MathematicsWhat values of x and y satisfies the equations x6+y15=4 and x3-y12=434 ?

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

Given,

\frac{x}{6} + \frac{y}{15} = 4

and

\frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4}

.
Consider the equation

\frac{x}{3} - \frac{y}{12} = 4 \frac{3}{4}

.
⇒

4 x - y = 57

⇒

y = 4 x - 57 . . . . (1)

Substitute the value of

y

in equation

\frac{x}{6} + \frac{y}{15} = 4

.
⇒

15 x + 6 y = 360

.
⇒

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

⇒

15 x + 24 x - 342 = 360

⇒

39 x = 702

⇒

x = 18

Substituting this value of

x

(1), we get,
⇒

y = 72 - 57

⇒

y = 15

Hence, the solution is

x = 18, y = 15

.
Hence, the correct option is 2.

JEE