(a)

(b)

m = $\frac{\mathrm{2x}}{\mathrm{1\; -}{\mathrm{x}}^2}$ and n = $\frac{\mathrm{2x}}{\mathrm{1\; +\; x}}$

2m - n = 2 x $\frac{\mathrm{2x}}{\mathrm{1\; -}{\mathrm{x}}^2}$ - $\frac{\mathrm{2x}}{\mathrm{1\; +\; x}}$

1 - x² is a difference of two squares.

1 - x² = (1 + x)(1 - x)

2m - n = $\frac{\mathrm{4x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$ - $\frac{\mathrm{2x}}{\mathrm{1\; +\; x}}$

2m - n = $\frac{\mathrm{1\; x\; 4x\; -\; (1\; -\; x)\; x\; 2x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

2m - n = $\frac{\mathrm{4x\; -}\left(\mathrm{2x\; -\; 2}{\mathrm{x}}^2\right)}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

2m - n = $\frac{\mathrm{4x\; -\; 2x\; +\; 2}{\mathrm{x}}^2}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

2m - n = $\frac{\mathrm{2x\; +\; 2}{\mathrm{x}}^2}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

Factorize 2x out.

2m - n = $\frac{\mathrm{2x(1\; +\; x)}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

(1 + x) cancel each other.

2m - n = $\frac{\mathrm{2x}}{\mathrm{(1\; -\; x)}}$

(a)

p = 2x$\sqrt{\frac{\mathrm{q}\left(\mathrm{1\; +}\frac{{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}}$

1 + $\frac{{\mathrm{r}}^2}{{\mathrm{y}}^2}$ = $\frac{\mathrm{1\; x}{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}$

1 + $\frac{{\mathrm{r}}^2}{{\mathrm{y}}^2}$ = $\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}$

p = 2x$\sqrt{\frac{\mathrm{q}\left(\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}}$

Get rid of the square root by squaring both sides of the equation.

p² = (2x)²$\frac{\mathrm{q}\left(\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}$

p² = (2² x x²)$\frac{\mathrm{q}\left(\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}$

p² = 4x²$\frac{\mathrm{q}\left(\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}$

p² = 4x²q$\frac{\left(\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}\right)}{\mathrm{s}}$

Get rid of the fraction by multiplying both sides by the denominator s.

p² x s = 4x²q x $\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}$

Divide both sides by 4x²q

$\frac{{\mathrm{p}}^2\mathrm{s}}{4{\mathrm{x}}^2\mathrm{q}}$ = $\frac{{\mathrm{y}}^2+{\mathrm{r}}^2}{{\mathrm{y}}^2}$

Cross multiply.

p²s x y² = (y² + r²)(4x²q)

p²y²s = 4x²y²q + 4x²r²q

Group the y².

p²y²s - 4x²y²q = 4x²r²q

Factorize y² out.

y²(p²s - 4x²q) = 4x²r²q

Divide both sides by (p²s - 4x²q)

y² = $\frac{4{\mathrm{x}}^2{\mathrm{r}}^2\mathrm{q}}{{\mathrm{p}}^2\mathrm{s}-4{\mathrm{x}}^2\mathrm{q}}$

Take square root of both sides to get rid of the square.

y = $\sqrt{\frac{4{\mathrm{x}}^2{\mathrm{r}}^2\mathrm{q}}{{\mathrm{p}}^2\mathrm{s}-4{\mathrm{x}}^2\mathrm{q}}}$

(b)

$\frac{\mathrm{2m}{\mathrm{n}}^2\mathrm{x}}{\mathrm{3m\; -\; n}}$

When m = 3, n = -2 and x = -1

$\frac{\mathrm{2m}{\mathrm{n}}^2\mathrm{x}}{\mathrm{3m\; -\; n}}$ = $\frac{\mathrm{2\; x\; (3)\; x}{\mathrm{(-2)}}^2\mathrm{x\; (-1)}}{\mathrm{3\; x\; 3\; -\; (-2)}}$

$\frac{\mathrm{2m}{\mathrm{n}}^2\mathrm{x}}{\mathrm{3m\; -\; n}}$ = $\frac{\mathrm{6\; x}4\mathrm{x\; (-1)}}{\mathrm{9\; +\; 2}}$

Note:
1. -- = + or -(-) = +
2. --2 = + 2 or -(-2) = + 2

$\frac{\mathrm{2m}{\mathrm{n}}^2\mathrm{x}}{\mathrm{3m\; -\; n}}$ = $\frac{-24}{11}$

$\frac{\mathrm{2m}{\mathrm{n}}^2\mathrm{x}}{\mathrm{3m\; -\; n}}$ = -2$\frac{2}{11}$

(i)

t = m$\sqrt{{\mathrm{n}}^2+\; 4r}$

Get rid of the square root by squaring both sides of the equation.

t² = m² x (n² + 4r)

Divide both sides by m²

$\frac{{\mathrm{t}}^2}{{\mathrm{m}}^2}$ = n² + 4r

$\frac{{\mathrm{t}}^2}{{\mathrm{m}}^2}$ - 4r = n²

Every number is being divided by 1

4r = $\frac{\mathrm{4r}}{1}$

$\frac{{\mathrm{t}}^2}{{\mathrm{m}}^2}$ - $\frac{\mathrm{4r}}{1}$ = n²

$\frac{\mathrm{1\; x\; }{\mathrm{t}}^2-{\mathrm{m}}^{2}x\mathrm{4r}}{{\mathrm{m}}^2}$ = n²

$\frac{{\mathrm{t}}^2-{\mathrm{m}}^2\mathrm{4r}}{{\mathrm{m}}^2}$ = n²

Get rid of the square by taking square root of both sides.

n = $\sqrt{\frac{{\mathrm{t}}^2-{\mathrm{m}}^2\mathrm{4r}}{{\mathrm{m}}^2}}$

(ii)

n = $\sqrt{\frac{{\mathrm{t}}^2-{\mathrm{m}}^2\mathrm{4r}}{{\mathrm{m}}^2}}$

When t = 25, m = 5 and r = 4

n = $\sqrt{\frac{{25}^2-5^2\mathrm{ \; x\; 4\; x \; 4}}{5^2}}$

n = $\sqrt{\frac{625-400}{25}}$

n = $\sqrt{\frac{225}{25}}$ = $\sqrt{9}$ = 3