Question

Solve the following pair of equations:

$3x-y=23$, $\frac{x}{3}+\frac{y}{4}=4$

• A. $x=2;y=5$
• B. $x=7;y=9$
• C. $x=0;y=5$
• D. $x=9;y=4$

Answer 1

The correct option is D $x=9;y=4$

The given equation is:

$3x-y=23.........\left(1\right)$

The equation $\frac{x}{3}+\frac{y}{4}=4$ can be solved as:

$\frac{x}{3}+\frac{y}{4}=4\Rightarrow \frac{4x+3y}{12}=4\Rightarrow 4x+3y=48.........\left(2\right)$

Multiply the equation 1 by $4$ and equation 2 by $3$ to make the coefficients of $x$ equal. Then we get the equations:

$12x-4y=\mathrm{92.........}\left(3\right)$

$12x+9y=\mathrm{144.........}\left(4\right)$

Subtract Equation 4 from equation 3 to eliminate $x$, because the coefficients of $x$ are the same. So, we get

$(12x-12x)+(-4y-9y)=92-144$

i.e. $-13y=-52$

i.e. $y=4$

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

$3x-4=23$

i.e. $3x=27$

i.e. $x=9$

Hence, the solution of the equations is $x=9,y=4$.