Chapter 10: Algebraic Expressions

Solution:-

(c) 4x

We know that, perimeter of the square = 4 × side

From the question it is given that, side length of the top of square table is x.

Then, perimeter = 4 × x

= 4x

Solution:-

(a) 2y

From the question it is given that, length of scarf is half metre = ½ m

Then, the number of scarfs can be made from y metres of cloth = y/(½)

= 2y

Solution:-

(d) 10x²y

123x²y – 138x²y = (123 – 138) x²y

= -15 x²y

Therefore, -15x²y is a like term of 10x²y, because both contain x²y.

Solution:-

(a) 1

From the question it is given that, value of x = 1

Substitute the value of x in the expression 3x² – 5x + 3

= (3 × (1)²) – (5 × 1) + 3

= 3 – 5 + 3

= 6 – 5

= 1

Solution:-

(d) n – 3

There are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is (n-3).

Solution:-

(d) 8x + 12

We know that, perimeter of the square = 4 × side

From the question it is given that, side length of the top of square table is 2x + 3.

Then, perimeter = 4 × (2x + 3)

= (4 × 2x) + (4 × 3)

= 8x + 12

In questions 17 to 32, fill in the blanks to make the statements true.

Solution:-

Sum or difference of two like terms is a like term.

Let us consider the two like terms = 2y and 3y

Sum of two like terms = 2y + 3y

= 5 y

Difference of two like terms = 2y – 3y

= -y

Solution:-

In the formula, area of circle = πr², the numerical constant of the expression πr² is π.

Solution:-

3a²b and –7ba² are like terms.

The terms having the same algebraic factors are called like terms.

Solution:-

–5a²b and –5b²a are unlike terms.

The terms having different algebraic factors are called unlike terms.