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# Find all pairs \$x+y=10\$ of real numbers such that \$x + y = 10\$ and \$x^2 + y^2 = 56\$. For example, to enter the solutions and ,&n

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Find all pairs \$x+y=10\$  of real numbers such that \$x + y = 10\$ and \$x^2 + y^2 = 56\$.
For example, to enter the solutions  and , you would enter "(2,4),(-3,9)" (without the quotation marks).

I reposted this since the last response was incorrect

May 15, 2021

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btw i accidentally left this part out its "to enter the solutions (2,4) and (-3,9), you-"

May 15, 2021
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x + y  =  10     --->    (x + y)2  =  (10)2     --->     x2 + 2xy + y2  =  100

Combining this with  x2 + y2  =  56     --->     x2 + 2xy + y2  =  100

x2           + y2  =    56

subtracting:                                                               2xy  =  44

xy  =  22

Since  x + y  =  10     --->     y  =  10 - x.

Combining these two equations:  xy  =  22     --->     x(10 - x)  =  22

10x - x2  =  22

x2 - 10x + 22  =  0

Using the quadratic equation:  x  =  5 + sqrt(3)     --->     y  =  10 - ( 5 + sqrt(3) )  =  5 - sqrt(3)

Also,  x  =  5 - sqrt(3)     and    y  = 5 + sqrt(3)

May 15, 2021