Question

Solve the following pair of simultaneous equations:

$3x+\frac{1}{y}=13;\frac{2}{y}-x=5$

• A. $\frac{2}{5}$ and $4$
• B. $3$ and $\frac{1}{4}$
• C. $7$ and $\frac{1}{7}$
• D. $3$ and $\frac{3}{4}$

Answer 1

Answer: (B)

$3$ and $\frac{1}{4}$

The equation $3x+\frac{1}{y}=13$ can be solved as:

$3x+\frac{1}{y}=13\Rightarrow \frac{3xy+1}{y}=13\Rightarrow 3xy+1=13y.........\left(1\right)$

The equation $\frac{2}{y}-x=5$ can be solved as:

$\frac{2}{y}-x=5\Rightarrow \frac{2-xy}{y}=5\Rightarrow 2-xy=5y.........\left(2\right)$

Multiply the equation 2 by $3$. Then we get:

$-3xy+6=15y.........\left(3\right)$

Add equations 1 and 3 to eliminate $xy$, because the coefficients of $xy$ are opposite. So, we get

$(3xy-3xy)+(1+6)=13y+15y$

i.e. $28y=7$

i.e. $y=\frac{1}{4}$

Substituting this value of $y$ in the equation 1, we get

$3x\left(\frac{1}{4}\right)+1=13\left(\frac{1}{4}\right)\Rightarrow \frac{3}{4}x+1=\frac{13}{4}\Rightarrow \frac{3}{4}x=\frac{13}{4}-1\Rightarrow \frac{3}{4}x=\frac{9}{4}\Rightarrow 3x=9\Rightarrow x=3$

Hence, the solution of the equations is $x=3,y=\frac{1}{4}$.