2n - 5 = $\frac{1}{2}$n

Get rid of the fraction by multiplying both sides of the equation by the denominator (2).

2 x 2n - 2 x 5 = 2 x $\frac{1}{2}$n

4n - 10 = 1 x n
4n - 10 = n
4n - n = 10
3n = 10

Divide both sides by 3 so that n can stand alone

n = $\frac{10}{3}$ = 3$\frac{1}{3}$

Get rid of the denominator (2) by multiplying each part of the equation by 2

2 x (x-1) = 1(x+2)

Expand the bracket

2 x x - 1 x 2 = 1 x x + 1 x 2

2x - 2 = x + 2

2x - x = 2+2

x = 4

Change the decimal to fraction by multiplying both the numerator and the denominator by 1n0s where n0s means number of zeros to attach to 1 for the multiplication

n is the number of times to move the decimal point to get a whole number

To change 10.5 to 105, you will realize we have to move the decimal to the right once (1 time)

Hence we can multiply 10.5 by 10/10

10/10 = 1 so doesn't affect the figure but for the sake of simplication that's why we put it so

10.5 = 10.5 x 10/10 = 105/10

The updated equation will now be:

Now cross multiply

n x 10 = 105 x 100

Divide both sides by 10 to make n the subject

10 cancels each other and the zeros cancel each other

n = 105 x 10 = 1050

Since the number outside the bracket are the same in both sides of the equation, the values inside the bracket are also the same

Thus if a x b = a x c, implies that b = c

12+18 = 15+k

30 = 15 + k

30 - 15 = k

15 = k

Solve for the value of x (Make x the subject)

2x – 1 = 5
2x = 5 + 1
2x = 6

Divide both sides by 2

2x/2 = 6/2

x = 3

NOTE: When negative crosses the equal to sign, it becomes positive, hence the -1 becomes +1 when it moves to the right hand side.

Solve: (1-x)÷3<4

Get rid of the denominators by multiplying both sides of the equation by L.C.M. Every number is divisible by 1.

The L.C.M of 3 and 1 is 3

Divide both sides by -1.

NOTE: When dividing x by negative, the sign must change x > -11

Alternatively you can take the x to the right hand side and the numbers to the left.

Which reads -11 is less than x

Hence if -11 is less than x means x is also greater than – 11.

Hence x > -11

Let’s test

Hence x > -11 is correct

Testing x < -11 is option too. Numbers less than -11 are -12, -13, -14, -15 etc.

when x = -12

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

(x-3)² = 16

√(x-3)² = √16

√(x-3)² = x-3

√16 = 4

NOTE: √a = a^½

⇒ √(x-3)² = (x-3)^2x½ = (x-3)¹ = (x-3) = x-3

(x-3)² = 16

⇒ x-3 = 4

⇒ x = 4+3 = 7

Get rid of the fraction by multiplying each side of the equation by the LCM of the denominators (3,1 and 1)

Thus 3(w-1) = 3(w-1)/1 (every number is being divided by 1 which has no effect on the number)

-1 = -1/1

Hence the denominators are 3,1 and 1 and the LCM is 3

Multiply both sides of the equation by 3

3 x w/3 = 3 x 3(w-1)-3x1

3 cancels 3 in 3 x w/3 = w

3 x w/3 = 3 x 3(w-1)-3x1

⇒ w = 9(w-1)-3

Applying the concept, a(b+c) = axb+axc, a(b-c) = axb-axc

⇒ 9(w-1) = 9xw - 9x1 = 9w-9

⇒ w = 9w-9-3

Grouping the letters (w) at one side and the numbers at the other side of the equation

Concept, when negative moves to the other side of the equation, becomes positive and positive becomes negative

Option 1, grouping the w at the left side of equation

w = 9w-9-3

⇒ w - 9w = -9-3

- 8w = -12

Divide both sides by -8

-8w/-8 = -12/-8

NOTE: The negative cancels each other

w = 12/8 = 3/2, thus 4 goes into 12, 3 times and into 8, 2 times

Option 2, grouping the w at the right side of equation

w = 9w-9-3

⇒ 9+3 = 9w-w

12 = 8w

Divide both sides by 8

12/8 = 8w/8

w = 12/8 = 3/2, thus 4 goes into 12, 3 times and into 8, 2 times

Concept

- x - = + (When a negative number [-] multiplies another negative number[-], the sign becomes positive [+])

- x + = - (When a negative number [-] multiplies a positive number [+], the sign becomes negative [-])

a² = a x a

2y = 1-3x²+4x

When x = -1, substitute x by -1

2y = 1-3(-1)²+4(-1)

(-1)² = -1 x -1 = + 1

4(-1) = 4 x -1 = -4

2y = 1-3x²+4x

⇒ 2y = 1-3(1)-4

2y = 1 -3 - 4

2y = 1 - 7

2y = - 6

Divide both sides by 2 to get rid of the 2 multiplying y

⇒ y = -6 / 2 = -3