Question

Which of the following substitution can convert given equations into linear equations with two variables?
$\frac{(x+y{)}^2}{x^{2}y+x{y}^2}=2;\frac{x^2-y^2}{x^{2}y+x{y}^2}=6$

• A. $\frac{1}{x}=u$
• B. $\frac{1}{y}=v$
• C. $\frac{1}{xy}=u$
• D. $\frac{x}{y}=u$

Answer 1

Answer: (A), (B)

$\frac{1}{y}=v$

Given equations are $\frac{(x+y{)}^2}{x^{2}y+x{y}^2}=2;\frac{x^2-y^2}{x^{2}y+x{y}^2}=6$

Equation 1:
$\frac{(x+y{)}^2}{x^{2}y+x{y}^2}=2\Rightarrow \frac{(x+y{)}^2}{xy(x+y)}=\frac{x+y}{xy}=\frac{1}{y}+\frac{1}{x}=2$

Substitute $\frac{1}{x}=u\&\frac{1}{y}=v$
We get, $\frac{1}{y}+\frac{1}{x}=\underset{–}{v+u=2}$

Equation 2:
$\frac{x^2-y^2}{x^{2}y+x{y}^2}=6\Rightarrow \frac{(x+y)(x-y)}{xy(x+y)}=\frac{x-y}{xy}=\frac{1}{y}-\frac{1}{x}=6$

Substitute $\frac{1}{x}=u\&\frac{1}{y}=v$
We get, $\frac{1}{y}-\frac{1}{x}=\underset{–}{v-u=6}$