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Solve the following pairs of equations:

(i) x+ y = 3.3 ; $\frac{0\cdot 6}{3x-2y}= -1,3x-2y\neq 0$
(ii) $\frac{x}{3}+\frac{y}{4}= 4$ , $\frac{5x}{6}-\frac{y}{8}= 4$
(iii) $4x+\frac{6}{y}= 15;6x-\frac{8}{y}= 14,y\neq 0$
(iv) $\frac{1}{2x}-\frac{1}{y}= -1;\frac{1}{x}+\frac{1}{2y}= 8,x,y\neq 0$
(v)43x + 67y = – 24 ; 67x + 43y = 24
(vi) $\frac{x}{a}+\frac{y}{b}= a+b;\frac{x}{a_{2}}+\frac{y}{b^{2}}= 2 \left ( a,b\neq 0 \right )$

(vii) $\frac{2xy}{xy}= \frac{2}{3};\frac{xy}{2x-y}= \frac{-3}{10};x+y\neq 0,2x-y\neq 0$

Answers (1)

(i)Solution:
Given equations are x + y = 3.3 … (1)
0.6 = 2y – 3x …(2)
Using elimination method in equation (1) and (2) we get
2x + 2y = 6.6 [multiply eq. (1) by 2]
–3x + 2y = 0.6
+ – –
5x = 6
x = $\frac{6}{5}$ = 1.2
Put x = 1.2 in (1) we get
1.2 + y = 3.3
y = 3.3 – 1.2
y = 2.1

(ii)Answer: y = 6 ; y = 8
Solution:
Given equations are :
$\frac{x}{3}+\frac{y}{4}= 4$ ………(1)

$\frac{5x}{6}-\frac{y}{8}= 4$ ……….(2)

Multiply eq. (1) by 12 and eq. (2) by 24 and then add them.
We get
4x + 3y + 20x -3y = 48 + 96
24x = 144

$x= \frac{144}{24}= 6$
x = 6
Put x = 6 in eq. (1) we have
$\frac{6}{3}+\frac{y}{4}= 4$
$2+\frac{y}{4}= 4$
$\frac{y}{4}= 4-2$
$\frac{y}{4}= 2$
y = 8

(iii)Solution:
Equation are: $4 x+\frac{6}{y}=15 \\$ … (1)

$6 x-\frac{8}{y}=14$ … (2)

Multiply equation (1) by 6 and equation (2) by 4 and then subtract equation (2) from (1)

24x + $\frac{36}{y}$ = 90

24x – 32/y = 56

– + –

$\frac{36}{y}+\frac{32}{y}= 34$

$\frac{36+32}{y}= 34$

68 = 34y

y = $\frac{68}{34}= 2$
y = 2
Put y = 2 in equation (1) we get

4x = 15 – 3
4x = 12
x = $\frac{12}{4}$ = 3
x = 3
(iv)Solution:
Let $\frac{1}{x}$ = u and $\frac{1}{y}$ = v and put in the given equations

u/2 – v = – 1 … (1)

u + $\frac{v}{2}$ = 8 … (2)
Multiply eq. (1) by 2 and subtract from eq. (2) we get
$\frac{v}{2}+2v= 8+2$
v + 4v = 10 × 2
5v = 20

v = $\frac{20}{+5}= 4$
v = 4
v = 1/y = 4 $\Rightarrow$
y = $\frac{1}{4}$
Put v = 4 in equation (1) we get
u/2= –1 + 4
u = 6
$\frac{1}{x} = 6$
(because 1/x = u)

$\Rightarrow x= \frac{1}{6}$
(v)Solution:

Equations are 43x + 67y = –24 … (1)
67x + 43y = 24 … (2)
Multiply equation (1) by 67 and equation (2) by 43 and then subtract equation (2) from (1)
2881x + 4489y = –1608
2881x + 1849y = 1032
– – –
2640y = –2640
y = –1
Put y = –1 in (1) we get
43x + 67(–1) = –24
43x = –24 + 67
43x = 43
x = 1
(vi)Solution:
Given equation are:
$\frac{x}{a}+\frac{y}{b}= a+b$
$\frac{x}{a^{2}}+\frac{y}{b^{2}}= 2$
$\frac{x}{a}+\frac{y}{b}-\left ( a+b \right )= 0$
$\frac{x}{a^{2}}+\frac{y}{b^{2}}-2= 0$
Using cross multiplication, we have

$\frac{x}{\frac{-2}{b}+\frac{a}{b^{2}}+\frac{1}{b}}= \frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{b^{2}}}= \frac{1}{\frac{1}{ab^{2}}-\frac{1}{ba^{2}}}$
$\frac{x}{\frac{a-b}{b^{2}}}= \frac{-y}{\frac{-a+b}{a^{2}}}= \frac{1}{\frac{1}{ab^{2}}-\frac{1}{b^{2}}}$
$\frac{x}{\frac{a-b}{b^{2}}}=\frac{1}{\frac{+\left ( a-b \right )}{a^{2}b^{2}}}$
$x= \frac{\left ( a-b \right )\times a^{2} b^{2}}{b^{2}\left ( a-b \right )}$
x = a²
$\frac{-y}{\frac{-\left ( a-b \right )}{a^{2}}}= \frac{1}{\frac{\left ( a-b \right )}{a^{2}b^{2}}}$
y = $\frac{\left ( a-b \right )a^{2}b^{2}}{a^{2}\left ( a-b \right )}$
y = b²
(vii)Solution: