Answer:
has only one solution: [tex]x=8, y=3[/tex]
Step-by-step explanation:
[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {\frac{x}{4} +\frac{5y}{3} =7(equation 2)}} \right\\[/tex]
substract [tex]\frac{1}{5}[/tex]×(equation 1) from equation2
[tex]\left \{ {{\frac{5x}{4} -\frac{2y}{3}=8(equation 1)} \atop {0x +\frac{9y}{5} =\frac{27}{5}(equation 2)}} \right\\[/tex]
from equation 2 we obtain
[tex]y=\frac{27}{9} =3[/tex]
and we replace 3 in equation 1
[tex]\frac{5x}{4} -\frac{2(3)}{3} =8\\\\\frac{5x}{4} =10\\\\x=\frac{40}{5} =8[/tex]
