Question

Solve the following pair of simultaneous equations:

$\frac{x}{2}-\frac{y}{3}=2;\frac{x}{5}+\frac{y}{3}=15$

• A. $x=24\frac{2}{7},y=30\frac{3}{7}$
• B. $x=12\frac{1}{3},y=10\frac{2}{3}$
• C. $x=16\frac{7}{6},y=13\frac{4}{3}$
• D. $x=2\frac{22}{7},y=12\frac{13}{17}$

Answer 1

The correct option is A $x=24\frac{2}{7},y=30\frac{3}{7}$

The equation $\frac{x}{2}-\frac{y}{3}=2$ can be solved as:

$\frac{x}{2}-\frac{y}{3}=2\Rightarrow \frac{3x-2y}{6}=2\Rightarrow 3x-2y=12.........\left(1\right)$

The equation $\frac{x}{5}+\frac{y}{3}=15$ can be solved as:

$\frac{x}{5}+\frac{y}{3}=15\Rightarrow \frac{3x+5y}{15}=15\Rightarrow 3x+5y=225.........\left(2\right)$

Subtract Equation 2 from equation 1 to eliminate $x$, because the coefficients of $x$ are the same. So, we get

$(3x-3x)+(-2y-5y)=12-225$

i.e. $-7y=-213$

i.e. $y=\frac{213}{7}=30\frac{3}{7}$

Substituting this value of $y$ in the equation 1, we get

$3x-2\left(\frac{213}{7}\right)=12\Rightarrow 3x-\frac{426}{7}=12\Rightarrow 21x-426=84\Rightarrow 21x=510\Rightarrow x=\frac{510}{21}=\frac{170}{7}=24\frac{2}{7}$

Hence, the solution of the equations is $x=24\frac{2}{7},y=30\frac{3}{7}$.