Answer:
[tex]\huge\boxed{x=\dfrac{4}{3};\ y=\dfrac{2}{3}\to\ \left(\dfrac{4}{3};\ \dfrac{2}{3}\right)}[/tex]
Step-by-step explanation:
[tex]\underline{+\left\{\begin{array}{ccc}\dfrac{1}{4}x+y=1\\\dfrac{3}{2}x-y=\dfrac{4}{3}\end{array}\right}\qquad|\text{add both sides of the equations}\\\dfrac{1}{4}x+\dfrac{3}{2}x=1+\dfrac{4}{3}\\\\\dfrac{1}{4}x+\dfrac{3\cdot2}{2\cdot2}x=\dfrac{3}{3}+\dfrac{4}{3}\\\\\dfrac{1}{4}x+\dfrac{6}{4}x=\dfrac{3+4}{3}\\\\\dfrac{1+6}{4}x=\dfrac{7}{3}\\\\\dfrac{7}{4}x=\dfrac{7}{3}\qquad|\text{multiply both sides by}\ \dfrac{4}{7}[/tex]
[tex]\dfrac{4\!\!\!\!\diagup}{7\!\!\!\!\diagup}\cdot\dfrac{7\!\!\!\!\diagup}{4\!\!\!\!\diagup}x=\dfrac{4}{7\!\!\!\!\diagup}\cdot\dfrac{7\!\!\!\!\diagup}{3}\\\\x=\dfrac{4}{3}[/tex]
Put it to the first equation
[tex]\dfrac{1}{4\!\!\!\!\diagup}\cdot\dfrac{4\!\!\!\!\diagup}{3}+y=1\\\\\dfrac{1}{3}+y=1\qquad|\text{subtract}\ \dfrac{1}{3}\ \text{from both sides}\\\\y=\dfrac{3}{3}-\dfrac{1}{3}\\\\y=\dfrac{3-1}{3}\\\\y=\dfrac{2}{3}[/tex]
