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Find all pairs (x,y) of real numbers such that x + y = 10 and x^2 - y^2 = 56.

 Jan 19, 2021

Best Answer 

 #1
avatar+9182 
+1

x + y  =  10     Subtract  y  from both sides of this equation to get     x  =  10 - y

 

Now since   x = 10 - y   we can substitute  10 - y  in for  x  in the second equation.

 

x2  -  y2   =   56

 

(10 - y)2  -  y2   =   56

 

(10 - y)(10 - y)  -  y2   =   56

 

100 - 20y + y2  -  y2   =   56

 

100 - 20y   =   56

 

-20y   =   -44

 

y   =   2.2

 

Using this value of  y  and the fact that   x  =  10 - y   we can find the value of  x

 

x   =   10 - y   =   10 - 2.2   =   7.8

 

So there is only one ordered pair of real numbers that satisfies the system of equations, and it is  (7.8, 2.2)

 

Check:  https://www.desmos.com/calculator/abco8fdboi

 Jan 19, 2021
 #1
avatar+9182 
+1
Best Answer

x + y  =  10     Subtract  y  from both sides of this equation to get     x  =  10 - y

 

Now since   x = 10 - y   we can substitute  10 - y  in for  x  in the second equation.

 

x2  -  y2   =   56

 

(10 - y)2  -  y2   =   56

 

(10 - y)(10 - y)  -  y2   =   56

 

100 - 20y + y2  -  y2   =   56

 

100 - 20y   =   56

 

-20y   =   -44

 

y   =   2.2

 

Using this value of  y  and the fact that   x  =  10 - y   we can find the value of  x

 

x   =   10 - y   =   10 - 2.2   =   7.8

 

So there is only one ordered pair of real numbers that satisfies the system of equations, and it is  (7.8, 2.2)

 

Check:  https://www.desmos.com/calculator/abco8fdboi

hectictar Jan 19, 2021

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