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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

 May 24, 2019

Best Answer 

 #1
avatar+195 
0

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.

 May 24, 2019
 #1
avatar+195 
0
Best Answer

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.

power27 May 24, 2019

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