**Problem**

Solve for x and y from the given system of equations.

$x + 2y = 6$ ← Equation (1)

$\sqrt{x} + \sqrt{y} = 3$ ← Equation (2)

**Solution**

From Equation (1)

$x = 6 - 2y$ ← Equation (3)

$x = 6 - 2y$ ← Equation (3)

Substitute x = 6 - 2y to Equation (2)

$\sqrt{6 - 2y} + \sqrt{y} = 3$

$\sqrt{6 - 2y} = 3 - \sqrt{y}$

Square both sides

$6 - 2y = 9 - 6\sqrt{y} + y$

$6\sqrt{y} = 3 + 3y$

$2\sqrt{y} = 1 + y$

Square again both sides

$4y = 1 + 2y + y^2$

$y^2 - 2y + 1 = 0$

$(y - 1)^2 = 0$

$y = 1 ~ \text{ twice}$

From Equation (3)

$x = 6 - 2(1) = 4$

Hence, x = 4 and y = 1 *answer*