Question

What values of x and y satisfies the equations $\frac{x}{6}+\frac{y}{15}=4$ and $\frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$

• A. $(6,15)$
• B. $(18,15)$
• C. $(15,18)$
• D. $(12,30)$

Answer 1

The correct option is B $(18,15)$

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation $\frac{x}{3}-\frac{y}{12}=4\frac{3}{4}$ or $4x-y=57$ and write it as

$y=4x-\mathrm{57...........}\left(1\right)$

Substitute the value of $y$ in Equation $\frac{x}{6}+\frac{y}{15}=4$ or $15x+6y=360$. We get

$15x+6(4x-57)=360$

i.e. $15x+24x-342=360$

i.e. $39x=702$

i.e. $x=18$

Therefore, $x=18$

Substituting this value of $x$ in equation 1, we get

$y=72-57$

i.e. $y=15$

Hence, the solution is $x=18,y=15$.