Question

Solve the following pair of equations:

$2x-3y-3=0$
$\frac{2x}{3}+4y+\frac{1}{2}=0$

• A. $x=\frac{1}{2};y=-\frac{7}{6}$
• B. $x=\frac{21}{20};y=-\frac{3}{10}$
• C. $x=\frac{14}{5};y=-\frac{6}{11}$
• D. $x=\frac{18}{19};y=-\frac{16}{10}$

Answer 1

The correct option is B $x=\frac{21}{20};y=-\frac{3}{10}$

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

Let us consider the equation $2x-3y-3=0$ and write it as

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

Substitute the value of $y$ in the equation $\frac{2x}{3}+4y+\frac{1}{2}=0$. We get

$\frac{2x}{3}+4y+\frac{1}{2}=0\Rightarrow \frac{2x}{3}+4\left(\frac{2x-3}{3}\right)+\frac{1}{2}=0\Rightarrow \frac{2x}{3}+\frac{8x-12}{3}+\frac{1}{2}=0\Rightarrow \frac{2x+8x-12}{3}+\frac{1}{2}=0\Rightarrow \frac{10x-12}{3}+\frac{1}{2}=0\Rightarrow \frac{2(10x-12)+3}{6}=0\Rightarrow \frac{20x-24+3}{6}=0\Rightarrow 20x=21\Rightarrow x=\frac{21}{20}$

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

$y=\frac{2\left(\frac{21}{20}\right)-3}{3}=\frac{42-60}{60}=-\frac{18}{60}=-\frac{3}{10}$

Hence, the solution is $x=\frac{21}{20},y=-\frac{3}{10}$.