Try these, Recap Exercise, Student Activities, Think and answer

Recap Exercise

Answer the following questions :

1. Write the number of terms in the following expressions

(i) x + y + z − xyz [Answer: 4 terms]

(ii) m2n2c [Answer: 1 term]

(iii) a2b2c − ab2c2 + a2bc2 + 3abc [Answer: 4 terms]

(iv) 8x2 − 4xy + 7xy2 [Answer: 3 terms]

2. Identify the numerical co-efficient of each term in the following expressions.

(i) 2x2 − 5xy + 6 y2 + 7x −10 y + 9

(ii) x/3 + 2y/5 − xy + 7

(i) 2x2 − 5xy + 6y2 + 7x − 10y + 9

Answer:

Numerical co efficient in 2x2 is 2

Numerical co efficient in −5xy is −5

Numerical co efficient in 6y2 is 6

Numerical co efficient in 7x is 7

Numerical co efficient in −10y is − 10

Numerical co−efficient in 9 is 9

(ii) x/3 + 2y/5 − xy + 7

Answer:

Numerical co efficient in x/3 is 1/3

Numerical co efficient in 2y/5 is 2/5

Numerical co efficient in −xy is −1

Numerical co efficient in 7 is 7

3. Pick out the like terms from the following:

Solution:

4. Add : 2x , 6 y, 9x − 2y

Solution:

2x + 6y + 9x − 2y = 2x + 9x + 6y − 2y = (2 + 9)x + (6 − 2)y = 11x + 4y

5. Simplify : (5x 3 y3 − 3x 2 y2 + xy + 7) + (2xy + x 3 y3 − 5 + 2x 2 y2 )

Solution:

(5x3y3 − 3x2y2 + xy + 7) + (2xy + x3y3 − 5 + 2x2y2)

= 5x3y3 + x3y3 – 3x2y2 + 2x2y2 + xy + 2xy + 7 − 5

= (5 + 1) x3y3 + (−3 + 2) x2y2 + (1 + 2) xy + 2

= 6 x3y3 − x2y2 + 3xy + 2

6. The sides of a triangle are 2x − 5y + 9, 3y + 6x − 7 and −4 x + y + 10 . Find the perimeter of the triangle .

Solution:

Perimeter of the triangle = Sum of three sides

= (2x − 5y + 9) + (3y + 6x −7) + (−4x + y + 10)

= 2x − 5y + 9 + 3y + 6x − 7 − 4x + y + 10

= 2x + 6x − 4x − 5y + 3y + y + 9 − 7 + 10

= (2 + 6 − 4)x + (−5 + 3 + l) y + (9 − 7 + 10)

= 4x –y + 12

∴  Perimeter of the triangle = 4x –y + 12 units.

7. Subtract −2mn from 6mn.

Solution:

6 mn − (−2 mn) = 6 mn + (+2 mn)

= (6 + 2) mn = 8 mn

8. Subtract 6a 2 − 5ab + 3b2 from 4a 2 − 3ab + b2 .

Solution:

(4a2 − 3ab + b2) − (6a2 − 5ab + 3b2)

= (4a2 − 6a2) + (− 3ab − (−5 ab)] + (b2 − 3b2)

= (4 − 6) a2 + [−3ab + (+5 ab)] + (1 − 3) b2

= [4 + (− 6)] a2 + (−3 + 5) ab + [1+ (−3)] b2

= −2a2 + 2ab − 2b2

9. The length of a log is 3a + 4b − 2 and a piece (2a − b) is removed from it. What is the length of the remaining log?

Solution:

Length of the log = 3a + 4b − 2

Length of the piece removed = 2a − b

Remaining length of the log = (3a + 4b −2) − (2a − b)

= (3a – 2a) + [4b − (− b)] − 2

= (3 − 2)a + (4 + 1)b − 2

= a + 5b – 2

10. A tin had x litres of oil. Another tin had (3x2 + 6x − 5) litres of oil.

The shopkeeper added (x+7) litres more to the second tin. Later, he sold (x2+6) litres of oil from the second tin. How much oil was left in the second tin?

Solution:

Quantity of oil in the second tin = 3x2 + 6x − 5 litres.

Quantity of oil added = x + 7 litres

∴  Total quantity of oil in the second tin

= (3x2 + 6x − 5) + (x + 7) litres

= 3x2 + (6x + x) + (−5 + 7) = 3x2 + (6 + 1 )x + 2

= 3x2 + 7x + 2 litres

Quantity of oil sold = x2 + 6 litres

∴ Quantity of oil left in the second tin

= (3x2 + 7x + 2) − (x2 + 6) = (3x2 − x2) + 7x + (2 − 6)

= (3 − 1) x2 + 7x + (−4) = 2x2 + 7x − 4

Quantity of oil left = 2x2 + 7x − 4 litres

Think

Every algebraic expression is a polynomial. Is this statement true? Why?

Solution:

No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.

Eg. 2y2 + 5y−1 − 3 is a an algebraic expression. But not a polynomial.

Try these

Find the product of

(i) 3ab2 , −2a2b3

(ii) 4xy, 5 y2 x,(−x2 )

(iii) 2m, −5n, −3p

Solution:

(i) (3ab2) × (−2a2b3) = (+) × (−) × (3 × 2) × (a × a2) × (b2 × b3) = − 6a3 b5

(ii) (4xy) × (5y2x) × (−x2) = (+) × (+) × (−) × (4 × 5 × 1) × (x × x × x2) × (y × y2)

= −20 x4y3

(iii) (2m) × (−5n) × (−3p) = (+) × (−) × (−) × (2 × 5 × 3) × m × n × p

 = + 30 mnp = 30 mnp

Think

Why 3+(4x–7y) ≠ 12x−21y ?

Solution:

Addition and multiplication are different 3 + (4x − 7y) = 3 + 4x − 7y

We can add only like terms.

Try these

Multiply

(i) (5 x2 + 7x – 3) by –4x2

(ii) (10x – 7y + 5z) by 6xyz

(ii) (ab+3bc –5ca) by 3a2bc

(iv) (4m2 – 3m + 7) by –5m3

Solution:

(i) (5x2 + 7x − 3) by – 4x2

(5x2 + 7x − 3) × (– 4x2) = 5x2 (− 4x2) + 7x (− 4x2) − 3(− 4x2)

= − 20 x4 − 28x3 + 12x2

(ii) (10x −7y + 5z) by 6xyz

(10x −7y + 5z) × 6xyz = 6xyz (10x − 7y + 5z) [∵ Multiplication is commutative]

= 6xyz(10x) + 6xyz(−7y) + 6xyz(5z)

= (6 × 10) (x × x × y × z) + (6 × −7) + (x × y × y × z ) + (6 × 5)(x × y × z × z)

= 60x2yz + (−42xy2z) + 30xyz2

= 60x2yz − 42xy2z + 30xyz2

(iii) (ab + 3bc − 5ca) by 3a2 bc

 (ab + 3bc − 5ca) × (3a2bc) = ab(3a2bc) + 3bc (3a2bc) − 5ca (3a2bc)

= 3a2b2c + 9a2b2c2 − 15a3bc2

(iv) (4m2 − 3m + 7) by − 5m3

 (4m2 − 3m + 7) × (− 5m3) = 4m2 (− 5m3) − (3m) (− 5m3) + 7(− 5m3)

= − 20m5 + 15m4 − 35m3

Try these

Multiply

(i) (a − 5) and (a + 4)

(ii) (a + b) and (a − b)

(iii) (m4 + n4 ) and (m − n)

(iv) (2x + 3)(x + 4)

(v) (3x + 7)( x −5)

(vi) (x − 2)(6x − 3)

Solution:

(i) (a − 5) and (a + 4)

 (a − 5) (a + 4) = a(a + 4) − 5(a + 4)

= (a × a) + (a × 4) + (−5 × a) + (−5 × 4)

= a2 + 4a − 5a − 20 = a2 − a − 20

(ii) (a + b) and (a − b)

 (a + b) (a − b) = a(a − b) + b(a − b)

= (a × a) + (a × −b) + (b × a ) + b(− b)

= a2 − ab + ab − b2 = a2 − b2

(iii) (m4 + n4) and (m − n)

(m4 + n4) (m − n) = m4 (m − n) + n4 (m − n)

= (m4 × m) + (m4 × (− n)) + (n4 × m) + (n4 × (−n))

= m5 − m4 n + mn4 – n5

(iv) (2x + 3) (x + 4)

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

= (2x2 × x) + (2x × 4) + (3 × x) + (3 × 4)

= 2x2 + 8x + 3x + 12 = 2x2 + 11x + 12 

(v) (3x + 7) (x − 5)

 (3x + 7) (x − 5)  = x(3x + 7) − 5(3x + 7)

= (x × 3x) + (x × 7) + (−5 × 3x) + (−5 × 7)

= 3x2 + 7x − 15x − 35

= 3x2 − 8x – 35

(vi) (x − 2) (6x − 3)

 (x − 2) (6x − 3)  = x(6x − 3) − 2(6x − 3)

= (x × 6x) + (x × (−3) − (2 × 6x) − (2 × 3)

= 6x2 − 3x − 12x + 6

= 6x2 − 15x + 6

Think

(i) In 3x2 (x4–7x3+2), what is the highest power in the expression?

Solution:

3x2 (x4 − 7x3 + 2)  = (3x2) (x4) + 3x2 (− 7x3) +  (3x2)2

= 3x2  − 21x5 + 6x2

Highest power is 6 in x6.

(ii) Is –5y2 +2y–6 = –(5y2 +2y–6)? If not, correct the mistake.

Solution:

No, – 5y2 + 2y – 6 = − (5y2 − 2y + 6 )

Think

Are the following correct?

(i) x3/x8 = x8-3 = x5

(ii) 10m4 / 10m4 = 0

(iii) When a monomial is divided by itself, we will get 1?

Solution:

(i) x3 / x8 = x8 – 3 = x5

 x3 / x8 = x3 – 8 = x−5 (or) x3 / x8 = 1 / x8 – 3 = 1 / x5

∴  The given answer is wrong.

(ii) 10m4 / 10m4 = 0

10m4 / 10m4 = [10/10] m4−4 = 1 m0 = 1     [ ∵ m0 = 1]

∴  The given answer is not correct

(iii) When a monomial is divided by itself, we will get 1?

When a monomial is divided by itself, we will get 1.

Eg. x/x = x1 – 1 = x0 = 1

∴ The given statement is correct

Try these

Divide

(i) 12x 3y2 by x2y (ii) −20a5b2 by 2a3b7 (iii) 28a4 c2 by 21ca2

(iv) (3 x 2 y)3 √6x2y3 (v) 64m4 (n2)3 ÷ 4m2n2 (vi) (8 x 2 y 2 )3 ÷ (8x 2 y2 )2

(vii) 81 p2q4 ÷ √[81p2q4 ]

(vii) ( 4x2 y3 )0 ÷ [( x3)2]/x6

Solution:

Solution:

(i) 12x3y2 / x2y = 12x3−2 y2−1  = 12x1y1 = 12 xy

(ii) −20a5b2 / 2a3b7 = −20a5 − 3 / 2b7 − 2  = −10a2 / b5

(iii) 28 a4c2 / 21ca2 = [28 / 21] a4 – 2 c2 – 1 = 4/3 a2c1 = 4/3 a2c

(iv) (3x2y)3 / 6x2y3 = 27 x6y3 / 6 x2y3  = [9/2] x6–2 = 9/2 x4

(v) 64m4(n2)3 / 4m2n2 = 64m4n6 / 4m2n2 = 16 m4−2 n6−2 = 16 m2n4

(vi) (8x2y2)3 / (8x2y2)2 = 512 x6y6 / 64 x4y4

= 8x6−4 y6−4

= 8x2y2

(vii) 81p2q4 / √81p2q4 = 81p2q4 / 9pq2 = 9 p2−1q4–2 = 9 pq2

(viii) (4x2y3)0 /  [ (x3)2 / x6 ] = 1 / [ x6 / x6 ] = 1 / 1 = 1

Think

Are the following divisions correct?

Solution:

(i) [4y + 3] / 4 = y + 3

 [4y + 3] / 4 = [4y / 4] + [3 / 4] = y + [3/4] is the correct answer.

∴  The given answer is not correct.

(ii) [5m2 + 9] / 9 = 5m2

[5m2 + 9] / 9 = [5m2 / 9] + [9 / 9] = [ 5/9 m2 ] + 1 is the correct answer

∴  The given answer is not correct.

(iii) [2x2 + 8] / 4 = 2x2 + 2 If not, correct it.

[2x2 + 8] / 4 = [(2x2) / 4 ] + [ 8 / 4 ] = (1/2) x2 + 2 is the correct answer

∴  The given answer is not correct.

Try these

(i) (16 y5 − 8y2 ) ÷ 4 y

(ii) ( p5 q2 + 24 p3 q −128q3 ) ÷ 6q

(iii) (4m2n + 9n2m + 3mn) ÷ 4mn

Solution:

(i) (16y5 − 8y2) ÷ 4y

(16y5 − 8y2) / 4y = [16y5 / 4y ]− [ 8y2 / 4y ] = 4y5 – 1 − 2y2 – 1 = 4y4 – 2y

(ii) ( p5q2 + 24p3q − 128 q3 ) ÷ 6q

( p5q2 + 24p3q − 128 q3 ) / 6q  = [ p5q2 / 6q ] + [ 24p3q / 6q ] – [ 128q3 / 6q ]

= (1/6) p5q2 – 1 + 4p3q1 – 1 – (64/3)q3 – 1

= (1/6)p5q1 + 4p3q0 – (64/3)q2 = (1/6)p5q + 4p3 – (64/3)q2

(iii) ( 4m2n + 9n2m + 3mn ) ÷ 4mn  

[ 4m2n + 9n2 m + 3mn ] / 4mn = [ 4m2n / 4mn ] + [ 9n2m / 4mn ] + [ 3mn / 4mn ]

= m2−1n1−1 + (9/4) m1−1n2−1 + (3/4)m1−1n1−1

= m1n0 + (9/4)m0 n1 + (3/4)m0n0 = m + (9/4)n + 3/4.    [ ∵ n0 =1]

Try these

Expand the following

(i) ( p + 2)2 =  p2 + 2(p) (2) + 22 =  p2 + 4p + 4

(ii) (3 − a)2 =  32 − 2(3) (a) + a2 = 9 − 6a + a2

(iii) (62 − x2 ) = (6 + x) (6 − x)

(iv) (a + b)2 − (a − b)2 = a2 + 2ab + b2 − (a2 − 2ab + b2)

= a2 + 2ab + b2 − a2 + 2ab − b2

= (1 − 1) a2 + (2 + 2) ab + (+1 −1) b2 = 4ab

(v) (a + b)2 = (a + b) × (a + b)

(vi) (m + n)(m − n) = m2 − n2

(vii) (m + 7)2 = m2 + 14m + 49

(xiii) (k2 - 49) = (k+ 7 )(k - 7 )

(ix) m2 − 6m + 9 = (m − 3)2

(x) (m − 10)(m + 5) = m2 + (−10 + 5)m + (−10)(5) = m2 − 5m − 50

Think

Which is corrcet? (3a)2 is equal to

(i) 3a2 (ii) 32a (iii) 6a2 (iv) 9a2

[Answer: (iv) 9a2]

Solution: (3a)2 = 32a2 = 9a2

Try these

Expand using appropriate identities.

(i) (3 p + 2q)2

(ii) (105)2

(iii) (2x − 5d)2

(iv) (98)2

(v) ( y − 5)( y + 5)

(vi) (3x)2 − 52

(vii) (2m + n)(2m + p)

(viii) 203 ×197

(ix) Find the area of the square whose side is (x − 2) units.

(x) Find the area of the rectangle whose length and breadth are ( y + 4) units and ( y − 3) units.

Solution:

1) (3p + 2q)2

Comparing (3p + 2q)2 with (a + b)2, we get a = 3p and b = 2q.

(a + b)2 = a2 + 2ab + b2

(3p + 2q)2 = (3p)2 + 2(3p) (2q) + (2q2) = 9p2 + 12pq + 4q2

2) (105)2 = (100+ 5)2

Comparing (100 + 5)2 with (a + b)2, we get a = 100 and b = 5.

(a + b)2 = a2 + 2ab + b2

(100 + 5)2 = (100)2 + 2 (100) (5) + 52 = 10000+ 1000 + 25

1052 = 11,025

3. (2x − 5d)2

Comparing with (a – b)2, we get a = 2x  b = 5d.

(a − b)2 = a2 − 2ab + b2

(2x − 5d)2 = (2x)2 − 2 (2x) (5d) + (5d)2

= 22x2 − 20xd + 52d2 = 4x2 − 20xd + 25d2

4. (98)2 = (100 − 2)2

Comparing (100− 2)2 with (a − b)2 we get

a = 100, b = 2

(a − b)2 = a2 − 2ab + b2

(100 − 2)2 = 1002 − 2 (100) (2) + 22

= 10000 − 400 + 4 = 9600 + 4 = 9604

5. (y − 5) (y + 5)

Comparing (y − 5) (y + 5) with (a − b) (a + b) we get

a = y ; b = 5

(a − b) (a + b) = a2 − b2

(y − 5)(y + 5) = y2 − 52 = y2 – 25

6. (3x)2 − 52

Comparing (3x)2 − 52 with a2 − b2 we have

a = 3x ; b = 5

(a2 − b2) = (a + b) (a − b)

(3x)2 − 52 = (3x + 5) (3x − 5) = 3x (3x − 5) + 5 (3x − 5)

= (3x) (3x) − (3x) (5) + 5 (3x) − 5 (5)

= 9x2 − 15x + 15x − 25 = 9x2 – 25

7. (2m + n) (2m + p)

Comparing (2m + n) (2m + p) with (x + a) (x + b) we have

x = 2n ; a = n ; b = p

(x − a)(x + b) = x2 + (a + b) x + ab

(2m + n) (2m + p) = (2m2) + (n + p)(2m) + (n) (p)

= 22m2 + n (2m) + p (2m) + np

= 4m2 + 2mn + 2mp + np

8. 203 × 197 = (200 + 3) (200 − 3)

Comparing (a + b) (a − b) we have

a = 200, b = 3

(a + b) (a − b) =  a2 − b2

(200 + 3) (200 − 3) = 2002 − 32

203 × 197 = 40000 − 9

203 × 197 = 39991

9. Side of a square = x − 2

∴ Area = Side × Side

= (x − 2) (x − 2) = x (x − 2) −2 (x − 2)

= x(x) + (x) (−2) + (−2)(x) + (−2) (−2)

= x2 − 2x − 2x + 4

= x2 − 4x + 4 units square

10. Length of the rectangle = y + 4

breadth of the rectangle = y − 3

Area of the rectangle = length × breadth

= (y + 4)(y − 3) = y2 + (4 + (−3)) y + (4) (−3)

= y2 + y – 12

Activity

You can visualize the geometrical proof of (a+b)3 with the help of your teacher.

Try these

Expand : (i) ( x + 5)3 (ii) ( y − 2)3 (iii) (x + 1)(x + 4)(x + 6)

Solution:

(i) Comparing (x + 5)3 with (a + b)3, we have a = x and b = 4.

(a + b)3 = a3 + 3a2 b + 3ab2 + b3

(x + 5)3 = x3 + 3x2 (5) + 3 (x) (5)2 + 53

. = x3 + 15x2 + 75x + 125

(ii) Comparing (y − 2)3 with (a − b)3 we have a = y b = z

(a − b)3 = a3 − 3a2 b + 3ab2 − b3

(y − 2)2 = y3 – 3y2(2) + 3y(2)2 + 23

= y3 − 6y2 + 12y + 8 

(iii) Comparing (x + 1) (x + 4) (x + 6) with (x + a) (x + b) (x + c) we have

a = 1 b =  4 and c = 6

(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc + ca) x + abc

= x3 + (1 + 4 + 6) x2 + ((1) (4) + (4) (6) + (6) (l))x + (1) (4) (6)

= x3 + 11x2 + (4 + 24 + 6) x +24

= x3 + 11x2 + 34x + 24

Try these

Find the factors

Think

 x2–4(x–2) = (x2–4)(x–2)

Is this correct? If not correct it.

Solution:

Not correct

(3a)2 = 32a2 − 9a2

x2 − 4(x − 2) = x2 – 4x + 8 

Try these

Factorise the following :

1) 3y + 6 

2) 10x 2 +15y2 

3) 7m(m − 5) + 1(5 − m) 

4) 64 − x2 

5) x2–3x+2 

6) y2–4y–32 

7) p2+2p–15 

8) m2+14m+48 

9) x2–x–90 

10) 9x2–6x–8

Solution:

1. 3y + 6

3y + 6 = 3 × y + 2 × 3

Taking out the common factor 3 from each term we get 3 (y + 2)

∴ 3y + 6 = 3(y + 2)

2. 10x2 + 15y2

10x2 + 15y2 = (2 × 5 × x × x) + (3 × 5 × y × y)

Taking out the common factor 5 we have

10x2 + 15y2  = 5(2x2 + 3y2)

3. 7m (m − 5) + (15 – m)

7m (m − 5) + (15 – m) = 7 m (m − 5) + (−1 )(−5 + m)

= 7m (m − 5) − 1(m − 5)

Taking out the common binomial factor (m − 5) = (m − 5) (7 m − 1)

4. 64 − x2

64 − x2 = 82 − x2

This is of the form a2 − b2

Comparing with a2 − b2 we have a = 8, b = x

a2− b2 = (a + b) (a − b)

64 − x2 = (8 + x) (8 − x)

5. x2 – 3x + 2

x2 – 3x + 2 = x2 − 2x − x + 2

= x (x − 2) − (x − 2)

= (x − 2) (x − l)

6. y2 – 4y – 32

 y2 – 4y – 32 = y2 – 8y + 4y − 32

= y(y − 8) + 4(y − 8)

= (y − 8) (y + 4)

7. p2 + 2p – 15

p2 + 2p – 15 =  p2 + 5p −3p − 15

= p(p + 5) −3 (p + 5)

= (p + 5) (p − 3)

8. m2 + 14 m + 48

m2 + 14 m + 48 = m2 + 8 m + 6 m + 48

= m (m + 8) + 6 (m + 8)

= (m + 6) (m + 8)

9. x2 − x – 90

x2 − x – 90 = x2 − 10x + 9x − 90

= x (x − 10) + 9 (x − 10)

= (x + 9) (x − 10)

10. 9x2 − 6x – 8

 9x2 − 6x – 8 = 9x2 − 12x + 6x − 8

= 3x (3x − 4) + 2(3x − 4)

= (3x + 2) (3x − 4)

Try these

Identify which among the following are linear equations.

(i) 2 + x = 19 − Linear as degree of the variable x is 1

(ii) 7x2 – 5 = 3 − not linear as highest degree of x is 2

(iii) 4p3 = 12 − not linear as highest degree of p is 3

(iv) 6m+2 − Linear, but not an equation

(v) n=10 −  Linear equation as degree of n is 1

(vi) 7k – 12= 0 − Linear equation as degree of k is 1

(vii) 6x/8 + y = 1 − Linear equation as degree of x & y is 1

(viii) 5 + y = 3x − Linear equation as degree of y & x is 1

(ix) 10p+2q=3 − Linear equation as degree of p & q is 1

(x) x2–2x–4 − not linear equation as highest degree of x is 2

Think

(i) Is t(t – 5)=10 a linear e quation? Why?

Solution:

t(t − 5) = 10

= (t × t) – (5 × t) = 10

= t2 − 5t = 10

This is not a linear equation as the highest degree of the variable ‘t’ is 2 

(ii) Is x2 =2x , a linear equation? Why?

Solution:

x2 = 2x

= x2 − 2x = 0

This is not a linear equations as the highest degree of the variable ‘x’ is 2

Try these

Convert the following statements into linear equations:

• On subtracting 8 from the product of 5 and a number, I get 32.

Solution:

Convert to linear equations:

Given that on subtracting 8 from product of 5 and a, we get 32

∴  5 × x − 8 = 32

∴  5x − 8 = 32

• The sum of three consecutive integers is 78.

Solution:

Sum of 3 consecutive integers is 78

Let 1st integer be ‘x’

∴  x + (x + 1) + (x + 2) = 78

∴  x + x + l + x + 2 = 78

3x + 3 = 78 

• Peter had a Two hundred rupee note. After buying 7 copies of a book he was left with ₹60.

Solution:

Let cost of one book be ‘x’

∴ Given that 200 −7 × x = 60

∴ 200 − 7x = 60

• The base angles of an isosceles triangle are equal and the vertex angle measures 80°.

Solution:

Let base angles each be equal to x & vertex bottom angle is 80°. Applying triangle property, sum of all angles is 180°

∴  x + x + 80 = 180°

∴  2x + 80 = 180

• In a triangle ABC, ∠A is 10o more than ∠B. Also ∠C is three times ∠A. Express the equation in terms of angle B.

Solution:

Let ∠B  = b

Given ∠A = 10° + ∠B = 10 + b

Also given that ∠C = 3 × ∠A = 3 × (10 + b) = 30 + 3b

Sum of the angles = 180°

∠A + ∠B + ∠C = 180°

10 + b + b + 30 + 3b = 180°

∴  5b + 40 = 180°

Note

Polynomial

A polynomial is an expression containing two or more algebraic terms. In a polynomial all variables are raised to only whole number powers.

a2 + 2ab + b2

4x2 + 3x – 7

A polynomial cannot contain :

1) Division by a variable. Eg. 4x2 – (5/1+x) is not a polynomial.

2) Negative exponents. Eg. 7x–2 + 5x – 6 is not a polynomial.

3) Fractional exponents. Eg. 3x3 + 4x1/2 + 5 is not a polynomial.

Monomial

An expression which contains only one term is called a monomial. Examples:

4x, 3x2 y, −2 y2 .

Binomial

An expression which contains only two terms is called a binomial.

Examples: 2x + 3, 5y2 + 9 y, a2b2 + 2b

Trinomial

An expression which contains only three terms is called a trinomial.

Examples : 2a2b −8ab + b2 , m2 −n 2 + 3

Note

Product of two terms is represented by the symbols ( ), dot (.) or × .

For example,

multiplying 4x2 and xy can be written in any one of the following ways.

Note

Distributive law

If a is a constant, x and y are variables then 

 a(x + y) = ax + ay

For example, 5(x + y) = 5x + 5 y