1 5x 1 6y 12 1 3x 3 7y 8

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

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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}.


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