Answer: (-1, 2) Step-by-step explanation: Let’s start by solving for x in [tex]-5x+2y=9\\-5x+2y+-2y=9+-2y\\-5x=9+-2y\\\frac{-5x}{-5} =\frac{9+-2y}{-5} \\x=\frac{2}{5} y+\frac{-9}{5}[/tex] We can now sub this value of x in 3x+5y=7 and solve [tex]3x+5y=7\\3(\frac{2}{5} y+\frac{-9}{5})+5y=7\\(3)(\frac{2}{5} y)+(3)(\frac{-9}{5})+5y=7\\\frac{6}{5} y+\frac{-27}{5}+5y=7\\\frac{6}{5} y+\frac{25}{5}y +\frac{-27}{5}=7\\\frac{31}{5} y+\frac{-27}{5}=7\\\frac{31}{5} y+\frac{-27}{5}+\frac{27}{5}=7+\frac{27}{5}\\\frac{31}{5} y=\frac{62}{5} \\\frac{\frac{31}{5} y}{\frac{31}{5}} =\frac{\frac{62}{5} }{\frac{31}{5}} \\y=2[/tex] Now we can sub in 2 for y in one of the equations [tex]3x+5(2)=7\\3x+10=7\\3x+10-10=7-10\\3x=-3\\\frac{3x}{3} =\frac{-3}{3} \\x=-1[/tex] x=-1 and y=2 giving us the point (-1, 2) Log in to Reply

Answer: (-1, 2)Step-by-step explanation:Let’s start by solving for x in

[tex]-5x+2y=9\\-5x+2y+-2y=9+-2y\\-5x=9+-2y\\\frac{-5x}{-5} =\frac{9+-2y}{-5} \\x=\frac{2}{5} y+\frac{-9}{5}[/tex]

We can now sub this value of x in 3x+5y=7 and solve

[tex]3x+5y=7\\3(\frac{2}{5} y+\frac{-9}{5})+5y=7\\(3)(\frac{2}{5} y)+(3)(\frac{-9}{5})+5y=7\\\frac{6}{5} y+\frac{-27}{5}+5y=7\\\frac{6}{5} y+\frac{25}{5}y +\frac{-27}{5}=7\\\frac{31}{5} y+\frac{-27}{5}=7\\\frac{31}{5} y+\frac{-27}{5}+\frac{27}{5}=7+\frac{27}{5}\\\frac{31}{5} y=\frac{62}{5} \\\frac{\frac{31}{5} y}{\frac{31}{5}} =\frac{\frac{62}{5} }{\frac{31}{5}} \\y=2[/tex]

Now we can sub in 2 for y in one of the equations

[tex]3x+5(2)=7\\3x+10=7\\3x+10-10=7-10\\3x=-3\\\frac{3x}{3} =\frac{-3}{3} \\x=-1[/tex]

x=-1 and y=2 giving us the point (-1, 2)