ADD SUBTRACT AND MULTIPLY LINEAR EXPRESSIONS

Example 1 :

Add (6a + 3) and (4a - 2).

Solution :

= (6a + 3) + (4a - 2)

 = 6a + 3 + 4a - 2

= 10a + 1

Example 2 :

Add (5y + 8 + 3z) and (4y - 5).

Solution :

= (5y + 8 + 3z) + (4y - 5)

 = 5y + 8 + 3z + 4y - 5

= 9y + 3z + 3

Example 3 :

Subtract (6a - 3b) from (-8a + 9b).

Solution :

= (-8a + 9b) - (6a - 3b)

 = -8a + 9b - 6a + 3b

= -14a + 12b

Example 4 :

Subtract (2x + 3y - 5z) from (5x - 4y - 5z).

Solution :

= (5x - 4y - 5z) - (2x + 3y - 5z)

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

= 3x - 7y

Example 5 :

Multiply (3x - 7) and (7x - 3).

Solution :

= (3x - 7)(7x - 3)

= 21x² - 9x - 49x + 21

= 21x² - 58x + 21

Example 6 :

Multiply (3a - 2b) and (2a + 3b).

Solution :

= (3a - 2b)(2a + 3b)

= 6a² + 9ab - 4ab - 6b²

= 6a² + 9ab - 4ab - 6b²

= 6a² + 5ab - 6b²

Example 7 :

Multiply (p + q + r) and (p + q - r).

Solution :

= (p + q + r)(p + q - r)

Let x = p + q.

= (x + r)(x - r)

Using Algebraic Identity a² - b² = (a + b)(a - b),

= x² - r²

Substitute x = p + q.

= (p + q)² - r²

Using Algebraic Identity (a + b)² = a² + 2ab + b².

= p² + 2pq + q² - r²

Example 8 :

Divide (5x + 20) by (5x + 35).

Solution :

= (5x + 20)/(5x + 35)

= [5(x + 4)]/[5(x + 7)]

= (x + 4)/(x + 7)

Example 9 :

Divide (5x - 25) by (3x - 15).

Solution :

= (5x - 25)/(3x - 15)

= 5(x - 5)/3(x - 5)

= 5/3

Example 10 :

Divide (2 - x) by (4x - 8).

Solution :

= (2 - x)/(4x - 8)

= (2 - x)/4(x - 2)

= -(x - 2)/4(x - 2)

= -1/4

Example 11 :

The subtraction of 5 times of y from x is

(a) 5x – y    (b) y – 5x    (c) x – 5y    (d) 5y – x

Solution :

5 times of y = 5y

Subtract this from x, so x - 5y

Option c is correct.

Example 12 :

– b – 0 is equal to

(a) –1 × b      (b) 1 – b – 0     (c) 0 – (–1) × b    (d) – b – 0 – 1

Solution :

– b – 0 = -b

Option a :

-1 x b = -b

Which matches. Option a is correct.

Example 13 :

The side length of the top of square table is x. The expression for perimeter is:

(a) 4 + x    (b) 2x    (c) 4x    (d) 8x

Solution :

Side length of square table = x

Perimeter = 4(side length)

= 4x

Option c is correct.

Example 14 :

The number of scarfs of length half metre that can be made from y metres of cloth is :

(a) 2y    (b) y/2     (c) y + 2   (d)  y + 1/2

Solution :

= y / (1/2)

= y (2/1)

= 2y

So, option a is correct.

Example 15 :

The expression for the number of diagonals that we can make from one vertex of a n sided polygon is:

(a) 2n + 1   (b) n – 2   (c) 5n + 2   (d) n – 3

Solution :

Consider triangle,

A triangle has three sides.

In a triangle, there is no diagonal

So, the number of diagonals is zero.

From the option,

1) 2n + 1

Put n = 3,

2n + 1 = 2(3) + 1

= 7

Therefore, option a is incorrect

2) n - 2

Put n = 3,

n - 2 = 3 - 2

= 1

Therefore, option b is incorrect.

3) 5n + 2

Put n = 3,

5n + 2 = 5(3) + 2

= 15 + 2

= 17

Therefore, option c is incorrect.

4) n - 3

Put n = 3,

n - 3 = 3 - 3

= 0

Therefore, option d is correct.

Example 16 :

The length of a side of square is given as 2x + 3. Which expression represents the perimeter of the square?

(a) 2x + 16    (b) 6x + 9    (c) 8x + 3    (d) 8x + 12

Solution :

Side length of square = 2x + 3

Perimeter of square = 4(side length)

= 4(2x + 3)

= 8x + 12

So, option d is correct.

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