Question 973174
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x^1/2 + y = 7
x + y^1/2 = 11
Find the value of x and y
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{{{matrix(2,1, " ", x^(1/2) + y = 7)}}}____{{{sqrt(x) + y = 7}}} 
                                   {{{sqrt(x) = 7  -  y}}}
                              {{{(sqrt(x))^2 = (7  -  y)^2}}}
                                     {{{x = 49  -  14y + y^2}}} ----- eq (i)

                              {{{x + sqrt(y) = 11}}}
                                     {{{x = 11 - sqrt(y)}}} ----- eq (ii)

We then get: {{{49  -  14y + y^2 = 11 - sqrt(y)}}}
         {{{y^2 + sqrt(y) - 14y + 49 - 11 = 0}}}
                {{{y^2 + sqrt(y) - 14y + 38 = 0}}}

                                 Let {{{sqrt(y) = t}}}
              Then: {{{system((sqrt(y))^2 = t^2, y = t^2, y^2 = (t^2)^2 = t^4)}}}
                        {{{y^2 + sqrt(y) - 14y + 38 = 0}}} then becomes: 
                        {{{t^4 + t - 14t^2 + 38 = 0}}}
                        {{{t^4 - 14t^2 + t + 38 = 0}}}
Using the RATIONAL ROOT THEOREM, we find that a root of the above equation is: t = 2, which makes its
FACTOR, t - 2. When divided by t - 2, using LONG DIVISION of POLYNOMIALS, or using SYNTHETIC DIVISION,
the other factor of {{{t^4  -  14t^2 + t + 38}}}, besides t - 2, is: {{{t^3 + 2t^2 - 10t - 19}}}.
From this, we find another REAL solution being approximately 3.13131. The other 2 are negative (< 0) and
so, MUST be REJECTED/IGNORED, since {{{sqrt(y) = t}}} CANNOT have a negative (< 0) value for t. 

I will continue with the REAL INTEGER value, 2.
     {{{sqrt(y) = t = 2}}} ---- Back-substituting t = 2 for {{{sqrt(y)}}}
{{{(sqrt(y))^2 = y = t^2 = 2^2 = 4}}}

       {{{x = 11 - sqrt(y)}}} ----- eq (ii)
        x = 11 - 2 ----- Substituting 2 for {{{sqrt(y)}}} in eq (ii)
        x = 9

So, the ONLY INTEGER-solution set is: (x, y) = (9, 4). I'll let you substitute the other REAL VALUE, 3.13131
for t, to determine the other SOLUTION-SET.</font></font></font></b></pre>