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

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

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

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

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

(3 + x)(1 - x) > 9 - x²

Expand the bracket

3(1 - x) + x(1 - x) > 9 - x²

3 - 3x + x - x² > 9 - x²

3 - 2x - x² > 9 - x²

- x² cancels each other at both sides of the equation.

3 - 2x > 9
- 2x > 9 - 3
- 2x > 6

Divide both sides by - 2. When dividing by a negative, the sign changes direction.

x < $\frac{6}{-2}$

x < -3

$\frac{1}{2}$(x-1)-$\frac{1}{3}$($\frac{1}{2}$x-1)

Expand the brackets

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

Note: - x - = +

-$\frac{1}{3}$ x - 1 = $\frac{1}{3}$

The Least Common Multiples (L.C.M) of the denominators (2,3 and 6) is 6

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

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

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

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

$\frac{1}{2}$(x-1)-$\frac{1}{3}$($\frac{1}{2}$x-1) = $\frac{1}{6}$(2x - 1)

Method I

4x + 6y = 5 --- eqn (1)
2x + 4y = 3 --- eqn (2)

Multiply equation 2 by 2 to eliminate x

2x x 2 + 4y x 2 = 3 x 2
4x + 8y = 6 --- eqn (3)

Equation 3 - Equation 1

2y = 1

Divide both sides by 2

y = $\frac{1}{2}$

Subtitute y = $\frac{1}{2}$ in equation 1

4x + 6 x $\frac{1}{2}$ = 5

4x + 3 = 5
4x = 5 - 3
4x = 2

Divide both sides by 4

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

x = $\frac{1}{2}$

x + 2y = $\frac{1}{2}$ + 2 x $\frac{1}{2}$

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

x + 2y = $\frac{3}{2}$

x + 2y = 1$\frac{1}{2}$

Method II

Express 2x + 4y = 3 to be in the form (x + 2y) and make (x + 2y) the subject

2x + 4y = 3

Factorize 2 out.

2(x + 2y) = 3

Divide both sides by 2 to make (x + 2y) the subject.

(x + 2y) = $\frac{3}{2}$

Change the improper fraction to mixed fraction.

(x + 2y) = 1$\frac{1}{2}$

$\frac{\mathrm{9\; -}{\mathrm{x}}^2}{\mathrm{3x\; -}{\mathrm{x}}^2}$

9 - x² is a difference of two square.

a² - b² = (a+b)(a-b)

9 = 3 x 3 = 3²

9 - x² = 3² - x²

3² - x² = (3+x)(3-x)

$\frac{\mathrm{9\; -}{\mathrm{x}}^2}{\mathrm{3x\; -}{\mathrm{x}}^2}$ = $\frac{\mathrm{(3+x)(3-x)}}{\mathrm{3x\; -}{\mathrm{x}}^2}$

Factorize x out in the denominator.

$\frac{\mathrm{9\; -}{\mathrm{x}}^2}{\mathrm{3x\; -}{\mathrm{x}}^2}$ = $\frac{\mathrm{(3+x)(3-x)}}{\mathrm{x(3-x)}}$

(3-x) cancels each other

$\frac{\mathrm{9\; -}{\mathrm{x}}^2}{\mathrm{3x\; -}{\mathrm{x}}^2}$ = $\frac{\mathrm{3+x}}{\mathrm{x}}$

$\frac{\mathrm{x\; +\; y}}{5}$ = $\frac{\mathrm{x\; -\; y}}{3}$

Cross multiply. The numerator at the left side multiplies the denominator at the right then equal to then the numerator at the right also multiplies the denominator at the left.

3 x (x + y) = 5 x (x - y)
3x + 3y = 5x - 5y
3y + 5y = 5x - 3x
8y = 2x

Divide both sides by y

8 = $\frac{\mathrm{2x}}{\mathrm{y}}$

Divide both sides by 2

$\frac{8}{2}$ = $\frac{\mathrm{x}}{\mathrm{y}}$

Division is :

8:2 = x:y

2 can divide itself once and 8, 4 times.

4:1 = x:y

p = x - $\frac{1}{\mathrm{x}}$

Every number is divisible by 1

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

p = $\frac{\mathrm{x}}{1}$ - $\frac{1}{\mathrm{x}}$

p = $\frac{\mathrm{x\; x\; x\; -\; 1}}{\mathrm{x}}$

p = $\frac{{\mathrm{x}}^2\mathrm{-\; 1}}{\mathrm{x}}$

Square both sides of the equation.

p² = ($\frac{{\mathrm{x}}^2\mathrm{-\; 1}}{\mathrm{x}}$)²

(x² - 1)² = (x² - 1) x (x² - 1)
(x² - 1)² = x²(x² - 1) - 1(x² - 1)
(x² - 1)² = x² x x² - 1 x x² - 1 xx² - 1 x -1
(x² - 1)² = x⁴ - x² - x² + 1
(x² - 1)² = x⁴ - 2x² + 1

p² = $\frac{{\mathrm{x}}^4\mathrm{-\; 2}{\mathrm{x}}^2\mathrm{+\; 1}}{{\mathrm{x}}^2}$

Divide each of the expressions at the right by the denominator (x²).

p² = $\frac{{\mathrm{x}}^4}{{\mathrm{x}}^2}$ - $\frac{2{\mathrm{x}}^2}{{\mathrm{x}}^2}$ + $\frac{1}{{\mathrm{x}}^2}$

p² = x² - 2 + $\frac{1}{{\mathrm{x}}^2}$

p² = x² + $\frac{1}{{\mathrm{x}}^2}$ - 2

But q = x² + $\frac{1}{{\mathrm{x}}^2}$

p² = q - 2

p² + 2 = q

4x + 2y = 16 ----- eqn 1

6x -2y = 4 ----- eqn 2

Get rid of y. eqn 1 + eqn 2

10x + 0 = 20

10x = 20

Divide both sides by 10

x = $\frac{20}{10}$ = 2

Substitute x = 2 into ----- eqn 1

4 x 2 + 2y = 16

8 + 2y = 16

2y = 16 - 8

2y = 8

Divide both sides by 2.

y = $\frac{8}{2}$ = 4

y - x = 4 - 2 = 2

3x - 2y = 10 ----------- eqn 1

x + 3y = 7 ------------ eqn 2

In order to get rid of x, multiply ----- eqn 2 by 3

3x + 9y = 21 ----------- eqn 3

Subtract ---- eqn 1 from ---- eqn 3

9y - - 2y = 11

Note: - - = + or -(-) = +

9y + 2y = 11

11y = 11

Divide both sides by 11

y = $\frac{11}{11}$ = 1

Substitute y = 1 into ------- eqn 1.

3x - 2(1) = 10

3x - 2 = 10

3x = 10 + 2

3x = 12

Divide both sides by 3

x = $\frac{12}{3}$ = 4

$\frac{1}{3}$(5 - 3x) < $\frac{2}{5}$(3 - 7x)

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (3 and 5).

The L.C.M of 3 and 5 is 15

15 x $\frac{1}{3}$(5 - 3x) < 15 x $\frac{2}{5}$(3 - 7x)

5(5 - 3x) < 3 x 2(3 - 7x)

25 - 15x < 6(3 - 7x)

25 - 15x < 18 - 42x

- 15x + 42x < 18 - 25

27x < -7

Divide both sides by 27

x < $\frac{-7}{27}$

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{\mathrm{a\; x\; a}-\mathrm{b\; x\; b}-\mathrm{a\; x\; c}}{\mathrm{ab}}$

$\frac{\mathrm{a}}{\mathrm{b}}$ - $\frac{\mathrm{b}}{\mathrm{a}}$ - $\frac{\mathrm{c}}{\mathrm{b}}$ = $\frac{{\mathrm{a}}^2-{\mathrm{b}}^2-\mathrm{ac}}{\mathrm{ab}}$

$\frac{\mathrm{y\; +\; 1}}{2}$ - $\frac{\mathrm{2y\; -\; 1}}{3}$ = 4

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

6 x $\frac{\mathrm{y\; +\; 1}}{2}$ - $\frac{\mathrm{2y\; -\; 1}}{3}$ x 6 = 4 x 6

3(y + 1) - 2(2y - 1) = 24

3y + 3 - 4y + 2 = 24

3y - 4y + 5 = 24

-y = 24 - 5

-y = 19

Divide both sids by -1

y = $\frac{19}{-1}$ = -19

3p = 4q ----- eqn 1

9p = 8q - 12 ---- eqn 2

From ---- eqn 1

Make p the subject by dividing both sides by 3.

p = $\frac{\mathrm{4q}}{3}$ ------- eqn 3

Substitute p into ---- eqn 2

9 x $\frac{\mathrm{4q}}{3}$ = 8q - 12

3 x 4q = 8q - 12

12q = 8q - 12

12q - 8q = - 12

4q = -12

Divide both sides by 4

q = $\frac{-12}{4}$ = -3

Substitute q = -3 into ---- eqn 3.

p = $\frac{\mathrm{4\; x\; -3}}{3}$

p = -4

pq = p x q

pq = -4 x -3

pq = 12

t - $\frac{9}{5}$ = -1$\frac{1}{15}$

Change the mixed fraction to improper fraction.

1$\frac{1}{15}$ = $\frac{\mathrm{1\; x\; 15\; +\; 1}}{15}$

1$\frac{1}{15}$ = $\frac{\mathrm{15\; +\; 1}}{15}$

1$\frac{1}{15}$ = $\frac{16}{15}$

t - $\frac{9}{5}$ = -$\frac{16}{15}$

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 5 and 15. The L.C.M is 15

15 x t - 15 x $\frac{9}{5}$ = -$\frac{16}{15}$ x 15

15t - 3 x 9 = -16

15t - 27 = -16

15t = 27 - 16

15t = 11

Divide both sides by 15

t = $\frac{11}{15}$

Concept

(a + b)² = a² + 2 x a x b + b² = a² + 2ab + b²
(a - b)² = a²-2 x a x b + b² = a² -2ab + b²

Alternatively
(a + b)² = (a + b)x(a + b) [Expand and simplify]
(a -b)² = (a - b)x(a - b) [Expand and simplify]

(x - 2)² = x² - 2 x x x 2 + 2² = x² -4x + 4
(x + 1)² = x² + 2 x x x 1 + 1² = x² + 2x + 1

(x - 2)² + 3 = (x + 1)² - 6
x² - 4x + 4 + 3 = x² + 2x + 1 - 6
x² - 4x + 7 = x² + 2x - 5
x² - x² - 4x - 2x = -5 - 7
-6x = -12

Divide both sides by -6

x = $\frac{-12}{-6}$ = 2

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z(z\; -\; 3)\; -\; (z\; -\; 2)(z\; +\; 2)}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{z\; x\; z\; -\; 3\; x\; z\; -\; [z(z\; +\; 2)\; -\; 2(z\; +\; 2)]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [z\; x\; z\; +\; z\; x\; 2\; -\; 2\; x\; z\; -2\; x\; 2]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [}{\mathrm{z}}^2\mathrm{+\; 2z\; -\; 2z\; -\; 4]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -\; [}{\mathrm{z}}^2\mathrm{-\; 4]}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

Notes:
1. -[+] = - x + = -
2. -[z²] = - z²
3. -[-] = - x - = +
4. -[-4] = + 4

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-\; 3z\; -}{\mathrm{z}}^2\mathrm{+\; 4}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{{\mathrm{z}}^2\mathrm{-}{\mathrm{z}}^2\mathrm{4\; -\; 3z}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

$\frac{\mathrm{z}}{\mathrm{z\; +\; 2}}$ - $\frac{\mathrm{z\; -\; 2}}{\mathrm{z\; -\; 3}}$ = $\frac{\mathrm{4\; -\; 3z}}{\mathrm{(z\; +\; 2)(z\; -\; 3)}}$

x = $\frac{\mathrm{mn}}{3}$ ------ eqn 1

Substitute m = $\frac{\mathrm{v}}{\mathrm{y}}$ into ------ eqn 1

x = $\frac{\mathrm{v}}{\mathrm{y}}$ x n ÷ 3

x = $\frac{\mathrm{v\; x\; n}}{\mathrm{y}}$ ÷ 3

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ ÷ 3

Every number is being divided by 1 but it is not written.

3 = $\frac{3}{1}$

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ ÷ $\frac{3}{1}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

x = $\frac{\mathrm{vn}}{\mathrm{y}}$ x $\frac{1}{3}$

x = $\frac{\mathrm{vn\; x\; 1}}{\mathrm{y\; x\; 3}}$

x = $\frac{\mathrm{vn}}{\mathrm{3y}}$

Change the mixed fractions to improper fractions.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

2$\frac{2}{3}$ = $\frac{\mathrm{3\; x\; 2\; +\; 2}}{3}$ = $\frac{\mathrm{6\; +\; 2}}{3}$ = $\frac{8}{3}$

$1\frac{1}{2}2\frac{2}{3}$ = $\frac{\mathrm{x}}{96}$

$\frac{3}{2}$ ÷ $\frac{8}{3}$ = $\frac{\mathrm{x}}{96}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{3}{2}$ x $\frac{3}{8}$ = $\frac{\mathrm{x}}{96}$

$\frac{\mathrm{3\; x\; 3}}{\mathrm{2\; x\; 8}}$ = $\frac{\mathrm{x}}{96}$

$\frac{9}{16}$ = $\frac{\mathrm{x}}{96}$

Cross multiply.

16 x x = 9 x 96

Divide both sides by 16

x = $\frac{\mathrm{9\; x\; 96}}{16}$ = 54

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{9\; x\; (2x\; +\; 1)\; -\; 2\; x\; (3x\; -\; 7)\; -\; 1\; x\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; +\; 9\; -\; 6x\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; -\; 6x\; +\; 9\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{12x\; +\; 18}}{18}$

Factorize 6 out at numerator.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{6(2x\; +\; 3)}}{18}$

Note: 6 divides itself 1 time and 18, 3 times.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{1\; x\; (2x\; +\; 3)}}{3}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 3}}{3}$