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# Help with algebra

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Find all pairs x+y=10 of real numbers such that x + y = 10 and x^2 + y^2 = 52.
For example, to enter the solutions and (2,4) and (-3,9), you would enter "(2,4),(-3,9)" (without the quotation marks).

Jun 13, 2021

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$$x+y=10$$

$$x^2+y^2+2xy=100$$

$$2xy=100-52$$

$$2xy = 48$$

$$xy=24$$

$$x+{24\over x}=10$$

$$x^2-10x+24=0$$

$$x=6,4$$

1st Case : when x = 6 ⇒y = 4

2nd Case : when x = 4 ⇒y = 6

∴ All pairs of solutions are (6,4) and (4,6)

Jun 14, 2021