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# help??

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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

#1
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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
+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.

power27 May 24, 2019