Question

Solve the following pair of equations by reducing them to a pair of linear equations:$\frac{4}{x}+3y=14,\frac{3}{x}-4y=23$

• A. $x=7,y=1$
• B. $x=-3,y=-1$
• C. $x=\frac{1}{5},y=-2$
• D. $x=\frac{7}{2},y=7$

Answer 1

The correct option is C $x=\frac{1}{5},y=-2$

Given

$\frac{4}{x}+3y=14$

$\frac{3}{x}-4y=23$

Let$\frac{1}{x}=a$

$4a+3y=\mathrm{14.......}eq1$
$3a-4y=\mathrm{23........}eq2$

multiply eq1 by 4 and eq2 by 3 we get
$16a+12y=\mathrm{56......}eq3$
$9a-12y=\mathrm{69.........}eq4$

Adding eq3 and eq4 we get
$25a=125$
$a=5$

Substituting $a=5$ in eq1 we get
$4\times 5+3y=14$
$3y=14-20$
$3y=-6$
$y=-2$

Hence $\frac{1}{x}=5$ then $x=\frac{1}{5},y=-2$