How to solve this system of equations

sqrt(7x+y)+sqrt(x+y)=6
sqrt(x+y)-y+x=2

√(7x+y)+√(x+y)=6
√(x+y)-y+x=2

√(7x+y)+√(x+y)=6
x+y≥0
7x+y≥0

√(x+y)-y+x=2
√(x+y)=2+y-x
2+y-x≥0

√(7x+y)+√(x+y)=6
√(7x+y)+(2+y-x)=6
√(7x+y)=4-y+x
4-y+x≥0

System
√(7x+y)=4-y+x
√(x+y)=2+y-x

|7x+y|=(4-y+x)²
|x+y|=(2+y-x)²

7x+y=(4-(y-x))²
x+y=(2+(y-x))²

7x+y=4²-2*4(y-x)+(y-x)²
x+y=2²+2*2(y-x)+(y-x)²

7x+y-(x+y)=4²-2*4(y-x)+(y-x)² -(2²+2*2(y-x)+(y-x)²)
6x=16-8(y-x)+(y-x)² -4 -4(y-x)-(y-x)²
6x=12 -12(y-x)
x=2 -2(y-x) => y=1+x/2

System
√(x+y)-y+x=2
y=1+x/2

√(x+(1+x/2))-(1+x/2)+x=2
√(1+3x/2)=3-x/2
3-x/2≥0 => 6≥x
1+3x/2=(3-x/2)²
1+3x/2=9-2*3x/2+(x/2)²
1+3x/2=9-2*3x/2+(x/2)²
x²/4 -9x/2 +8=0
x₁=2
x₂=16 >6

y=1+2/2=2

Answer (2,2)

√(7*2+2)+√(2+2)=4+2=6
√(2+2)-2+2=2
http://www.wolframalpha.com/input/?i=%E2…

Use the second equation to find sqrt(x+y):

sqrt(x+y) - y + x = 2
sqrt(x+y) = 2 + y - x

Then substitute into the first equation and solve for sqrt(7x+y):

sqrt(7x+y) + 2 + y - x = 6
sqrt(7x+y) = 4 - y + x

Then square both sides to remove the sqrt in both the first and second equations:

x + y = (2 + y - x)^2
x + y = 4 + 2y - 2x + 2y + y^2 - xy - 2x - xy + x^2
x + y = 4 + 4y - 4x - 2xy + y^2 + x^2
5x - 3y - 4 = -2xy + y^2 + x^2

7x + y = (4 - y + x)^2
7x + y = 16 - 4y + 4x - 4y + y^2 - xy + 4x - xy + x^2
7x + y = 16 - 8y + 8x - 2xy + y^2 + x^2
9y - x - 16 = -2xy + y^2 + x^2

Then equate the two equations:

5x - 3y - 4 = 9y - x - 16
6x + 12 = 12y
x + 2 = 2y
x = 2y - 2

Then substitute this equation for x back into the second equation and find y:

sqrt(2y-2+y) - y + 2y - 2 = 2
sqrt(3y-2) = 4 - y
3y - 2 = (4 - y)^2
3y - 2 = 16 - 8y + y^2
y^2 - 11y + 18 = 0
(y - 9)(y - 2) = 0
so y = 9 and y = 2 are solutions of the equation.

Then use the values of y to find x:

x = 2y - 2 = 18 - 2 = 16 for y = 9
x = 2y - 2 = 4 - 2 = 2 for y = 2

so the possible solutions of the equations are (x,y) = (2,2) and (x,y) = (16,9)

Check solutions with the first equation:
sqrt(7*2+2) + sqrt(2+2) = sqrt(16) + sqrt(4) = 4 + 2 = 6
sqrt(7*16+9) + sqrt(16+9) = sqrt(121) + sqrt(25) = 11 + 5 = 16 != 6

so the solution is (x,y) = (2,2)

i can't be bothered to actually do it but i think the best approach would be to square both equations twice (i.e. square once, then make the sqrt term the subject of the equation and square again) before equating them as usual...

hope this helps