Answer:
[tex]\large\boxed{\left(7,\ \dfrac{13}{3}\right)}[/tex]
Step-by-step explanation:
You must use the algebraic method: substitution or elimination.
I will solve a given system of equations the substitution method.
[tex]\left\{\begin{array}{ccc}y=\dfrac{1}{3}x+2&(1)\\\\y=\dfrac{4}{3}x-5&(2)\end{array}\right\qquad\text{substitute (2) to (1)}\\\\\dfrac{4}{3}x-5=\dfrac{1}{3}x+2\qquad\text{multiply both sides by 3}\\\\3\!\!\!\!\diagup^1\cdot\dfrac{4}{3\!\!\!\!\diagup_1}x-(3)(5)=3\!\!\!\!\diagup^1\cdot\dfrac{1}{3\!\!\!\!\diagup_1}x+(3)(2)\\\\4x-15=x+6\qquad\text{add 15 to both sides}\\\\4x=x+21\qquad\text{subtract x from both sides}\\\\3x=21\qquad\text{divide both sides by 3}\\\\x=7\\\\\text{put the value of x to the first equation}\\\\y=\dfrac{1}{3}\cdot7+2\\\\y=\dfrac{7}{3}+\dfrac{6}{3}\\\\y=\dfrac{13}{3}[/tex]
