SOLUTION: If x+y=1 and {{{x^3+y^3=19}}}, find the value of {{{x^2+y^2}}}

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Question 1111874: If x+y=1 and , find the value of
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


Solve the first equation for y and substitute in the second equation:


The x^3 terms cancel, leaving you with a quadratic equation you can solve easily, leading easily to the answer to the question.

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.

            Let me show you the  RIGHT WAY  solving this problem.

We are given

x + y = 1.


Cube both sides. You will get

1 =  =  + .


Now replace  by 19 ( ! given !)  and replace (x+y) by 1 in the right side.  You will get

1 = 19 + 3xy,    which implies  3xy = 1 - 19 = -18.


Hence, xy =  = -6.


Now the last step is 

 =  = 1 - 2xy = 1 - 2*(-6) = 1 + 12 = 13.

Answer.   x^2 + y^2 = 13.

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            It is how this problem MUST be solved,  and it is how  IT  IS  EXPECTED  you will solve it   (if you know the way !)

            Any other way is the way to  N O W H E R E.

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If you want to see other similar solved problems, look into the lesson

    - HOW TO evaluate expressions involving  ,  ,    and  
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Evaluation, substitution".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

==============

Again:   This problem is to check your competency:  whether you know the way . . .

             It is very good for an interview . . . (on a mathematical position . . . )


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