x(3x + 4) = 4

Expand the bracket.

3x x x + 4 x x = 4

3x² + 4x = 4

3x² + 4x - 4 = 0

Notes:

1. a x c = 3 x -4 = -12

2. Split b (4) such that two numbers when you multiply you will get a x c (-12) and when you add you will get b(4)

3. The numbers are 6 and - 2

3x² + 6x - 2x - 4 = 0

3x(x + 2) - 2(x + 2) = 0

(x + 2)(3x - 2) = 0

When x + 2 = 0, x = -2

When 3x - 2 = 0, x = $\frac{2}{3}$

$\frac{\mathrm{x\; -\; 2}}{4}$ - $\frac{\mathrm{2x\; -\; 4}}{3}$ = $\frac{5}{6}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (4, 3 and 6). The L.C.M is 12.

12 x $\frac{\mathrm{x\; -\; 2}}{4}$ - 12 x $\frac{\mathrm{2x\; -\; 4}}{3}$ = 12 x $\frac{5}{6}$

3 x (x - 2) - 4(2x - 4) = 2 x 5

3(x - 2) - 4(2x - 4) = 10

3x - 6 - 8x + 16 = 10

Note:
1. - x - = +
2. -4 x -4 = + 16

3x - 8x - 6 + 16 = 10

-5x + 10 = 10

-5x = 10 - 10

-5x = 0

Divide both sides by -5

x = $\frac{0}{-5}$ = 0

$\frac{2}{3}$(x + 5) = 1 - $\frac{\mathrm{x\; -\; 7}}{2}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3 and 2. The L.C.M is 6.

6 x $\frac{2}{3}$(x + 5) = 1 x 6 - $\frac{\mathrm{x\; -\; 7}}{2}$ x 6

2 x 2(x + 5) = 6 - 3(x - 7)

4(x + 5) = 6 - 3x + 21

Notes:
1. - x - = +
2. -3 x -7 = + 21

4x + 20 = 27 - 3x

4x + 3x = 27 - 20

7x = 7

Divide both sides by 7

x = $\frac{7}{7}$ = 1

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$

When m = 5, n = 3 and r = 2

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{(5^2-3^2)+2^2}{5^2+(3^2-2^2)}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{\mathrm{25\; -\; 9\; +\; 4}}{\mathrm{25\; +\; 9\; -\; 4}}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{20}{30}$

$\frac{({\mathrm{m}}^2-{\mathrm{n}}^2)+{\mathrm{r}}^2}{{\mathrm{m}}^2+({\mathrm{n}}^2-{\mathrm{r}}^2)}$ = $\frac{2}{3}$