WBBSE Solutions For Class 8 Maths Algebra Chapter 1 Simplify Revision Of Old Lessons

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Algebra Chapter 1 Revision Of Old Lessons

Necessary Formulae:

  1. (a + b)2 = a2 + 2ab + b2 = (a – b)2 + 4ab
  2. (a – b)2 =a2 – 2ab + b2 = (a + b)2 – 4ab
  3. (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
  4. a2 + b2 = (a + b)2 – 2ab = (a – b)2 + 2ab
  5. a2 – b2 = (a + b) (a – b)
  6. 2 (a2+ b2) = (a + b)2 + (a – b)2
  7. 4ab = (a + b)2 – (a – b)2
  8. ab = \(\left(\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\)

WBBSE Solutions For Class 8 Maths Algebra Chapter 1 Simplify Revision Of Old Lessons

Algebra Chapter 1 Revision Of Old Lessons Examples

Example 1. Simplify: – 3 – [-5 – 6- {3 – 5 – (2 – \(\overline{3-a}\))}]

Solution:

Given That:

f(x) = – 3 – [- 5 – 6 – {3 – 5 – (2 – \(\overline{3-a}\)}]

= – 3 – [- 5 – 6 – {3 – 5 – (2 – 3 + a)}]

= – 3 – [- 5 – 6 – {3 – 5 – (1 + a)}]

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= – 3- [- 5 – 6 – {3 – 5 + 1 – a}]

= – 3 – [- 5 – 6 – {-1 – a}]

= – 3 – [- 5 – 6 + 1 + a]

= – 3 – [10 + a]

= – 3 + 10 – a

f(x) = – 3 – [- 5 – 6 – {3 – 5 – (2 – \(\overline{3-a}\)}] = 7 – a

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Example 2. Find the summation of 3x2 – 7xy – 5y2, x2 + 2xy + 7y2, 9x2 – 10xy – 15y2

Solution:

Given That:

let us consider equations as f(x),f(y),f(z).

f(x) = (3x2 – 7xy – 5y2)

f(y) = (x2 + 2xy + 7y2)

f(z) = (9x2 – 10xy – 15y2)

Here summation means adding i.e., adding of equation 1 and equation 2 and Equation 3.

f(x + y + z) = (3x2 – 7xy – 5y2) + (x2 + 2xy + 7y2) + (9x2 – 10xy – 15y2)

f(x + y + z) = 3x2 -7xy – 5y2 + x2 + 2xy + 7y2 + 9x2 – 10xy – 15y2

f(x + y + z) = 3x2 + x2 + 9x2 – 7xy + 2xy – 10xy – 5y2 + 7y2 – 15y2

f(x + y + z) = 13x2 – 15xy – 13y2

Example 3. Subtract: (2x – 3y + 4z) from (-3x + 8y – 8z)

Solution:

Given That

f(x) = (- 3x + 8y – 8z)_______ eq (1)
f(y) = (2x – 3y + 4z)  _______ eq (2)

Now substraction of Equation 2 from equation 1

f(x – y) = (- 3x + 8y – 8z) – (2x – 3y + 4z)

f(x – y)  = – 3x + 8y – 8z – 2x + 3y – 4z

f(x – y)  = – 3x – 2x + 8y + 3y – 8z – 4z

f(x – y)  = 5x + 11y – 12z

Example 4. Simplify: (x − y) (x2 + xy + y2) + (y − z) (y2 + yz + z2) + (z − x) (z2 + zx + x2)

Solution:

Given That :

f(x) = (x – y) (x2 + xy + y2) + (y − z) (y2 + yz + z2) + (z – x) (z2 + zx + x2)

= x (x2 + xy + y2) − y (x2 + xy + y2) + y (y2 + yz + z2) − z (y2 + yz + z2) + z (z2 + zx + x2) − x (z2 + zx + x2)

= x3 + x2y + xy2 – x2y – xy2 – y3 + y3 + y2z + yz2 -y2z – yz2 – z3 + z3 + xz2 + x2z – xz2 – 2z – x3

f(x)  = 0

(x − y) (x2 + xy + y2) + (y − z) (y2 + yz + z2) + (z − x) (z2 + zx + x2) = 0

Example 5. Divide: (15 a3b2c – 10 ab2c – 25 a2bc2) by (- 5 abc)

Solution: \(\frac{15 a^3 b^2 c-10 a b^2 c-25 a^2 b c^2}{-5 a b c}\)

= \(-\frac{15 a^3 b^2 c}{5 a b c}+\frac{10 a b^2 c}{5 a b c}+\frac{25 a^2 b c^2}{5 a b c}\)

= \(-3 a^{3-1} b^{2-1} c^{1-1}+2 a^{1-1} b^{2-1} c^{1-1}+5 a^{2-1} b^{1-1} c^{2-1}\)

= \(-3 a^2 b^1 c^0+2 a^0 b^1 c^0+5 a^1 b^0 c^1\)

= \(-3 a^2 b+2 b+5 a c\)  [a0 =1, b0 =1, c0 =1]

(15 a3b2c – 10 ab2c – 25 a2bc2) by (- 5 abc) = \(-3 a^2 b+2 b+5 a c\)  [a0 =1, b0 =1, c0 =1]

Example 6. Find the square of a + 2b – c + 3d

Solution: (a + 2b – c + 3d)2

= {(a + 2b) – (c-3d)}2

= (a + 2b)2 – 2 (a + 2b) (c – 3d) + (c – 3d)2

= a2 + 2.a.2b + (2b)2 – 2 (ac – 3ad + 2bc – 6bd) + c2 – 2.c.3d + (3d)2

= a2 + 4ab + 4b2 -2ac + 6ad – 4bc + 12bd + c2 – 6cd + 9d2

= a2 + 4b2 + c2 + 9d2 + 4ab – 2ac + 6ad – 4bc + 12bd – 6cd

The square of a + 2b – c + 3d = a2 + 4b2 + c2 + 9d2 + 4ab – 2ac + 6ad – 4bc + 12bd – 6cd

Example 7. Express as perfect square and find the value. 25a2 – 30 (b – 2c) + 9 (b – 2c)2 where a = 1, b = -2, c = 3

Solution: 25a2 – 30 (b – 2c) + 9 (b – 2c)2

= (5a)2 – 2.5a.3 (b – 2c) + {3 (b- 2c)}2

= {5a – 3 (b – 2c)}2

= (5a – 3b + 6c)2

= (5.1 – 3(-2) + 6.3)2

= (5 + 6 + 18)2 = (29)2 = 841

25a2 – 30 (b – 2c) + 9 (b – 2c)2  = 841

Example 8. If a + b = 5 and ab= 6, then find the value of (a2 – b2)

Solution: a + b = 5, ab = 6

(a – b)2 = (a + b)2 – 4ab

= (5)2 – 4 x 6 = 25 – 24 = 1

= a – b ± √1 = ±1

a2 – b2 = (a + b) (a – b) = 5 x (± 1) = ± 5

The value of (a2 – b2) = ± 5

Example 9. If x + y = 12 and x – y = 4, then find the value of 3xy (x2 + y2).

Solution: x + y = 12, x – y = 4

3xy (x2 + y2)

= \(\frac{3}{8}\).4xy.2 (x2 + y2)

= \(\frac{3}{8}\){(x + y2 – (x − y)2} {(x + y)2 + (x − y)2}

= \(\frac{3}{8}\) x ((12)2– (4)2) ((12)2 + (4)2)

= \(\frac{3}{8}\) x (144 – 16) (144 + 16)

= \(\frac{3}{8}\) x128 x 160 = 7680

The value of 3xy (x2 + y2) = 7680

Example 10. If \(\left(\frac{a^2}{b^2}+t a+\frac{b^2}{25}\right)\) is a perfect square, then what will be the value of t where
a ≠ 0.

Solution: \(\frac{a^2}{b^2}+t a+\frac{b^2}{25}\)

= \(\left(\frac{a}{b}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{t b}{2}+\left(\frac{t b}{2}\right)^2-\left(\frac{t b}{2}\right)^2+\frac{b^2}{25}\)

= \(\left(\frac{a}{b}+\frac{t b}{2}\right)^2-\frac{t^2 b^2}{4}+\frac{b^2}{25}\)

As, given expression is a perfect square

so \(-\frac{t^2 b^2}{4}+\frac{b^2}{25}=0 \Rightarrow-\frac{t^2 b^2}{4}=-\frac{b^2}{25} \Rightarrow \frac{t^2 b^2}{4}=\frac{b^2}{25}\)

⇒ \(t^2=\frac{b^2}{25} \times \frac{4}{b^2} \Rightarrow t^2=\frac{4}{25} \Rightarrow t= \pm \sqrt{\frac{4}{25}} \Rightarrow t= \pm \frac{2}{5}\)

Example 11. Resolve into Factors:

  1. a2 – b2 + 2bc – c2
  2. 64ax2 – 49a (x – 2y)2
  3. 2a2b2 + 2b2c2 + 2c2a2 – a4 – b4 – c4
  4. x4 + x2 + 1
  5. x2 – 2 (a2 + b2) x + (a2 – b2)2

Solution:

1. \(a^2-b^2+2 b c-c^2\)

= \(a^2-\left(b^2-2 b c+c^2\right)\)

= \(a^2-(b-c)^2\)

= \((a+b-c)(a-b+c)\)

2. \(64 a x^2-49 a(x-2 y)^2\)

= \(a\left[64 x^2-49(x-2 y)^2\right]\)

= \(a\left[8 x^2-7(x-2 y)^2\right]\)

= \(a\left[(8 x)^2-(7 x-14 y)^2\right]\)

= \(a(8 x+7 x-14 y)(8 x-7 x+14 y)\)

= \(a(15 x-14 y)(x+14 y)\)

3. \(2 a^2 b^2+2 b^2 c^2+2 c^2 a^2-a^4-b^4-c^4\)

= \(4 a^2 b^2-2 a^2 b^2+2 b^2 c^2+2 c^2 a^2-a^4-b^4-c^4\)

= \(4 a^2 b^2-\left(a^4+b^4+c^4+2 a^2 b^2-2 b^2 c^2-2 c^2 a^2\right)\)

= \((2 a b)^2-\left(a^2+b^2-c^2\right)^2\)

= \(\left(2 a b+a^2+b^2-c^2\right)\left(2 a b-a^2-b^2+c^2\right)\)

= \(\left\{\left(a^2+2 a b+b^2\right)-c^2\right\}\left\{c^2-\left(a^2-2 a b+b^2\right)\right\}\)

= \(\left\{(a+b)^2-c^2\right\}\left\{c^2-(a-b)^2\right\}\)

= (a + b + c)(a + b – c)(c + a – b)(c – a + b)

4. \(x^4+x^2+1\)

= \(\left(x^2\right)^2+2 x^2 \cdot 1+(1)^2-x^2\)

= \(\left(x^2+1\right)^2-x^2\)

= \(\left(x^2+1+x\right)\left(x^2+1-x\right)\)

5. \(x^2-2\left(a^2+b^2\right) x+\left(a^2-b^2\right)^2\)

= \(x^2-2\left(a^2+b^2\right) x+(a+b)^2(a-b)^2\)

= \(x^2-\left\{(a+b)^2+(a-b)^2\right\} x+(a+b)^2(a-b)^2\)

= \(x^2-(a+b)^2 x-(a-b)^2 x+(a+b)^2(a-b)^2\)

= \(x\left\{x-(a+b)^2\right\}-(a-b)^2\left\{x-(a+b)^2\right\}\)

= \(\left\{x-(a+b)^2\right\}\left\{x-(a-b)^2\right\}\)

Example 12. Solve:

  1. 4x – 6 {8 – (x – 5)+ 7x} – 50 = 72
  2. \(\frac{1}{4}\)(x + 4) + \(\frac{1}{5}\)(x + 5) = \(\frac{1}{6}\)(x + 6) + \(\frac{1}{7}\)( x + 7)
  3. \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3 a-3 b}{a+b}=0\)
  4. 2 (x – 3) – (5 – 3x) = 3(x + 1)-5 (2 + x)

Solution:

1. 4x – 6 {8- (x – 5) + 7x} – 50 = 72

⇒ 4x – 6 {8x + 5 + 7x} = 72 + 50

⇒ 4x – 6 {6x + 13} = 122

⇒ 4x – 36x – 78 = 122

⇒ – 32x = 122 + 78

⇒ -32x = 200

⇒ x = –\(\frac{200}{32}\)

⇒ x = –\(\frac{25}{4}\)

2. \(\frac{1}{4}\)(x + 4) + \(\frac{1}{5}\)(x + 5) = \(\frac{1}{6}\)(x + 6) + \(\frac{1}{7}\)( x + 7)

⇒ \(\frac{x}{4}+1+\frac{x}{5}+1=\frac{x}{6}+1+\frac{x}{7}+1\)

⇒ \(\frac{x}{4}+\frac{x}{5}-\frac{x}{6}-\frac{x}{7}=1+1-1-1\)

⇒ \(\frac{105 x+84 x-70 x-60 x}{420}=0\)

⇒ \(\frac{59 x}{420}=0 \Rightarrow x=0\)

3. \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3 a-3 b}{a+b}=0\)

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x-3(a+b)}{a+b}=0\)

⇒ \(\frac{x-a}{b}+\frac{x-b}{a}+\frac{x}{a+b}-3=0\)

⇒ \(\left(\frac{x-a}{b}-1\right)+\left(\frac{x-b}{a}-1\right)+\left(\frac{x}{a+b}-1\right)=0\)

⇒ \(\frac{x-a-b}{b}+\frac{x-b-a}{a}+\frac{x-a-b}{a+b}=0\)

⇒ (x – a – b)\(\left(\frac{1}{b}+\frac{1}{a}+\frac{1}{a+b}\right)\)

⇒ x – a – b = 0  [\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}\right)\) ≠ 0]

⇒ x = a + b

4. 2 (x-3) – (5 – 3x) = 3(x + 1)- 5(2 + x)

2x – 6 – 5 + 3x = 3x + 3 – 10 – 5x

→ 2x + 3x – 3x + 5x = 3 – 10 + 6 + 5

⇒ 7x = 4

⇒ x = \(\frac{4}{7}\)

Example 13. Ten’s digit of a number of two digits is greater by 7 than the unit’s digit. Sum of the digits is \(\frac{1}{9}\) of the number. Find the number.

Solution: Let the digit at unit’s place be x. Then digit at ten’s place is (x + 7).

Number = 10 (x + 7) + x = 10x + 70 + x = 11x + 70

According to question,

x + (x + 7) = \(\frac{1}{9}\)(11x + 70)

⇒ 9(2x+7)= 11x + 70

⇒ 18x + 63 = 11x + 70

⇒ 18x – 11x = 70 – 63 7x = 7

⇒ x = \(\frac{7}{7}\) = 1

∴ The required number is (11 x 1 + 70) = 81

Example 14. Choose the correct answer:

1. If x + \(\frac{1}{x}\) = 4, then the value of \(x^2+\frac{1}{x^2}\)

  1. 14
  2. 16
  3. 2
  4. 4

Solution: x + \(\frac{1}{x}\) = 4

\(x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 \cdot x \cdot \frac{1}{t}=(4)^2-2=16-2=14\)

∴ The correct answer is 1. 14

2. If 4x2 + tx + 9 is a perfect square, then the value of t is

  1. ± 12
  2. ± 6
  3. ± 8
  4. ± 36

Solution: 4x2 + tx +9

= (2x)2 + 2·2x.3 + (3)2 ∓ 2·2x.3 + tx = (2x ± 3)2 ∓ 12x + tx

∴ ∓ 12x + tx = 0

⇒ tx = ±12x

⇒ t = ± 12

∴ So the correct answer is 1. ± 12

3. The value of (2:31)2 – (1.69)2 is

  1. 2.48
  2. 4.24
  3. 2.24
  4. 4.48

Solution: (2:31)2 – (1.69)2 = (2.31 + 1.69) (2.31 – 1.69) = 4 x 0.62 = 2.48

∴ The correct answer is (a)

Example 15. Write ‘True’ or ‘False’:

1. If a + b = 0, then a2 – b2 ≠ 0

Solution: a2 – b2 = (a + b) (a – b) = (0) (a – b) = 0

∴ The statement is false.

2. x = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

Solution: x = x.1 = \(\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2\)

∴ The statement is true.

3. If \(\frac{a}{b}-{b}{a}\) = 4, then the value of \(\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)\) is 18

Solution: \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=\left(\frac{a}{b}\right)^2+\left(\frac{b}{a}\right)^2=\left(\frac{a}{b}-\frac{b}{a}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{b}{a}=(4)^2+2=18\)

∴ The statement is true

Example 16. Fill in the blanks:

1. (a – 2) (a + 2) (a2 + 4) = ________

Solution: (a− 2) (a + 2) (a2 + 4) = (a2 – 22) (a2 + 4)

= (a2 – 4) (a2 + 4) = (a2)2 – (4)2 = a4 – 16

(a – 2) (a + 2) (a2 + 4) = a4 – 16

2. The value of \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\) is ________

Solution: \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\)

= \(\frac{(3.19)^2-(1.81)^2}{3.19-1.81}=\frac{(3.19+1.81)(3.19-1.81)}{(3.19-1.81)}\)

= 5.00 = 5

The value of \(\frac{3 \cdot 19 \times 3 \cdot 19-1 \cdot 81 \times 1 \cdot 81}{3 \cdot 19-1 \cdot 81}\) is 5.

3. If (x + 3) (x − p) = x2 – 9 then the value of p is _________

Solution: (x + 3) (x − p) ⇒ x2 – 9

⇒ (x + 3) (x − p) = x2 – 32

⇒ (x + 3) (x − p)(x + 3) (x – 3) ⇒ x – p = x -3

⇒ – p = -3

⇒ p = 3

The value of p is 3.

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