Question

Find the values of $x$ and $y$ respectively, for the following equations:
i) $\dfrac{1}{3}x + y = \dfrac{{10}}{3}$
ii) $2x + \dfrac{1}{4}y = \dfrac{{11}}{4}$

A) $3,1$
B) $3, - 1$
C) $1,3$
D) $ - 1, - 3$

Answer 1

Hint:We will first convert the given equations in linear form by multiplying both the equations by some constants. Now we will just multiply one of the equations and make the coefficients equal so that we can solve the equations to calculate the values of the variables.

Complete step-by-step answer:
The given equations are:
i) $\dfrac{1}{3}x + y = \dfrac{{10}}{3}$
ii) $2x + \dfrac{1}{4}y = \dfrac{{11}}{4}$
Note that both the equations are linear equations in two variables therefore, we can find the solutions for both the variables.
We just need to put them in a way that both the equations have some coefficient common so that we can eliminate one coefficient and find the other.
Then we will use the obtained coefficient to find the remaining coefficient.
We will multiply the first equation by $3$ we get,
$x + 3y = 10$ … (iii)
Similarly, we will multiply the second equation by $4$ we get,
$8x + y = 11$ … (iv)
Now we will multiply the equation (iv) by $3$ we get,
$24x + 3y = 33$ … (v)
Now, subtract equation (iii) from equation (v),
$23x = 23$
Therefore, $x = 1$ .
Substituting in equation (iii) we get,
$1 + 3y = 10$
Solving for $y$ we get,
$y = 3$

So, the correct answer is “Option C”.

Note:In the present problem, both the equations are linear and in two different variables, therefore the equations are solvable. We can also use the method where we express one variable in another and then substitute it to solve the problem. But this method is more efficient.