Question

1/5x + 1/6y = 12 1/3x - 3/7y = 8 solve

Answer 1

GIVEN :

The equations are $\frac{1}{5x}+\frac{1}{6y}=12$ and $\frac{1}{3x}-\frac{3}{7y}=8$

TO FIND :

The values of x and y in the given equation.

SOLUTION :

Given equations are $\frac{1}{5x}+\frac{1}{6y}=12\hfill (1)$ and $\frac{1}{3x}-\frac{3}{7y}=8\hfill (2)$

Equation (1) becomes ,

$\frac{1}{5x}+\frac{1}{6y}=12$

⇒ $\frac{6y+5x}{30xy}=12$

$5x+6y=360xy\hfill (3)$

and Equation (2) becomes ,

$\frac{1}{3x}-\frac{3}{7y}=8$

⇒ $\frac{7y-9x}{21xy}=18$

$-9x+7y=168xy\hfill (4)$

Solving equations (3) and (4) by Elimination Method.

Multiplying the equation (3) into 9 we get

$45x+54y=3240xy\hfill (5)$

Multiplying the equation (4) into 5 we get

$-45x+35y=840xy\hfill (6)$

Now adding the equations (5) and (6)

45x+54y=3240xy

-45x+35y=840xy

_______________

     89y=4080xy

4080xy=89y

$\frac{4080xy}{89y}=1$

$\frac{4080x}{89}=1$

$x=1\times \frac{89}{4080}$

∴ $x=\frac{89}{4080}$

Substituting the value of x in the equation (3) we get,

$5(\frac{89}{4080})+6y=360(\frac{89}{4080})y$

$\frac{89}{816}+6y=\frac{801}{102}y$

$6y-\frac{801}{102}y=\frac{-89}{816}$

$\frac{612y-801y}{102}=\frac{-89}{816}$

$\frac{-189y}{102}=\frac{-89}{816}$

$189y=\frac{89}{816}\times 102$

$189y=\frac{89}{8}$

$y=\frac{89}{8}\times \frac{1}{189}$

∴ $y=\frac{89}{1512}$

∴ the values are $x=\frac{89}{4080}$ and $y=\frac{89}{1512}$.

Answer 2

Answer:

GIVEN :

The equations are  and  

TO FIND :

The values of x and y in the given equation.

SOLUTION :

Given equations are  and  

Equation (1) becomes ,

⇒  

and Equation (2) becomes ,

⇒  

Solving equations (3) and (4) by Elimination Method.

Multiplying the equation (3) into 9 we get

Multiplying the equation (4) into 5 we get

Now adding the equations (5) and (6)

45x+54y=3240xy

-45x+35y=840xy

_______________

    89y=4080xy

4080xy=89y

∴  

Substituting the value of x in the equation (3) we get,

∴  

∴ the values are  and .

Step-by-step explanation: