Question

1/3x+1/5y=1/15 ; 1/2x + 1/3y = 1/12
solve this simultaneous equation ​

Answer 1

Required Answer:-

Given:

$\sf\left \{ {{ \dfrac{1}{3x} + \dfrac{1}{5y} = \dfrac{1}{15} } \atop { \dfrac{1}{2x} + \dfrac{1}{3y} = \dfrac{1}{12} }} \right.$

To Find:

• The values of x and y.

Answer:

• (x, y) = (2, -2)

Solution:

Given that,

$\sf \implies\dfrac{1}{3x} + \dfrac{1}{5y} = \dfrac{1}{15} \: - \: (i)$

$\sf \implies\dfrac{1}{2x} + \dfrac{1}{3y} = \dfrac{1}{12} \: - \: (ii)$

Let us assume that,

$\sf \implies a = \dfrac{1}{x} \: \: and \: \: b = \dfrac{1}{y}$

Therefore,

$\sf \implies\dfrac{a}{3} + \dfrac{b}{5} = \dfrac{1}{15}$

$\sf \implies15 \times \bigg(\dfrac{a}{3} + \dfrac{b}{5} \bigg)= 15 \times \dfrac{1}{15}$

$\sf \implies 5a+ 3b= 1 \: - \: (iii)$

Also,

$\sf \implies\dfrac{a}{2} + \dfrac{b}{3} = \dfrac{1}{12}$

$\sf \implies 12 \times \bigg( \dfrac{a}{2} + \dfrac{b}{3} \bigg) = 12 \times \dfrac{1}{12}$

$\sf \implies 6a + 4b = 1 \: - \: (iv)$

Multiplying both sides of equation (iii) by 4, we get,

$\sf \implies 20a+ 12b = 4 \: - \: (v)$

Multiplying both sides of equation (iv) by -3, we get,

$\sf \implies - 18a + - 12b = - 3 \: - \: (vi)$

Adding equations (v) and (vi), we get,

$\sf \implies 2a = 1$

$\sf \implies a = \dfrac{1}{2}$

Substituting the value of a in equation (iii), we get,

$\sf \implies 5 \times \dfrac{1}{2} + 3b= 1$

$\sf \implies 3b= 1 - \dfrac{5}{2}$

$\sf \implies 3b= \dfrac{2 - 5}{2}$

$\sf \implies 3b= \dfrac{ - 3}{2}$

$\sf \implies b= \dfrac{ - 3}{2} \times \dfrac{1}{3}$

$\sf \implies b= \dfrac{ - 1}{2}$

Therefore,

$\sf \implies (a,b) = \bigg( \dfrac{1}{2} , \dfrac{ - 1}{2} \bigg)$

Now,

$\sf \implies a= \dfrac{1}{x}$

$\sf \implies \dfrac{1}{2} = \dfrac{1}{x}$

$\sf \implies x = 2$

Again,

$\sf \implies b= \dfrac{1}{y}$

$\sf \implies \dfrac{ - 1}{2} = \dfrac{1}{y}$

$\sf \implies y = - 2$

Therefore,

$\sf \implies (x,y) = (2, - 2)$

Answer 2

QuestioN :

1/3x+1/5y=1/15 ; 1/2x + 1/3y = 1/12

solve this simultaneous equation ​

GiveN :

• 1/3x + 1/5y = 1/15
• 1/2x + 1/3y = 1/12

To FiNd :

• solve this simultaneous equation ​

ANswer :

(x, y) = (2, -2)

SolutioN :

Given that,

1/3x + 1/5y = 1/15      ---(i)

1/2x + 1/3y = 1/12      ---(ii)

Let us find that,

=> a = 1/x and b = 1/y

Therefore,

=> 1/3x + 1/5y = 1/15

=> 15 × ( a/3 + b/5 ) = 15 × 1/15

=> 5a +3b = 1           ---(iii)

Also,

=> a/2 + b/3 = 1/12

=> 12 × ( a/2 + b/3 ) = 12 × 1/12

=> 6a + 4b = 1          ---(iv)

Multiplying both sides of equation (iii) by 4, we get,

=> 20a + 12b = 4       ---(v)

Multiplying both sides of equation (iv) by -3, we get,

=> -18a + -12b = -3    ---(vi)

Adding equations (v) and (vi), we get,

=> 2a = 1

=> a = 1/2

Substituting the value of a in equation (iii), we get,

=> 5 × 1/2 + 3b = 1

=> 3b = 1 -5/2

=> 3b = 2-5/2

=> 3b = -3/2

=> b = -3/2 × 1/3

=> b = -1/2

∴ ( a,b ) = (1/2,-1/2)

Now,

=> a = 1/x

=> 1/2 = 1/x

=> x = 2

Again,

=> b = 1/y

=> -1/2 = 1/y

=> y = -2

∴(x, y) = (2, -2)