√x+y=7,x+√y=11.find the value of x and y.

Question

Very hard ,very hard ,very hard

Answer 1

√x+y=7

x+√y=11

√x=7-y

√y=11-x

the square root of a number is always the positive square root

therefore y can't be greater than 7 and x can't be greater than 11

look at the perfect squares 1, 4, 9, 16

assuming x and y are perfect squares and a little trial and error...

x=9  and y=4

 

Answer 2

 

(1)   (√x) + y = 7

(2)   x + √y  = 11

 

I did this: 

Let u = √x    and  v = √y

 

So (1) and (2) can be re-written as:

 

(1')     u + v² = 7

(2')     u² + v = 11

 

(1') —>   u = 7-v²

Substituting into (2') for u:

              (7-v²)² + v = 11

              (49 - 14v² + v⁴) + v = 11

              v⁴ - 14v² + v + 38 = 0

        (v-2)(v³+2v²-10v-19) = 0

        (v-2)(v-3.13131)(v+1.84813)(v+3.28319) = 0   (<<< from www.wolframalpha.com)

 

So possible answers are:

             v=2                        —> y=4                  —(from (1))—>  √x=3   —> x=9

             v=3.13131              —> y=9.80510         ——>    √x=-2.8051   —> x=7.86859

             v=-1.84813             —> y=3.41558        ——>    √x=3.58442    —> x=12.84807

             v=-3.28319             —> y=10.77934       ——>   √x=-3.77934    —> x=14.28341

 

Check:

       x=9, y=4:  

                (1) 3+4=7  (ok)

                (2) 9+2=11 (ok)

 

       x=7.86859, y=9.80510

                (1) -2.8051+9.80510=7 (ok)     (only if we take the negative root of x)

                (2) 7.86859+3.13131=11 (ok)

                 

       x=12.84807, y=3.41558 

                (1) 3.58442+3.41558=7 (ok)

                (2) 12.84807-1.84813=11 (ok)   (only if we take the negative root of y)

 

        x=14.28341, y=10.77934

                (1)  -3.77934+10.77934=7 (ok)   (only if we take the negative root of x)

                (2)  14.28341-3.28319=11  (ok)   (only if we take the negative root of y)

 

 

Given no other information, √a  is defined to be the positive square root of a only —>  so only x=9, y=4  is a proper answer.