Get live Maths 1-on-1 Classs - Class 6 to 12

Examples

Example 1

Example 2 Important

Example 3 You are here

Example 4 Important

Example 5 (i)

Example 5 (ii) Important

Example 6 (i)

Example 6 (ii) Important

Example 7

Example 8 (i)

Example 8 (ii)

Example 9 (i)

Example 9 (ii) Important

Example 10

Example 11 (i)

Example 11 (ii) Important

Example 12 (i)

Example 12 (ii) Important

Example 13 (i)

Example 13 (iii)

Example 13 (ii) Important

Example 14 (i)

Example 14 (ii) Important

Last updated at March 23, 2023 by Teachoo

Example 1 Add: 7xy + 5yz – 3zx, 4yz + 9zx – 4y , – 3xz + 5x – 2xy. Expressions are 7xy + 5yz – 3zx 4yz + 9zx – 4y – 3xz + 5x – 2xy We put like terms below like terms So, required sum = 5xy + 9yz + 3zx – 4y + 5x Example 2 Subtract 5x2 – 4y2 + 6y – 3 from 7x2 – 4xy + 8y2 + 5x – 3y. We have to subtract 1st expression from the 2nd expression. So, we write 2nd expression first. Putting like terms below like terms Example 3 Complete the table for area of a rectangle with given length and breadth. Length = 3𝒙 , Breadth = 5y Area of rectangle = Length × Breadth = 3𝑥 × 5y = 3 × 𝑥 × 5 × y = (3 × 5) × (𝑥 × y) = (15) × (𝑥y) = 15𝑥y (ii) Length = 9y , Breadth = 4y2 Area of rectangle = Length × Breadth = 9y × 4y2 = 9 × y × 4 × y2 = (9 × 4) × (y × y2) = 36 × y1 + 2 = 36 × y3 = 36y3 (ii) Length = 9y , Breadth = 4y2 Area of rectangle = Length × Breadth = 9y × 4y2 = 9 × y × 4 × y2 = (9 × 4) × (y × y2) = 36 × y1 + 2 = 36 × y3 = 36y3 (iii) Length = 4ab , Breadth = 5bc Area of rectangle = Length × Breadth = 4ab × 5bc = 4 × a × b × 5 × b × c = (4 × 5) × a × (b × b) × c = 20 × a × b2 × c = 20ab2c (iv) Length = 2l2m , Breadth = 3lm2 Area of rectangle = Length × Breadth = 2l2m × 3lm2 = 2 × l2 × m × 3 × l × m2 = (2 × 3) × (l2 × l) × (m × m2) = 6 × l3 × m3 = 6l3m3 So, our completed table looks like