Question

$$\begin{gathered} \frac{x}{6}+\frac{y}{15}=4, \\ \frac{x}{3}-\frac{y}{12}=\frac{19}{4}. \end{gathered}$$

Answer 1

The given equations may be written as:
$\frac{x}{6}+\frac{y}{15}-4=0$ ...(i)
$\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0$ ...(ii)
Here, $a_1=\frac{1}{6},b_1=\frac{1}{15},c_1=-4,a_2=\frac{1}{3},b_2=-\frac{1}{12}\mathrm{and}{c}_2=-\frac{19}{4}$
By cross multiplication, we have:

∴ $\frac{x}{\left[{\displaystyle \frac{1}{15}}\times \left(-{\displaystyle \frac{19}{4}}\right)-\left(-{\displaystyle \frac{1}{12}}\right)\times \left(-4\right)\right]}=\frac{y}{\left[\left(-4\right)\times {\displaystyle \frac{1}{3}}-\left({\displaystyle \frac{1}{6}}\right)\times \left(-{\displaystyle \frac{19}{4}}\right)\right]}=\frac{1}{\left[{\displaystyle \frac{1}{6}}\times \left(-{\displaystyle \frac{1}{12}}\right)-{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{1}{15}}\right]}$
⇒ $\frac{x}{\left(-{\displaystyle \frac{19}{60}}-{\displaystyle \frac{1}{3}}\right)}=\frac{y}{\left(-{\displaystyle \frac{4}{3}}+{\displaystyle \frac{19}{24}}\right)}=\frac{1}{\left(-{\displaystyle \frac{1}{72}}-{\displaystyle \frac{1}{45}}\right)}$
⇒ $\frac{x}{\left(-{\displaystyle \frac{39}{60}}\right)}=\frac{y}{\left(-{\displaystyle \frac{13}{24}}\right)}=\frac{1}{\left(-{\displaystyle \frac{13}{360}}\right)}$
⇒ $x=\left[\left(-\frac{39}{60}\right)\times \left(-\frac{360}{13}\right)\right]=18,y=\left[\left(-\frac{13}{24}\right)\times \left(-\frac{360}{13}\right)\right]=15$
Hence, x = 18 and y = 15 is the required solution.