Find all pairs of real numbers (x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. If you find more than one pair, then list your pairs in order by increasing x$ value, separated by commas. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(-3,9),(2,4)" (without the quotation marks
+247 x+y=6
y=6-x
Now (6-x) can be substituted for y
\(x^2+(-x+6)^2=28\)
Expanding this gets:
\(x^2+x^2-12x+36=28\)
\(2x^2-12x+8=0\) Combine like terms and set equation to 0
\(x^2-6x+4=0\) Divide by 2
This can't be factored so the quadratic equation needs to be used
\(x=\frac{6 \pm \sqrt{36-4(1)(4)}}{2(1)}\)
\(x=\frac{6 \pm \sqrt{20}}{2}\)
\(x=3+2\sqrt{5}\)
\(x=3-2\sqrt{5}\)
Plug these values into x+y=6 to get two points as the final answer.
+247 x+y=6
y=6-x
Now (6-x) can be substituted for y
\(x^2+(-x+6)^2=28\)
Expanding this gets:
\(x^2+x^2-12x+36=28\)
\(2x^2-12x+8=0\) Combine like terms and set equation to 0
\(x^2-6x+4=0\) Divide by 2
This can't be factored so the quadratic equation needs to be used
\(x=\frac{6 \pm \sqrt{36-4(1)(4)}}{2(1)}\)
\(x=\frac{6 \pm \sqrt{20}}{2}\)
\(x=3+2\sqrt{5}\)
\(x=3-2\sqrt{5}\)
Plug these values into x+y=6 to get two points as the final answer.
