Answer:The two solutions in exact form are:[tex](\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})[/tex]and[tex](\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})[/tex].If you prefer to look at approximations just put into your calculator:[tex](3.3303,-4.9909)[/tex]and[tex](-0.3303,5.9909)[/tex].Step-by-step explanation:I guess you are asked to find the solution the given system.I'm going to use substitution.This means I'm going to plug the second equation into the first giving me:[tex](-3x+5)^2+x^2=36[/tex] I replaced the 1st y with what the 2nd y equaled.Before we continue solving this I'm going to expand the [tex](-3x+5)^2[/tex] using the following:[tex](a+b)^2=a^2+2ab+b^2[/tex].[tex](-3x+5)^2=(-3x)^2+2(-3x)(5)+(5)^2[/tex][tex](-3x+5)^2=9x^2-30x+25[/tex]Let's go back to the equation we had:[tex](-3x+5)^2+x^2=36[/tex] After expansion of the squared binomial we have:[tex]9x^2-30x+25+x^2=36[/tex]Combine like terms (doing the [tex]9x^2+x^2[/tex] part:[tex]10x^2-30x+25=36[/tex]Subtract 36 on both sides:[tex]10x^2-30x+25-36=0[/tex]Simplify the 25-36 part:[tex]10x^2-30x-11=0[/tex]Compare this to [tex]ax^2+bx+c=0[/tex] which is standard form for a quadratic.We should see the following:[tex]a=10[/tex][tex]b=-30[/tex][tex]c=-11[/tex]The formula that solves this equation for the variable [tex]x[/tex] is:[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]Plugging in our values for [tex]a,b, \text{ and } c[/tex] give us:[tex]x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{2(10)}[/tex]Simplify the bottom; that is 2(10)=20:[tex]x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{20}[/tex]Put the inside of square root into the calculator; that is put [tex](-30)^2-4(10)(-11)[/tex] in the calculator:[tex]x=\frac{30 \pm \sqrt{1340}}{20}[/tex]Side notes before continuation:Let's see if 1340 has a perfect square. I know 1340 is divisible by 10 because it ends in 0.1340=10(134)134 is even so it is divisible by 2:1340=10(2)(67)1340=2(2)(5)(67)1340=4(5)(67)1340=4(335)4 is a perfect square so we can simplify the square root part further:[tex]\sqrt{1340}=\sqrt{4}\sqrt{335}=2\sqrt{335}[/tex].Let's go back to the solution:[tex]x=\frac{30 \pm \sqrt{1340}}{20}[/tex][tex]x=\frac{30 \pm 2 \sqrt{335}}{20}[/tex]Now I see all three terms contain a common factor of 2 so I'm going to divide top and bottom by 2:[tex]x=\frac{\frac{30}{2} \pm \frac{2 \sqrt{335}}{2}}{\frac{20}{2}}[/tex][tex]x=\frac{15 \pm \sqrt{335}}{10}[/tex]So we have these two x values:[tex]x=\frac{15+\sqrt{335}}{10} \text{ or } \frac{15-\sqrt{335}}{10}[/tex]Now we just need to find the corresponding y-coordinate for each pair of points.I'm going to use the easier equation [tex]y=-3x+5[/tex].Let's do it for the first x I mentioned:If [tex]x=\frac{15+\sqrt{335}}{10}[/tex] then[tex]y=-3(\frac{15+\sqrt{335}}{10})+5[/tex].Let's simplify:Distribute the -3 to the terms on top:[tex]y=\frac{-45-3\sqrt{335}}{10}+5[/tex]Combine the two terms; I'm going to do this by writing 5 as 50/10:[tex]y=\frac{-45-3\sqrt{335}+50}{10}[/tex]Combine like terms on top; the -45+50 part:[tex]y=\frac{5-3\sqrt{335}}{10}[/tex].So one solution point is:[tex](\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})[/tex].Let's find the other one for the other x that we got.If [tex]x=\frac{15-\sqrt{335}}{10}[/tex] then[tex]y=-3(\frac{15-\sqrt{335}}{10})+5[/tex].Let's simplify.Distribute the -3 on top:[tex]y=\frac{-45+3\sqrt{335}}{10}+5[/tex]I'm going to write 5 as 50/10 so I can combine the terms as one fraction:[tex]y=\frac{-45+3\sqrt{335}+50}{5}[/tex]Simplify the -45+50 part:[tex]y=\frac{5+3\sqrt{335}}{10}[/tex].So the other point of intersection is:[tex](\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})[/tex].The two solutions in exact form are:[tex](\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})[/tex]and[tex](\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})[/tex].If you prefer to look at approximations just put into your calculator:[tex](3.3303,-4.9909)[/tex]and[tex](-0.3303,5.9909)[/tex].