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Algebraic Expressions

    Introduction

    Algebaic term and Numeric term

    Coefficients

    Expressions

    Degree of Monomial

    Finding the Value of an Expression

    Addition of Algebraic Expressions

    Exercise 10.1

    Exercise 10.2

    Exercise 10.3

    Exercise 10.4

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

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Algebraic Expressions > Exercise 10.3

Exercise 10.3

Find the length of the line segment PR in terms of 'a'.

Looking at the figure, we can see:
  • P to Q is
  • Q to R is
To find PR, we need to add the lengths: PR = PQ + QR = +
Therefore, PR =
The length of line segment PR is 5a units.

2. Find the perimeter of the following triangle with sides 2x, 5x, and 6x.

(i)

Instruction

To find the perimeter, we need to add all sides of the triangle:
The three sides are:
  • First side =
  • Second side =
  • Third side =
Perimeter = sum of all sides = 2x + 5x + 6x =
Therefore, the perimeter of the triangle is 13x units.

(ii)

Find the perimeter of the following rectangle.

For a rectangle: (from the figure)
  • Length (l) =
  • Width (w) =
The perimeter of a rectangle = 2(l + w) = 2(2x + 3x) = 2(5x) =
Therefore, the perimeter of the rectangle is 10x units.

3. Subtract the second term from first term.

(i)

(i) 8x, 5x

First term = x

Second term = x

Subtracting: x - x = x

Therefore, 8x - 5x = 3x

(ii)

(ii) 5p, 11p

First term = p

Second term = p

Subtracting: p - p = p

Therefore, 5p - 11p = -6p

(iii)

(iii) 13m2, 2m2

First term = m2

Second term = m2

Subtracting: m2 - m2 = m2

Therefore, 13m2 - 2m2 = 11m2

4. Find the value of following monomials, if x = 1.

(i)

(i) -x

Given x =

Substituting: -x = -

Therefore, when x = 1, -x = -1

(ii)

(ii) 4x

Given x =

Substituting: 4x = 4 × =

Therefore, when x = 1, 4x = 4

(iii)

(iii) -2x^2

Given x =

Substituting: −2x2 = -2 × ^2 =

Therefore, when x = 1, −2x2 = -2

5. Simplify and find the value of 4x+x−2x2+x−1 when x = -1

First, let's combine like terms:
  • Terms with x: 4x+x+x =
  • Terms with x^2: -x2
  • Constants:
Simplified expression: −2x2+6x−1
Now substitute x = -1: −2−12+6−1−1 = -2(1) + (-) - =
Therefore, when x = -1, the value is -9

6. Write the expression 5x2−4−3x2+6x+8+5x−13 in its simplified form. Find its value when x = -2

Let's group like terms:
  • Terms with x^2: 5x2−3x2 = x2
  • Terms with x: 6x+5x = x
  • Constants: -4 + 8 - 13 =
Simplified expression: 2x2+11x−9
Now substitute x = -2: 2−22+11−2−9 = 2(4) + (-22) - 9 = 8 - 22 - 9 =
Therefore, when x = -2, the value is -23

7. If x = 1; y = 2 find the values of the following expressions

(i) 4x−3y+5

Given x = 1, y = 2
Substituting: 41−32+5 = - + 5 =
Therefore, 4x−3y+5 = 3

(ii) x2+y2

Given x = 1, y = 2
Substituting: 12+22 = + =
Therefore, x2+y2 = 5

(iii) xy+3y−9

Given x = 1, y = 2
Substituting: 12+32−9 = + - 9 =
Therefore, xy+3y−9 = -1

8. Area of a rectangle is given by A = l × b. If l = 9cm, b = 6cm, find its area?

Given:
  • Length (l) = cm
  • Breadth (b) = cm
Area = l × b = 9 × 6 =
Therefore, the area of the rectangle is 54 cm²

9. Simple interest is given by I = PTR/100. If P = ₹900, T = 2 years; and R = 5%, find the simple interest.

Given:
  • Principal (P) = ₹
  • Time (T) = years
  • Rate (R) = %
Simple Interest = PTR/100 = 900 × 2 × 5/100 =
Therefore, the Simple Interest is ₹90