Thanks for clarifying the problem.
There is a quick way to solve this and a slow, methodical way. I'll do both because the quick way cannot always be used.
The quick way is based on logic and an understanding of how square roots and irrational numbers work:- "x" is the radicand of a square root in the first equation. So "x" must be non-negative. "y" is the radicad of a square root in the second equation. So "y" must be non-negative, too. So we know that both "x" and "y" re non-negative.
- Square roots are non-negative.
- For two non-negatives to add up to 11, both numbers must be between 0 and 11.
- For two non-negatives to add up to 7, both numbers must be between 0 and 7.
- 11 and 7 are rational numbers.
- Both equations, then, say that the sum of two non-negatives is a rational number.
- Square roots are irrational unless the radicand is a perfect square.
- A sum involving an irrational number cannot add up to a rational number (except adding an irrational number and its opposite/negative results in zero).
- "x" and "y" must be perfect squares (or else their square roots would be irrational).
Putting this all together we know that "x" and "y" are both perfect squares less than or equal to 11. So the only possibilities are 0, 1, 4 and 9. It should not take long to figure out which numbers work.
The slow, methodical way is to use Algebra, Since this is a system of two equations and two variables, we can use the Substitution Method. This method starts with solving one equation for one variable. If we subtract the square root from each side of the first equation we get:
Now we can substitute for y in the second equation:
Next we solve this equation. The following procedure can be used for solving equations of one variable where the variable is in a square root:- Isolate a square root that has a variable in its radicand.
- Square both sides of the equation.
- If there is still a square root with a variable in its radicand, then repeat steps 1 and 2.
- There should not longer be any square roots with a variable in its radicand. Use appropriate techniques to solve this equation.
- Check your solution(s). This is not optional. Whenever you square both sides of an equation (which has been done at least once at step 2 so far), extraneious solutions may be introduced. Extraneous solutions are solutions which fit the squared equation but do not fit the original equation. Extraneous solutions can occur even if no mistakes have been made! So solutions must be checked and extraneous solutions, if any, must be rejected.
Let's see this in action...
1) Isolate a square root...
Subtracting x from each side we get:
2) Square both sides
The left side is easy to square. The right side requires using FOIL or the pattern. I prefer using the pattern:
which simplifies to:
3) There is still a square root with the variable in its radicand so we repeat steps 1 and 2.
1) Isolate a square root
Subtracting 11 from each side we get:
(The minus in front of the square root is not in the way. It will go away when we square both sides next. But if it bothers you, multiply both sides of the equation by -1.)
2) Square both sides
Again the left side is easy to square. The right side is more difficult because we can't use FOIL or that pattern. We just multiply each term of by each term of and then add like terms, if any:
which simplifies as follows:
There are no more square roots so we can go on to step 4.
4) Use appropriate techniques to solve the equation.
This is a 4th degree equation. So we will make one side zero and then try to factor the other side. Subtracting x from each side we get:
To factor this, the only method I see is to try possible rational roots. The possible rational roots of this polynomial are all the fractions, positive and negative, that can be formed using a factor of the constant term (at the end) over a factor of the leading coefficient (in front). For this polynomial the possible rational roots would be all the fractions, positive and negative, that can be formed using a factor 1444 ove the factors of 1. There are many factors of 1444 so ther ea re many possible rational roots and it could take a long time to find the actual roots. But I've already used the short way so I know which rational root works.
To test a rational root, it is easiest to use synthetic division:
4 | 1 -28 272 -1065 1444
---- 4 -96 704 -1444
-----------------------------
1 -24 176 -381 0
The zero in the lower right tells us that x-4 is a factor of and if x-4 is a factor then 4 is a root. So x = 4 is a solution to the equation.
You can try other possible rational roots (but you won't find any).
We can finally so step 5, check
Checking x = 4"
which simplifies as follows:
3 = 3 Check!
Now we can find y. Using one of the original equations:
Substituting 4 for x:
So the solution to your system is (4, 9).