Question

Based on equations reducible to linear equations
Solve for x and y: $9+25xy=53x$ and $27-4xy=x$

• A. $x=1,y=4$
• B. $x=2,y=7$
• C. $x=6,y=2$
• D. $x=3,y=2$

Answer 1

The correct option is D $x=3,y=2$

Given systems of equations are:
$9+25xy=53x$
$27-4xy=x$
Now using reducible method will get
Let convert this equation in $\frac{1}{x}and\frac{1}{y}$ form
Will take $xy$ as $L.C.M$ as $xy$ is common
From eq will get
$\Rightarrow \frac{9}{xy}+25=\frac{53}{y}$ ...3
$\frac{27}{xy}-4=\frac{1}{y}$ ...4
Now let $\frac{1}{xy}=vand\frac{1}{y}=u$
Now will get
$9v+25=53u$ and $27v-4=u$
Now Rearranging both the equation
$53u-9v=25...\left(5\right)$
$u-27v=-4....\left(6\right)$
Now multiplying eq5 by 3
$\Rightarrow 159u-27v=75...\left(7\right)$
Now subtracting eq 7 and eq6
$\Rightarrow 159u-27v=75$
$\Rightarrow -u+27v=4$
$\Rightarrow 158u=79$
$\Rightarrow u=\frac{1}{2}$
Now putting $u=\frac{1}{2}$ in eq 6
$\Rightarrow \frac{1}{2}-27\left(2\right)v=-4\left(2\right)$
$\Rightarrow 1-54v=-8$
$\Rightarrow 54v=9$
$\Rightarrow v=\frac{9}{54}$
$\Rightarrow v=\frac{1}{6}$
Now putting u and v values
$\frac{1}{y}=u$
$\Rightarrow \frac{1}{y}=\frac{1}{2}$
$\Rightarrow y=2$
$\frac{1}{xy}=v$
$\Rightarrow \frac{1}{xy}=\frac{1}{6}$
$\Rightarrow \frac{1}{2x}=\frac{1}{6}$
$\Rightarrow x=\frac{6}{2}$
$\Rightarrow x=3$
$\Rightarrow x=3,y=2$