Solution
- The solution steps to solve the system of equations by elimination is given below:
[tex]\begin{gathered} \frac{x}{3}+\frac{y}{2}=\frac{1}{2}\text{ \lparen Equation 1\rparen} \\ \\ \frac{x}{6}-\frac{y}{3}=\frac{5}{6}\text{ \lparen Equation 2\rparen} \\ \\ \text{ Multiply Equation 2 by 2} \\ 2\times(\frac{x}{6}-\frac{y}{3})=\frac{5}{6}\times2 \\ \\ \frac{x}{3}-\frac{2y}{3}=\frac{5}{3}\text{ \lparen Equation 3\rparen} \\ \\ \\ \text{ Now, }\frac{x}{3}\text{ is common to both Equations 1 and 3.} \\ \\ \text{ We can therefore subtract both equations to eliminate }x. \\ \text{ We have:} \\ \text{ Equation 1 }-\text{ Equation 3} \\ \\ \frac{x}{3}+\frac{y}{2}-(\frac{x}{3}-\frac{2y}{3})=\frac{1}{2}-\frac{5}{3} \\ \\ \frac{x}{3}-\frac{x}{3}+\frac{y}{2}+\frac{2y}{3}=\frac{1}{2}-\frac{5}{3}=\frac{3}{6}-\frac{10}{6} \\ \\ \frac{y}{2}+\frac{2y}{3}=-\frac{7}{6} \\ \\ \frac{3y}{6}+\frac{4y}{6}=-\frac{7}{6} \\ \\ \frac{7y}{6}=-\frac{7}{6} \\ \\ \therefore y=-1 \\ \\ \text{ Substitute the value of }y\text{ into any of the equations, we have:} \\ \frac{1}{3}x+\frac{1}{2}y=\frac{1}{2} \\ \frac{1}{3}x+\frac{1}{2}(-1)=\frac{1}{2} \\ \\ \frac{1}{3}x=\frac{1}{2}+\frac{1}{2} \\ \\ \frac{1}{3}x=1 \\ \\ \therefore x=3 \end{gathered}[/tex]Final Answer
The answer is:
[tex]\begin{gathered} x=3,y=-1 \\ \\ \therefore(x,y)=(3,-1) \end{gathered}[/tex]