Question

The solution of the equations $\frac{3x-y+1}{3}=\frac{2x+y+2}{5}=\frac{3x+2y+1}{6}$ is given by:

• A. $x=2,y=1$
• B. $x=1,y=1$
• C. $x=-1,y=-1$
• D. $x=1,y=2$

Answer 1

The correct option is B

$x=1,y=1$

Multiply both sides of the equation by a common denominator:

Consider the given system of the equations as:

$\frac{3x-y+1}{3}=\frac{2x+y+2}{5}=\frac{3x+2y+1}{6}$

The above equation can be written as:

$\begin{array}{rcl}\frac{3x-y+1}{3}& =& \frac{2x+y+2}{5}......\left(1\right)\\ \frac{2x+y+2}{5}& =& \frac{3x+2y+1}{6}.....\left(2\right)\end{array}$

Now, equation (1) can be solved as:

$\begin{array}{rcl}\frac{3x-y+1}{3}& =& \frac{2x+y+2}{5}\\ 15x-5y+5& =& 6x+3y+6\\ 15x-6x-5y-3y& =& 6-5\\ 9x-8y& =& 1.......\left(3\right)\end{array}$

Similarly, solving equation (2) as

$\begin{array}{rcl}\frac{2x+y+2}{5}& =& \frac{3x+2y+1}{6}\\ 12x+6y+12& =& 15x+10y+5\\ 3x+4y& =& 7......\left(4\right)\end{array}$

Step 2: Solve obtained system of equations.

Now, multiply equation (4) by 2 to get:

$6x+8y=14...\left(5\right)$

Now, adding equations (3) and (5), we get:

$\begin{array}{rcl}9x-8y+6x+8y& =& 1+14\\ 15x& =& 15\\ x& =& 1\end{array}$

Now, substitute $x=1$into equation (3) as:

$\begin{array}{rcl}9x-8y& =& 1\\ 9-8y& =& 1\\ -8y& =& -8\\ y& =& 1\end{array}$

Hence, the required answer is $x=1,y=1$. Thus, option (b) is correct.