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# Algebra

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Find all ordered pairs \$(x,y)\$ of real numbers such that \$x + y = 10\$ and \$x^2 + y^2 = 32 + 2xy\$.
For example, to enter the solutions \$(2,4)\$ and \$(-3,9)\$, you would enter "(2,4),(-3,9)" (without the quotation marks).

Jan 10, 2024

#1
+36919
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x+y = 10     =====>  y = 10 -x       sub this definition of y into the second equation

x^2 + (10-x)^2  - 2 x(10-x) -32 = 0      now solve for the value(s) of x

x^2 + x^2 - 20x + 100  - 20x + 2x^2 -32 = 0

4 x^2 -40x + 68 = 0            Quadratic Formula shows x =  7.828    and    2.172

using these values in the line equation shows            y =   2.172     and  7.828

Jan 11, 2024

#1
+36919
+1

x+y = 10     =====>  y = 10 -x       sub this definition of y into the second equation

x^2 + (10-x)^2  - 2 x(10-x) -32 = 0      now solve for the value(s) of x

x^2 + x^2 - 20x + 100  - 20x + 2x^2 -32 = 0

4 x^2 -40x + 68 = 0            Quadratic Formula shows x =  7.828    and    2.172

using these values in the line equation shows            y =   2.172     and  7.828

ElectricPavlov Jan 11, 2024
#2
+36919
+1

Here is perhaps a simler , exact - answer way:

x^2 + y^2 - 2xy = 32

(x-y)^2 = 32

x -y = +-  4 sqrt 2

y =   x - 4 sqrt 2    and     x +  4sqrt 2

then use the line quation:

x + y = 10

x + x-  4 sqrt 2 = 10

2x = 10 +4 sqrt 2

x = 5 +2 sqrt 2        and the other is x = 5 - 2 sqrt 2    Then the corresponding 'y' s are

y=  5 - 2 sqrt 2        and                          5 + 2 sqrt2