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## Solve the following simultaneous equations.${7 \over 2x+1}+{13 \over y+2}=27$ ;${13 \over 2x+1} + {7 \over y+2 } =33$

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Sol:

$7({1 \over 2x+1}) + 13({1 \over y+ 2})=27$     ...(I)

$13({1 \over 2x +1})+ {7({1 \over y+2})}=33$    ...(II)

Replace  $1\over 2x+1$  by a  and  $1 \over y+2$ by b in  equations (I) and (II)

we get,

7a + 13b = 27 ...(III)

13a + 7b = 33 ...(IV)

[In the two equations above, the coefficients of x and y are interchanged so use ASA]

7a + 13 b = 27

+ 13a +7 b = 33

20a  +20 b = 60

a + b = 3 ...(V)

Subtracting eq. (III) and (IV)

7a +13b = 27

13a +7 b = 33

(–)    (–)    (–)

–6a +6 b = –6

–a + b = –1     ...(VI)

a + b =3

+  –a + b = –1

2b = 2

b =1

Place b = 1 in eq. (V)

a + 1 = 3
∴ a = 3 – 1

∴ a =2

Resubstituting the values of a and b

${1 \over 2x+1 } =2$   and   ${1 \over y+2} = 1$

2x + 1 = $1 \over 2$   and   y + 2 = 1

$2x={1\over 2}-1$  and  y =1 – 2

∴ $2x ={-1\over 2}$   and ∴  y = –1

$x={-1\over 4}$

∴  (x, y) = $({-1 \over 4}, -1)$  is the required solutions.

Linear equation in two variables August 04 , 2018 0 Comments 189 views