{ 0.4}{0.2}$ =2

∴ x=2, y=3

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**(v) **$\sqrt2x + \sqrt3y$ =0 .................(1)
$\sqrt3x - \sqrt8y$ =0 .................(2)

from equation (1), we obtain
x= -$\frac{\sqrt3y}{\sqrt2}$ .....................(3)

Substituting this value in equation (2), we obtain
$\sqrt3(\frac{-\sqrt3y}{\sqrt2}) -\sqrt8y$=0

$\frac{-3y}{\sqrt2} -2\sqrt2y$ =0

y($\frac{-3y}{\sqrt2} -2\sqrt2y$) =0

y=0

Substituting this value in equation (3), we obtain
x = 0
∴ x = 0, y = 0

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**(vi) **$\frac{3}{2}x - \frac{5}{3}y$ = -2 ...............(1)
$\frac{x}{3} + \frac{y}{2 }$=$\frac{13}{6}$ ................(2)

From equation (1), we obtain
9x-10y =12
x=$\frac{-12+10y}{9}$ ...............(3)

Substituting this value in equation (2), we obtain
$\frac{-12+\frac{10y}{9}}{3} +\frac{y}{2}$ =$\frac{13}{6}$

-$\frac{12+10y}{27 }$+$\frac{y}{2}$ = $\frac{13}{6}$

$\frac{-24 +20y +27y}{54}$ = $\frac{13}{6}$

$47y$= $117 + 24$
$47y$ =$141$
$y$ = $3$

Substituting this value in equation (3), we obtain
x= $\frac{-12 +10 \times 3}{9 }$= $\frac{18}{9}$ =2

∴ x=2, y=3