Question

Solve the following pairs of equations by reducing them to a pair of linear equations by elimination method ​

Answer 1

$\sf 6x+3y=6xy ...(1)$

$\sf 2x+4y=5xy ...(2)$

Dividing equation 1 and 2 by xy we get,

$\sf \frac{6x}{xy}+\frac{3y}{6xy}=\frac{6xy}{xy}$

$\sf \frac{6}{y}+\frac{3}{x}=6 ...(3)$

$\sf \frac{2x}{xy}+\frac{4y}{6xy}=\frac{5xy}{xy}$

$\sf \frac{2}{y}+{4}{x}=5 ...(4)$

Let us assume $\frac{1}{x}=p \:\: and \:\: \frac{1}{y}=q$

$\sf 6q+3p=6 ...(5)$

$\sf 2q+4p=5 ...(6)$

Multiple equation 6 by 3 we get,

$\sf 3(2q+4p=5)=6q+12p=15 ...(7)$

Subtracting equation 5 from equation 7 we get,

6q+12p=15

- 6q+3p=6

(-) (-) (-)

___________

9p=9

___________

$\sf 9p=9$

$\sf\red{ p=1 }$

Substituting value of p in equation 5 we get,

$\sf 6q+3(1)=6$

$\sf 6q+3=6$

$\sf 6q=3$

$\sf q=\frac{3}{6}$

$\sf\red {q=\frac{1}{2} }$

$\sf p=\frac{1}{x}$

Value of p is 1 So,

$\sf 1=\frac{1}{x}$

$\sf\red{ x=1 }$

$\sf \frac{1}{y}=q$

Value of q is $\frac{1}{y}=2$

$\sf \frac{1}{y}=\frac{1}{2}$

$\sf\red{ y=2 }$