Answer:
[tex]\left(\dfrac{2}{3},\dfrac{1}{5}\right)[/tex]
Step-by-step explanation:
Given the system of two equations:
[tex]\left\{\begin{array}{l}\dfrac{1}{2}x-\dfrac{3}{4}y=\dfrac{11}{60}\\ \\\dfrac{2}{5}x+\dfrac{1}{6}y=\dfrac{3}{10}\end{array}\right.[/tex]
Multiply the first equation and the second equation by 60 to get rid of fractions:
[tex]\left\{\begin{array}{l}30x-45y=11\\ \\24x+10y=18\end{array}\right.[/tex]
Now multiply the first equation by 4 and the second equation by 5:
[tex]\left\{\begin{array}{l}120x-180y=44\\ \\120x+50y=90\end{array}\right.[/tex]
Subtract them:
[tex](120x-180y)-(120x+50y)=44-90\\ \\120x-180y-120x-50y=-46\\ \\-230y=-46\\ \\y=\dfrac{46}{230}=\dfrac{1}{5}[/tex]
Substitute it into the first equation:
[tex]30x-45\cdot \dfrac{1}{5}=11\\ \\30x-9=11\\ \\30x=11+9\\ \\30x=20\\ \\x=\dfrac{2}{3}[/tex]
The solution is [tex]\left(\dfrac{2}{3},\dfrac{1}{5}\right)[/tex]
