Solve the following pair of equations.x4+2y3=7; x6+3y5=11

Given equations are

\frac{x}{4} + \frac{2 y}{3} = 7; \frac{x}{6} + \frac{3 y}{5} = 11

.
Considering

\frac{x}{4} + \frac{2 y}{3} = 7

,
Taking L.C.M of

4

and

3

i.e.,

12

\Rightarrow \frac{3 x + 8 y}{12} = 7

\Rightarrow 3 x + 8 y = 84

\Rightarrow 3 x = 84 - 8 y

\Rightarrow x = \frac{84 - 8 y}{3}

Substitute the value of

x

\frac{x}{6} + \frac{3 y}{5} = 11

\Rightarrow \frac{\frac{84 - 8 y}{3}}{6} + \frac{3 y}{5} = 11

\Rightarrow \frac{84 - 8 y}{18} + \frac{3 y}{5} = 11

\Rightarrow \frac{42 - 4 y}{9} + \frac{3 y}{5} = 11

Take L.C.M of

9

and

5

i.e.,

45

\Rightarrow \frac{5 (42 - 4 y) + 9 (3 y)}{45} = 11

Simplifying further,

\Rightarrow 210 - 20 y + 27 y = 45 \times 11

\Rightarrow 210 - 20 y + 27 y = 495

\Rightarrow 7 y = 495 - 210

\Rightarrow y = \frac{285}{7}

\Rightarrow y = 40.71

Substitute

y = 40.71

x = \frac{84 - 8 y}{3}

\Rightarrow x = \frac{84 - 8 (40.71)}{3}

\Rightarrow x = \frac{84 - 325.68}{3}

\Rightarrow x = \frac{- 241.68}{3}

\Rightarrow x = - 80.56

There the value of

x = - 80.56

and

y = 40.71

.
Hence the correct option is 3.

Solve the following pair of equations. $\frac{x}{4} + \frac{2 y}{3} = 7; \frac{x}{6} + \frac{3 y}{5} = 11$