Students can Download Chapter 12 Algebraic Expressions Ex 12.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.

If m = 2, find the value of:

i) m – 2

2 – 2 = 0

ii) 3m – 5

3 × 2 – 5 = 6 – 5 = 1

iii) 9 – 5m

9 – 5 × 2 ÷ 9 – 10 = -1

iv) 3m^{2} – 2m – 7

= 3 × 2^{2} – 2 × 2 – 7

= 3 × 4 – 4 – 7

= 12 – 11 = 1

v)

Solution:

Question 2.

If p = -2, find the value of :

Solution:

i) 4p + 7

= 4x – 2 + 7

= -8 + 7 = -1

ii) -3p^{2} + 4p + 7

= -3 (-2)^{2} + 4 (-2) + 7

= 3 × 4 – 8 + 7

= 12 – 8 + 7

= -13

iii) -2p^{3} – 3p^{2} – 4p + 7

= -2 (-2)^{3} – 3(2)^{2} + 4 (-2) + 7

= -2(-8) – 3(4) – 8 + 7

= 16 – 12 – 8 + 7

= 16 + 7 – 12 – 8

= 23 – 20 = 3

Question 3.

Find the value of the following expressions, when x = -1:

i) 2x – 7

= 2(-1) – 7

= -2 – 7 = -9

ii) -x + 2

= -1 + 2 = 1

iii) x^{2} + 2x + 1

= 1 – 2 + 1 = 0

iv) 2x^{2} – x – 2

= 2(-1)^{2} – (-1) – 2

= 2 × 1 + 1 – 2

= 2 + 1 – 2 = 1

Question 4.

If a = 2, b = -2, find the value of:

i) a^{2} + b^{2}

= 2^{2} + (-2)^{2}

= 4 + 4 = 8

ii) a^{2} + ab + b^{2}

= 2^{2} + 2 × -2 + (-2)

= 4 – 4 + 4 = 4

iii) a^{2} – b^{2}

= (2)^{2} – (-2)^{2} = 4 – 4 = 0

Question 5.

When a = 0, b = -1, find the value of the given expressions :

i) 2a + 2b

= 2 × 0 + 2 × -1

= o – 2 = -2

ii) 2a^{2} + 2b^{2} + 1

= 2(0)^{2} + (-1)^{2} + 1

= 0 + 1 + 1 = 2

iii) 2a^{2}b + 2ab^{2} + ab

= 2 × 0^{2} × -1 + 2 × 0 × (-1)^{2} + 0 × -1

= 0 + 0 + 0 = 0

iv) a^{2} + ab + 2

= (0)^{2} + 0 × -1 + 2

= 0 + 0 + 2 = 2

Question 6.

Simplify the expressions and find the value if x is equal to 2

Solution:

i) x + 7 + 4 (x – 5)

= x + 7 + 4x – 20 (re-arranging the terms)

= 5x – 13 = 5(2) – 12

= 10 – 13 = -3

ii) 3(x + 2) + 5x – 7

= 3x + 6 + 5x – 1

= 8x – 1 = 8(2) – 1

= 16 – 1 = 15

iii) 6x + 5 + 5(x – 2)

= 6x+ 5x – 10

= 11x – 10 = 11(2) – 10

= 22 – 10 = 12

iv) 4(2x – 1) + 3x + 11

= 8x – 4 + 3x + 11

= 8x + 3x – 4 + 11 (re-arranging the terms)

= 11x + 7 = 11(2) + 7 = 22 + 7 = 29

Question 7.

Simplify these expressions and find their values if x = 3, a = -1, b = -2.

i) 3x – 5 – x + 9

= 3x – x + 9 – 5

= 2x + 4 = 2(3) + 4 = 6 + 4 = 10

ii) 2 – 8x + 4x + 4

= 2 + 4 – 4x

= -4x + 6

= -4 × 3 + 6

= -12 + 6 = -6

iii) 3a + 5 – 8a + 1

= 3a – 8a + 5 + 1

= -5a + 6

= -5(-1) + 6

= 5 + 6 = 11

iv) 10 – 3b – 4 – 5b

= -3b – 5b + 10 – 4

= -8b + 6

= -8 (-2) + 6

= 16 + 16 = 22

v) 2a – 2b – 4 – 5 + a

= 2a – 2b – 9

= 3a – 2b – 9

= 3(-1) – 2 (-2) – 9

= -3 + 4 – 9

= -12 + 4 = -8

Question 8.

i) If z = 10, find the value of z^{3} – 3(z – 10)

= (10)^{3} – 3 (10 – 10)

= 1000 – 3(0)

= 1000 – 0 = 1000

ii) If p = -10, find the value of p^{2} – 2p – 100

= (10)^{2} – 2(-10) – 100

= 100 + 20 – 100

= 120 – 100

= 20

Question 9.

What should be the value of a if the value of 2x^{2} + x – a equals to 5, when x = 0 ?

Solution:

= 2x^{2} + x – a

= 2(0)^{2} + 0 – a

= 5

= 0 + 0 – a = 5

∴ a = -5

Question 10.

Simplify the expression and find its value when a = 5 and b = -3, 2(a^{2} + ab) + 3 – ab

= 2a^{2} + 2ab + 3 – ab

= 2(5)^{2} + 2 × 5 × -3 + 3 – 5 × -3

= 2 × 25 – 30 + 3 + 15

= 50 – 30 + 3 + 15

= 50 + 3 + 15 – 30

= 68 – 30 = 38