MATHS CLASS 8 ALGEBRAIC EXPRESSIONS

 

MATHS CLASS 8 ALGEBRAIC EXPRESSIONS(IMPORTANT)

BY AARISH SIR

1. Identify the type of expressions:

(i) x²y + xy²

(ii) 564xy

(iii) -8x + 4y

(iv) x² + x + 7

(iv) xy + yz + zp + px + 9xy

Solution:

(i) x²y + xy² = Binomial

(ii) 564xy = Monomial

(iii) -8x + 4y = Binomial

(iv) x² + x + 7 = Trinomial

(iv) xy + yz + zp + px + 9xy = Polynomial

 

 

2. Using suitable algebraic identity, solve 1092²

Solution:

Use the algebraic identity: (a + b)² = a² + 2ab + b²

Now, 1092 = 1000 + 92

So, 1092² = (1000 + 92)²

(1000 + 92)² = ( 1000 )² + 2 × 1000 × 92 + ( 92 )²

= 1000000 + 184000 + 8464

Thus, 1092² = 1192464.

3. Identify terms and their coefficients from the following expressions:

(i) 6x²y² – 9x²y²z²+ 4z²

(ii) 3xyz – 8y

(iii) 6.1x – 5.9xy + 2.3y

Solution:

(i) 6x²y² – 9x²y²z²+ 4z²

Terms = 6x²y², -9x²y²z², and 4z²

Coefficients = 6, -9, and 4

(ii) 3xyz – 8y

Terms = 3xyz, and -8y

Coefficients = 3, and -8

(iii) 6.1x – 5.9xy + 2.3y

Terms = 6.1x, – 5.9xy, and 2.3y

Coefficients = 6.1, – 5.9 and 2.3

4. Find the area of a square with side 5x²y

Solution:

Given that the side of square = 5x²y

Area of square = side² = (5x²y)² = 25x⁴y²

5. Calculate the area of a rectangle whose length and breadths are given as 3x²y m and 5xy² m respectively.

Solution:

Given,

Length = 3x²y m

Breadth = 5xy² m

Area of rectangle = Length × Breadth

= (3x²y × 5xy²) = (3 × 5) × x²y × xy² = 15x³y³ m²

6. Simplify the following expressions:

(i) (x + y + z)(x + y – z)

(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)

(iii) 2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)

Solution:

Notes: “+” × “+” = “+”, “-” × “-” = “+”, and “+” × “-” = “-”.

(i) (x + y + z)(x + y – z)

= x² + xy – xz + yx + y² – yz + zx + zy – z²

Add similar terms like xy and yx, xz and zx, and yz and zy. Then simplify and rearrange.

= x² + y² – z² + 2xy

(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)

= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³

Now, add the similar terms and rearrange.

= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³

= 6x³ – x²y² – 2xy³

(iii) 2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)

= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x

= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x

= -x³ – x² – 16x

7. Add the following polynomials.

(i) x + y + xy, x – z + yx, and z + x + xz

(ii) 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy

(iii) -3a²b², (–5/2) a²b², 4a²b², and (⅔) a²b²

Solution:

(i) x + y + xy, x – z + yx, and z + x + xz

= (x + y + xy) + (x – z + yx) + (z + x + xz)

= x + y + xy + x – z + yx + z + x + xz

Add similar elements and rearrange.

= 2xy + xz + 3x + y

(ii) 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy

= (2x²y²– 3xy + 4) + (5 + 7xy – 3x²y²) + (4x²y² + 10xy)

= 2x²y² – 3xy + 4 + 5 + 7xy – 3x²y² + 4x²y² + 10xy

Add similar elements and rearrange.

= 3x²y² + 14xy + 9

(iii) -3a²b², (–5/2) a²b², 4a²b², and (⅔) a²b²

8. Subtract the following polynomials.

(i) (7x + 2) from (-6x + 8)

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

(iii) 2x²y²– 3xy + 4 from 4x²y² + 10xy

Solution:

(i) (7x + 2) from (-6x + 8)

= (-6x + 8) – (7x + 2)

= -6x + 8 – (7x + 2)

= -6x + 8 – 7x – 2

= -13x + 6

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

9. Calculate the volume of a cuboidal box whose dimensions are 5x × 3x² × 7x⁴

Solution:

Given,

Length = 5x

Breadth = 3x²

Height = 7x⁴

Volume of cuboid = Length × Breadth × Height

= 5x × 3x² × 7x⁴

Multiply 5, 3, and 7

= 105xx²x⁴

Use exponents rule: x^a × x^b = x^(a+b)

So, 105xx²x⁴ = 105x¹⁺²⁺⁴ = 105x⁷

10. Simplify 7x²(3x – 9) + 3 and find its values for x = 4 and x = 6

Solution:

7x²(3x – 9) + 3

Solve for 7x²(3x – 9)

= (7x² × 3x)  – (7x² × 9) (using distributive law: a(b – c) = ab – ac)

= 21x³ – 63x²

So, 7x²(3x – 9) + 3

= 21x³ – 63x² + 3

Now, for x = 4,

21x³ – 63x² + 3

= 21 × 4³ – 63 × 4² + 3

= 1344 – 1008 + 3

= 336 + 3 = 339

Now, for x = 6,

21x³ – 63x²

= 21 × 6³ – 63 × 6² + 3

= 2268 + 3

= 2271