48 = 3 x 16
6 = 3 x 2

$\sqrt{\mathrm{ab}}$ = $\sqrt{a}$ x $\sqrt{b}$

$\sqrt{\mathrm{48}}$ = $\sqrt{\mathrm{16\; x\; 3}}$ = $\sqrt{\mathrm{16}}$ x $\sqrt{3}$

$\sqrt{\mathrm{16}}$ = 4

$\sqrt{6}$ = $\sqrt{\mathrm{3\; x\; 2}}$ = $\sqrt{3}$ x $\sqrt{2}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{16}}\mathrm{x}\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{\sqrt{3}\mathrm{+\; 4}\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{5\sqrt{3}}{\sqrt{3}\mathrm{x}\sqrt{2}}$

$\sqrt{3}$ cancels each other.

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = $\frac{5}{\sqrt{2}}$

$\sqrt{2}$ = 2^½

$\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$ = a^-b

$\frac{\sqrt{3}\mathrm{+}\sqrt{\mathrm{48}}}{\sqrt{6}}$ = 5 x 2^-½

Law of multiplication of indices

a^m x aⁿ = a^{m + m}

5 x 2^-½ = 5 x 2^-1 x 2^½

5 x 2^-½ = $\frac{5}{2}$ x 2^½

2^½ = $\sqrt{2}$

5 x 2^-½ = $\frac{5}{2}$$\sqrt{2}$

$\sqrt{a}$ x $\sqrt{a}$ = a

Get rid of the denominator surds by multiplying the numerator and the denominator by the surd.

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ x $\frac{\sqrt{7}}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ x $\frac{\sqrt{\mathrm{21}}}{\sqrt{\mathrm{21}}}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = 2$\sqrt{7}$ - $\frac{14\sqrt{7}}{7}$ + $\frac{7\sqrt{\mathrm{21}}}{21}$

Every number is being divided by 1.

2$\sqrt{7}$ = $\frac{2\sqrt{7}}{1}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{2\sqrt{7}}{1}$ - $\frac{14\sqrt{7}}{7}$ + $\frac{7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{\mathrm{21\; x\; 2}\sqrt{7}-3x14\sqrt{7}+7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{42\sqrt{7}-42\sqrt{7}+7\sqrt{\mathrm{21}}}{21}$

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{7\sqrt{\mathrm{21}}}{21}$

Note:

1. $42\sqrt{7}$ - $42\sqrt{7}$ = 0

2. 7 dividies itself 1 time and 21, 3 times

2$\sqrt{7}$ - $\frac{14}{\sqrt{7}}$ + $\frac{7}{\sqrt{\mathrm{21}}}$ = $\frac{\sqrt{\mathrm{21}}}{3}$

Simplify the left hand side and compare with the right hand side to determine the values of x and y.

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$

Divide each of the expression at the numerator by the denominator $\sqrt{5}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}}{\sqrt{5}}$ + $\frac{\sqrt{5}}{\sqrt{5}}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}}{\sqrt{5}}$ + 1

Get rid of the denominator surd in $\frac{\sqrt{3}}{\sqrt{5}}$ by multiplying both the numerator and the denominator by $\sqrt{5}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{3}x\sqrt{5}}{\sqrt{5}x\sqrt{5}}$ + 1

Notes:

1. $\sqrt{a}$ x $\sqrt{a}$ = a

$\sqrt{5}$ x $\sqrt{5}$ = 5

2. $\sqrt{a}$ x $\sqrt{b}$ = $\sqrt{\mathrm{a\; x\; b}}$

$\sqrt{3}$ x $\sqrt{5}$ = $\sqrt{\mathrm{3\; x\; 5}}$ = $\sqrt{\mathrm{15}}$

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{\mathrm{3\; x\; 5}}}{5}$ + 1

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{\sqrt{15}}{5}$ + 1

$\frac{\sqrt{3}+\sqrt{5}}{\sqrt{5}}$ = $\frac{1}{5}$ x $\sqrt{\mathrm{15}}$ + 1

Comparing with x + y$\sqrt{\mathrm{15}}$,

x = 1 and y = $\frac{1}{5}$

x + y = 1 + $\frac{1}{5}$

x + y = 1$\frac{1}{5}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$

Express each surds to be in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{28}}$ = $\sqrt{\mathrm{4\; x\; 7}}$

$\sqrt{\mathrm{50}}$ = $\sqrt{\mathrm{25\; x\; 2}}$

$\sqrt{\mathrm{72}}$ = $\sqrt{\mathrm{36\; x\; 2}}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2$\sqrt{\mathrm{4\; x\; 7}}$ - 3$\sqrt{\mathrm{25\; x\; 2}}$ + $\sqrt{\mathrm{36\; x\; 2}}$

From $\sqrt{\mathrm{a\; x\; b}}$ = $\sqrt{a}$ x $\sqrt{b}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x $\sqrt{4}$ x $\sqrt{7}$ - 3 x $\sqrt{\mathrm{25}}$ x $\sqrt{2}$ + $\sqrt{\mathrm{36}}$ x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 2 x 2 x $\sqrt{7}$ - 3 x 5 x $\sqrt{2}$ + 6 x $\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 15$\sqrt{2}$ + 6$\sqrt{2}$

2$\sqrt{\mathrm{28}}$ - 3$\sqrt{\mathrm{50}}$ + $\sqrt{\mathrm{72}}$ = 4$\sqrt{7}$ - 9$\sqrt{2}$

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4-\sqrt{2}}{\sqrt{2}}$

Get rid of the denominator surd $\sqrt{2}$ by multiplying both the numerator and denominator by the denominator surd.

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{(4-\sqrt{2})x\sqrt{2}}{\sqrt{2}x\sqrt{2}}$

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4x\sqrt{2}-\sqrt{2}x\sqrt{2}}{\sqrt{2}x\sqrt{2}}$

Notes:

1. $\sqrt{a}$ x $\sqrt{a}$ = a

2. $\sqrt{2}$ x $\sqrt{2}$ = 2

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4\sqrt{2}-2}{2}$

The denominator affects each of the terms in the numerator.

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = $\frac{4\sqrt{2}}{2}$ - $\frac{2}{2}$

$\frac{4-\sqrt{2}}{\sqrt{2}}$ = 2$\sqrt{2}$ - 1

Express $\sqrt{\mathrm{72}}$ in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{72}}$ = $\sqrt{\mathrm{36\; x\; 2}}$ = $\sqrt{\mathrm{36}}$ x $\sqrt{2}$ = 6$\sqrt{2}$

Note:

$\sqrt{\mathrm{36}}$ = 6

Express $\sqrt{\mathrm{48}}$ in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{48}}$ = $\sqrt{\mathrm{16\; x\; 3}}$ = $\sqrt{\mathrm{16}}$ x $\sqrt{3}$ = 4$\sqrt{3}$

Note:

$\sqrt{\mathrm{16}}$ = 4

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{6\sqrt{2}}{4\sqrt{3}-\sqrt{3}}$

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{6\sqrt{2}}{3\sqrt{3}}$

Note: 3 divides itself 1 time and 6, 2 times.

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{2}}{\sqrt{3}}$

Multiply both the numerator and the denominator by $\sqrt{3}$ to get rid of the denominator root

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{2}x\sqrt{3}}{\sqrt{3}x\sqrt{3}}$

2$\sqrt{2}$ x $\sqrt{3}$ = 2$\sqrt{\mathrm{2\; x\; 3}}$ = 2$\sqrt{6}$

$\sqrt{3}$ x $\sqrt{3}$ = 3

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{6}}{3}$