Question

Find the values of x and y in the given pair of equations.

$\frac{2}{x}$ + $\frac{3}{y}$ = 13

$\frac{5}{x}$ - $\frac{4}{y}$ = -2

• A. $\frac{1}{2}$, $\frac{1}{3}$
• B. $\frac{1}{3}$, $\frac{1}{2}$
• C. 2, 3
• D. 3, 2

Answer 1

The correct option is A

$\frac{1}{2}$, $\frac{1}{3}$

In the above given pair of equations we see that the variables are in form of $\frac{1}{x}$ and $\frac{1}{y}$. If we start solving the equation as it is, we will find that after simplifying we end up in two equations that are not linear.

See what we will get

$2y+3x=13xy$

$5y-4x=-2xy$

The above two equations are no longer linear equations.

So, to get the above equations in linear form we can substitute another variables instead of $\frac{1}{x}$ and $\frac{1}{y}$ .

So we will substitute $\frac{1}{x}$ as ‘a’ and $\frac{1}{y}$ as ‘b’

We will get our equations as

$2a+3b=13$

$5a-4b=-2$

Now we can solve this pair of equation by using any of the three methods (elimination, substitution and cross multiplication) to solve the pair of linear equation.

If we use method of elimination

Multiply the first equation by 4 and second by 3. Equations can be written as

$8a+12b=52$

$15a-12b=-6$

Now adding the two equations

$23a=46$

$\Rightarrow a=2$

Now substitute the value of a in any of the two equations and we will find out that

$b=3$

Earlier we have assumed that $a=\frac{1}{x}$,

therefore $x=\frac{1}{2}$

Similarly, $y=\frac{1}{3}$