WBBSE Solutions For Class 7 Maths Algebra Chapter 5 formulae And Their Applications Exercise 5 Solved Example Problems

Algebra Chapter 5 formulae And Their Applications Exercise 5 Solved Example Problems

Formulae and Their Applications Introduction

The entire structure of algebra is based on some formulae. We can form some easy secting AD at G and BC at H where GD = 6. formulae by simple algebraic multiplication and we may also verify them geometrically.

Actually, every formula is an identity that is true for all values of the literals present in that identity.

The fundamental formulae are of great importance because all the harder formulae are derived from the fundamental formulae. We may define a formula as follows:

A general relation, expressed in terms of algebraic symbols is called an algebraic formula or in brief a formula.

Math Solution Of Class 7 Wbbse

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Formula for the square of the sum of two terms

(a + b)² = (a + b) (a + b)

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

= a² + ab + ab + b² = a² + 2 ab + b².

Thus we see that,

The square of the sum of two numbers is equal to the square of the first number plus twice the product of the two numbers plus the square of the second number.

Geometrical representation of the formula (a + b)²  = a² + 2ab + b².

In the diagram, ABCD is a square. The length of each side of ABCD = a + b.

Hence, the area of the square ABCD = (a + b)².

Now, draw a vertical line EF perpendicular to AB and intersecting AB at E and CD at F where BE = b.

Again draw a horizontal line GH perpendicular to AD and intersecting AD at G BC and H where GD = b.

Let EF and GH intersect at P.

WBBSE Class 7 Algebra Formulae Solutions

Now, area of AEPG = \(a^2\), Area of EBHP = ab, area of GPFD = ab, and area of PHCF = \(b^2\).

Thus, the sum of these four areas = \(a^2+a b+a b+b^2=a^2+2 a b+b^2\)

Since, the area of ABCD = area of (AEPG + EBHP + GPFD + PHCF)

Therefore, \((a+b)^2=a^2+2 a b+b^2\) (Proved).

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Formula for the square of the sum of three terms \((a+b+c)^2=\{(a+b)+c\}^2\)

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

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

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

Formula for the square of the difference of two terms

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

= a (a – b) -b (a – b)

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

Thus we see that,

The square of the difference of two numbers is equal to the square of the first number minus twice the product of the two numbers plus the square of the second number.

Geometrical representation of the formula \((a-b)^2=a^2-2 a b+b^2\)

Let ABCD be a square of side a.

Draw a horizontal line EF perpendicular to AD and intersecting AD at E and BC at F, where AE = b.

Next draw GH perpendicular to EF, where GF = b, and draw HP perpendicular to BC produced, where CP = b.

Let GH and DC intersect at Q.

Now, area of ABCD = \(a^2\), area of EGQD = \(\left(a-b^2\right)\),

area of ABFE = ab, area of GFPH = ab, area of QCPH = \(b^2\),

Now, area of EGQD = area of (ABCD + QCPH – ABFE – GFPH)

or, \((a-b)^2=a^2+b^2-a b-a b\)

i.e., \((a-b)^2=a^2-2 a b+b^2\) (Proved).

Some relations based on the square of the sum of two terms and the square of the difference of two terms.

1. Since \((a+b)^2=a^2+2 a b+b^2\) therefore, \(a^2+b^2=(a+b)^2-2\)
2. Since \((a-b)^2=a^2-2 a b+b^2\) therefore, \(a^2+b^2=(a-b)^2+2 a b\)
3. \((a+b)^2=a^2+2 a b+b^2=a^2-l a b+b^2+4 a b-(a-b)^2+4 a b\) Hence, \((a+b)^2=(a-b)^2+4\)
4. \((a-b)^2=a^2-2 a b+b^2=a^2+2 a b+b^2-4 a b=(a+b)^2-4 a b\) Hence, \((a-b)^2=(a+b)^2-4 a b\)
5. \((a+b)^2+(a-b)^2=a^2+2 a b+b^2+a^2-2 a b+b^2=a^2+a^2+b^2+b^2\)
   \(=2 a^2+2 b^2=2\left(a^2+b^2\right)\)
   Hence, \((a+b)^2+(a-b)^2=2\left(a^2+b^2\right) .\)
6. \((a+b)^2-(a-b)^2=\left(a^2+2 a b+b^2\right)-\left(a^2-2 a b+6^2\right)\)
   \(=a^2+2 a b+b^2-a^2+2 a b-b^2=4 a b\)
   Hence, \((a+b)^2-(a-b)^2=4 a b .\)

The formula for the product of the sum and the difference of two terms.

(a +b) (a – b) = (a – b) + 6 (a – b) = a² – ab + ab – b² = a² – b².

Thus we see that, The product of the sum and the difference of two numbers is equal to the difference of the squares of those two numbers.

Geometrical representation of the formula (a + b) (a – b) = a² – b².

Let ABCD be square of side a. Draw a vertical line EF perpendicular to AB intersecting AB at E where AE = b.

Next draw a horizontal line GH perpendicular to AD and intersecting AD at G where AG = b.

Now GH intersects EF at F, BC at Q.

Here QH = b, HP is drawn perpendicular from H on DC produced.

According to the construction, the area of EBQF = area of QHPC

Now, Area of GHPD = Area of (GQCD + QHPC)

= Area of (GQCD + EBQF)

= Area of (ABCD – AEFG)

or, (a +b) (a – b) = a² – b² (Proved).

Solved Examples for Class 7 Algebra Chapter 5

To express the product of two terms as the difference of two perfect squares,

⇒ \(a^2+2 a b+b^2=(a+b)^2\)

⇒ \(a^2-2 a b+b^2=(a-b)^2\)

By subtraction, \(4 a b=(a+b)^2-(a-b)^2\)

⇒ \(a b=\frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}\)

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

Algebra Chapter 5 formulae And Their Applications Exercise 5 Some Examples On (a ± b)² = a² ± 2ab + b².

Example 1. Find the square of (x + 2y).

Solution:

Given:

(x + 2y)

Square of (x + 2y)

= \((x+2 y)^2\)

= \((x)^2+2 \cdot x \cdot 2 y+(2 y)^2=x^2+4 x y+4 y^2\)

∴ \(x^2+4 x y+4 y^2\).

The square of (x + 2y) =  \(x^2+4 x y+4 y^2\).

Example 2. Find the square of 3ab + 2bc.

Solution:

Given:

3ab + 2bc

Square of (3ab + 2bc) = \((3 a b+2 b c)^2\)

⇒ \((3 a b)^2+2 \cdot 3 a b \cdot 2 b c+(2 b c)^2=9 a^2 b^2+12 a b^2 c+4 b^2 c^2 9 a^2 b^2+12 a \cdot b^2 e+4 b^2 c^2\).

The square of 3ab + 2bc = \((3 a b)^2+2 \cdot 3 a b \cdot 2 b c+(2 b c)^2=9 a^2 b^2+12 a b^2 c+4 b^2 c^2 9 a^2 b^2+12 a \cdot b^2 e+4 b^2 c^2\).

Example 3. Find the square of 101.

Solution:

Given:

101

Square of 101

= \((101)^2=(100+1)^2\)

= \((100)^2+2 \times 100 \times 1+(1)^2=10000+200+1=10201\)

∴ 10201.

The square of 101 is 10201.

Class 7 Maths Formulae Exercise Solutions

Example 4. Find the square of \(2 x^2-3 y^2\).

Solution:

Given:

⇒ \(2 x^2-3 y^2\)

Square of \(\left(2 x^2-3 y^2\right)\)

= \(\left(2 x^2-3 y^2\right)^2\)

= \(\left(2 x^2\right)^2-2 x 2 x^2 x 3 y^2+\left(3 y^2\right)^2=4 x^4-12 x^2 y^2+9 y^4\)

∴ \(4 x^4-12 x^2 y^2+9 y^4\)

The square of \(2 x^2-3 y^2\) = \(4 x^4-12 x^2 y^2+9 y^4\)

Example 5. Find the square of 498.

Solution:

Given: 498

Square of 498

= \((498)^2=(500-2)^2=(500)^2-2 \times 500 \times 2+(2)^2\)

= 250000 – 2000 + 4 = 250004 – 2000 =248004

∴ 248004.

The square of 498 is 248004.

Class Vii Math Solution Wbbse

Example 6. Simplify : \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\). 

Solution :

Given:

⇒ \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\)

Let 5a + 3b = x and 5a – 3b = y

Hence, the given expression = \(x^2+2 x y+y^2=(x+y)^2\)

= \((5 a+3 b+5 a-3 b)^2=(10 a)^2=100 a^2\)

∴ \(100 a^2\).

⇒ \((5 a+3 b)^2+2 \cdot(5 a+3 b)(5 a-3 b)+(5 a-3 b)^2\) = \(100 a^2\).

Example 7. Simplify : \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)(5x – 7y)² + 2(5x – 7y)(9y – 4x) + (9y – 4x)².

Solution:

Given:

⇒ \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)

Let, 5x -7y- a and 9y – 4x = 6

Then the given expression = \(a^2+2 a b+b^2=(a+b)^2\)

= \((5 x-7 y+9 y-4 x)^2=(x+2 y)^2=(x)^2+2 \cdot x \cdot 2 y+(2 y)^2\)

= \(x^2+4 x y+4 y^2\)

∴ \(x^2+4 x y+4 y^2\).

∴ \((5 x-7 y)^2+2(5 x-7 y)(9 y-4 x)+(9 y-4 x)^2\)(5x – 7y)² + 2(5x – 7y)(9y – 4x) + (9y – 4x)² = \(x^2+4 x y+4 y^2\).

Example 8. Simplify \((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\).

Solution:

Given:

\((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\)

Let, 3a + 4b = x and a + 4b = y

Then the given expression = \(x^2-2 x y+y^2\)

= \((x-y)^2\)

= \(\{(3 a+4 b)-(a+4 b)\}^2\)

= \((3 a+4 b-a-4 b)^2=(2 a)^2=4 a^2\)

∴ \(4 a^2\)4a²

\((3 a+4 b)^2-2(3 a+46)(a+4 b)+(a+4 b)^2\) = \(4 a^2\)4a²

Example 9. Simplify \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\).

Solution:

Given:

⇒ \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\)

Let 7x + 4y = a and 7x – 4y = b

Hence, the given expression = \(a^2-2 a b+b^2=(a-b)^2\)

= \(\{(7 x+4 y)-(7 x-4 y))^2\)

= \(\{7 x+4 y-7 x+4 y)^2\)

= \((8 y)^2=64 y^2\)

∴ \(64 y^2\)

⇒ \((7 x+4 y)^2-2 \cdot(7 x+4 y) x(7 x-4 y)+(7 x-A y)^2\) = \(64 y^2\)

West Bengal Board Class 7 Algebra Assistance

Example 10.  If x =-l then find the value of \(81 x^2-18 x+1\).

Solution:

Given:

x =-1 And \(81 x^2-18 x+1\)

⇒ \(81 x^2-18 x+1=(9 x)^2-2 \times 9 x \times 1+(1)^2\)81x² – 18x + 1 = (9x)² – 2 X 9x X 1 + (1)²

= \((9 x-1)^2=\{9(-1)-1\}^2=(-9-1)^2=(-10)^2=100\)

∴ 100.

Example 11. If p = 2, q = -1 and r = 3 then find the value of \(p^2 q^2+10 p q r+25 r^2\).

Solution:

Given:

p = 2, q = -1 and r = 3 And \(p^2 q^2+10 p q r+25 r^2\)

\(p^2 q^2+10 p q r+25 r^2\)

= \((\mathrm{pq})^2+2 \times \mathrm{pq} \times 5 r+(5 \mathrm{r})^2\)

=\((\mathrm{pq}+5 \mathrm{r})^2=\{2(-1)+5 \times 3\}^2=(-2+15)^2=(13)^2=169\)

∴ 169.

Example 12. If a = 1, b = 2 and c = -1 then find the value of \(4 a^2 b^2-20 a b c+25 c^2\).

Solution:

Given: 

a = 1, b = 2 and c = -1 And \(4 a^2 b^2-20 a b c+25 c^2\)

⇒ \(4 a^2 b^2-20 a b c+25 c^2\)

= \((2 a b)^2-2 \times 2 a b \times 5 c+(5 c)^2\)

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

= \(\{4+5\}^2=(9)^2=81\)

∴ 81.

∴ \(4 a^2 b^2-20 a b c+25 c^2\) = 81

Example 13. Find the product 201 x 201 with the help of the formula.

Solution:

⇒ \(201 \times 201=(201)^2=(200+1)^2\)

= \((200)^2+2 \times 200 \times 1+(1)^2\) = 40000 + 400 + 1 =40401

∴ 40401.

201 x 201 = 40401.

Example 14. Simplify 0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18.

Solution:

Let a = 0.82 and b =0.18

Then the given expression =a x a+2 x a x b+b x 6

= \(a^2+2 a b+b^2=(a+b)^2=(0.82+0.18)^2\)

= \((1.00)^2=(1)^2=1\)

∴ 1.

0.82 x 0.82 + 2 x 0.82 x 0.18 + 0.18 x 0.18 = 1.

WBBSE Class 7 Chapter 5 Formula Guide

Example 15. What is to be added with x² + 4x +\(\frac{3}{2}\) so that the sum is a perfect square?

Solution:

x² + 4x +\(\frac{3}{2}\)

= (x)² + 2.x.2 + (2)²  – (2)² +\(\frac{3}{2}\)

= (x + 2)² – 4 + \(\frac{3}{2}\)

= (x + 2)²– \(\frac{5}{2}\)

Hence, to make it a perfect square \(\frac{3}{2}\) is to be added.

Example 16. If x² – 2x + 1 = 0, then find the values of  \(x+\frac{1}{x} \quad x^{10}+\frac{1}{x^{10}} \quad x^n+\frac{1}{x^n}\)

Solution: \(x^2-2 x+1=0\)

or, \((x-1)^2=0\)

or, x – 1 = 0

∴ x = 1

⇒ \(x+\frac{1}{x}=1+\frac{1}{1}=1+1=2\)

⇒ \(x^{10}+\frac{1}{x^{10}}=(1)^{10}+\frac{1}{(1)^{10}}=1+1=2\)

⇒ \(x^n+\frac{1}{x^n}=(1)^n+\frac{1}{(1)^n}=1+1=2\)

∴ 1. 2, 2. 2, 3. 2.

⇒ \(x+\frac{1}{x} \quad x^{10}+\frac{1}{x^{10}} \quad x^n+\frac{1}{x^n}\) = 1. 2, 2. 2, 3. 2.

Chapter 5 formulae And Their Applications Exercise 5 Some Problems On Various Formulae Derived From (a ± b)² = a² ± 2ab + b².

Example 1. Find the square of x + 2y – 3z.

Solution:

Given:

x + 2y – 3z

Square of x + 2y – 3z.

= \((x+2 y-3 z)^2\)

= \((x+2 y)^2-2 \cdot(x+2 y) \cdot 3 z+(3 z)^2\)

= \((x)^2+2 \cdot x \cdot 2 y+(2 y)^2-6 z(x+2 y)+9 z^2\)

= \(x^2+4 x y+4 y^2-6 z x-12 y z+9 z^2\)

= \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

∴ \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

The square of x + 2y – 3z = \(x^2+4 y^2+9 z^2+4 x y-12 y z-6 z x\)

Applications of Algebraic Formulae for Class 7

Example 2. If a + 6 = 5 and ab = 5, find the value of a² + b².

Solution :

Given:

a + 6 = 5 and ab = 5

⇒ \(a^2+b^2=(a+b)^2-2 a b\)

= (5)²– 2.5 = 25- 10= 15

∴ 15.

∴ a² + b²= 15.

Example 3. If a – 6 = 5 and ab = 14, find the value of a² + b².

Solution:

Given:

a – 6 = 5 and ab = 14

⇒ \(a^2+b^2=(a-b)^2+2 a b=(5)^2+2.14=25+28=53\)

∴ 53.

a² + b² = 53.

Example 4. If x + y =2 and x-y = 1, then find the value of \(8 x y\left(x^2+y^2\right)\).

Solution:

Given:

x + y =2 and x-y = 1

⇒ \(8 x y\left(x^2+y^2\right)\)

= \(4 x y \cdot 2\left(x^2+y^2\right)\)

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

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

= (4-1) (4+1) = 3 x 5 = 15

∴15.

∴ \(8 x y\left(x^2+y^2\right)\) = 15.

Wbbse Class 7 Maths Solutions

Example 5. If 2x + 3y = 9 and xy = 3, find the value of \(4 x^2+9 y^2\).

Solution:

Given:

2x + 3y = 9 and xy = 3

⇒ \(4 x^2+9 y^2\).

= \((2 \cdot x)^2+(3 y)^2=(2 x+3 y)^2-2 \cdot 2 x \cdot 3 y\)

=\((2 x+3 y)^2-12 x y=(9)^2-12 \cdot 3=81-36=45\)

∴ 45.

∴ \(4 x^2+9 y^2\)= 45.

Example 6. Express 35 as the difference of two squares.

Solution:

35 = 7 x 5

⇒ \(\left(\frac{7+5}{2}\right)^2-\left(\frac{7-5}{2}\right)^2\)

⇒ \(\left(\frac{12}{2}\right)^2-\left(\frac{2}{2}\right)^2\)

= \((6)^2-(1)^2\)

∴ \(35=(6)^2-(1)^2\).

Example 7. Express 4xy as the difference of two perfect squares.

Solution:

4xy = 2.x.2.y

⇒ \(\left(\frac{2 x+2 y}{2}\right)^2-\left(\frac{2 x-2 y}{2}\right)^2\)

⇒ \(\left\{\frac{2(x+y)}{2}\right\}^2-\left\{\frac{2(x-y)}{2}\right\}^2\)

= \((x+y)^2-(x-y)^2\)

∴ \(4 x y=(x+y)^2-(x-y)^2\).

Example 8. Express \(2\left(a^2+b^2\right)\) as the sum of two perfect squares.

Solution:

⇒ \(2\left(a^2+b^2\right)\)

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

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

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

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

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

Wbbse Class 7 Maths Solutions

Example 9. Express 4a² + 1/4a² +1 as the difference of two perfect squares.

Solution:

⇒ \(4 a^2+\frac{1}{4 a^2}+1\)

= \((2 a)^2+2 \cdot 2 a \cdot \frac{1}{2 a}+\left(\frac{1}{2 a}\right)^2+1-2\)

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

Example 10. If x = a + \(\frac{1}{a}\) and y = a – \(\frac{1}{a}\) find the value  of \(x^4+y^4-2 x^2 y^2\).

Solution:

⇒ \(x^4+y^4-2 x^2 y^2\)

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

= \(\{(x+y)(x-y)\}^2=(x+y)^2(x-y)^2\)

⇒ \(\left\{\left(a+\frac{1}{a}\right)+\left(a-\frac{1}{a}\right)\right\}^2\left\{\left(a+\frac{1}{a}\right)-\left(a-\frac{1}{a}\right)\right\}^2\)

⇒ \(\left\{a+\frac{1}{a}+a-\frac{1}{a}\right\}^2\left\{a+\frac{1}{a}-a+\frac{1}{a}\right\}^2\)

⇒ \((2 a)^2 \cdot\left(\frac{2}{a}\right)^2=4 a^2 \cdot \frac{4}{a^2}=16\)

∴ 16

Example 11. If x + \(\frac{1}{4}\) = 4, then show that, x² + 1/x² = 14

Solution:

L.H.S. = \(x^2+\frac{1}{x^2}\)

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

= (4)² – 2 = 16 – 2 = 14 = R.H.S. (Proved).

Class 7 Maths Exercise 5 Solved Examples

Example 12. If 6x² – 1 = 4x, then show that, 36x² +1/x² = 28.

Solution:

6x² -1 = 4x

⇒ \((6 x)^2-2.6 x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^2=(4)^2\)

⇒ \(36 x^2-12+\frac{1}{x^2}=16\)

⇒ \(36 x^2+\frac{1}{x^2}=16+12\)

∴ \(36 x^2+\frac{1}{x^2}=28\)  (Proved).

Example 13. If \(\frac{p}{6}\) + \(\frac{1}{p}\) = \(\frac{7}{6}\) then prove that \(p^2+36 / p^2=37\)

Solution:

⇒ \(\frac{p}{6}\) or, p + \(\frac{6}{p}\) = 7

[Multiplying both sides by 6]

Squaring both sides,

⇒ \((p)^2+2 \cdot p \cdot \frac{6}{p}+\left(\frac{6}{p}\right)^2=(7)^2\)

or, \(p^2+12+\frac{36}{p^2}=49\)

⇒ \(p^2+\frac{36}{p^2}=49-12\)

∴ \(p^2+\frac{36}{p^2}=37\) (Proved).

Example 14. If a + b = 9 and a – b = 5 then what is the value of a+b²/2ab?

Solution:

⇒\(\frac{a^2+b^2}{2 a b}=\frac{2\left(a^2+b^2\right)}{4 a b}\)

= \(\frac{(a+b)^2+(a-b)^2}{(a+b)^2-(a-b)^2}\)

= \(\frac{(9)^2+(5)^2}{(9)^2-(5)^2}=\frac{81+25}{81-25}=\frac{106}{56}=\frac{53}{28}=1 \frac{25}{28}\)

Example 15. If x² + y² = 4xy then prove that at⁴ + y⁴ = 14x²y².

Solution:

⇒ \(x^2+y^2=4 x y\)

or, \(\frac{x^2}{x y}+\frac{y^2}{x y}=4 \text { or, } \frac{x}{y}+\frac{y}{x}=4\)

Squaring both sides we get,

⇒ \(\frac{x^2}{y^2}+2 \cdot \frac{x}{y} \cdot \frac{y}{x}+\frac{y^2}{x^2}=16\)

or, \(\frac{x^2}{y^2}+2+\frac{y^2}{x^2}=16\)

or, \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=14\)

or, \(\frac{x^4+y^4}{x^2 y^2}=14\)

or, \(x^4+y^4=14 x^2 y^2\)

Example 16. If 2a + \(\frac{1}{3a}\) = \(\frac{2}{3}\)

Solution:

2a+ \(\frac{1}{3a}\) = \(\frac{2}{3}\)

or, 3a + = \(\frac{1}{2a}\) = 1 [Multiplying both sides by \(\frac{3}{2}\)]

Squaring both sides,

⇒ \((3 a)^2+2 \cdot 3 a \cdot \frac{1}{2 a}+\left(\frac{1}{2 a}\right)^2=1\)

⇒ \(9 a^2+3+\frac{1}{4 a^2}=1 \text { or, } 9 a^2+\frac{1}{4 a^2}=1-3\)

∴ \(9 a^2+\frac{1}{4 a^2}=-2\)   (Proved).

Example 17. If a² + b² = 5ab then show that, a²/b² + b²/a²= 23.

Solution:

Given:

⇒ \(a^2+b^2=5 a b\)

or, \(\frac{a^2+b^2}{a b} \text { or, } \frac{a}{b}+\frac{b}{a}=5\)

Squaring both sides,

⇒ \(\left(\frac{a}{b}\right)^2+2 \cdot \frac{a}{b} \cdot \frac{b}{a}+\left(\frac{b}{a}\right)^2=(5)^2\)

or, \(\frac{a^2}{b^2}+2+\frac{b^2}{a^2}=25\)

or, \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=25-2\)

or, \(\frac{a^2}{b^2}+\frac{b^2}{a^2}=23\) (Proved).

Example 18. Find the continued product: \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\).

Solution:

The given expression = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^4-b^4\right)\left(a^4+b^4\right)\left(a^8+b^8\right)\)

= \(\left(a^8-b^8\right)\left(a^8+b^8\right)\)

∴ \(a^{16}-b^{16}\)

Example 19. Multiply with the help of the formula : \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\).

Solution:

Given:

⇒ \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\)

⇒ \(\left(a^2+a+1\right)\left(a^2-a+1\right)\left(a^4-a^2+1\right)\)

= \(\left\{\left(a^2+1\right)+a\right\}\left\{\left(a^2+1\right)-a\right\}\left\{\left(a^4-a^2+1\right)\right\}\)

= \(\left\{\left(a^2+1\right)^2-(a)^2\right\}\left(a^4-a^2+1\right)\)

= \(\left(a^4+2 a^2+1-a^2\right)\left(a^4-a^2+1\right)=\left(a^4+a^2+1\right)\left(a^4-a^2+1\right)\)

= \(\left\{\left(a^4+1\right)+a^2\right\}\left\{\left(a^4+1\right)-a^2\right\}\)

= \(\left(a^4+1\right)^2\left(a^2\right)^2\)

= \(a^8+2 a^4+1-a^4=a^8+a^4+1\)

∴ \(a^8+a^4+1\).

Example 20. If x + y = 5 and x – y = 1, then find the value of \(8 x y\left(x^2+y^2\right)\)

Solution:

Given:

x + y = 5 and x – y = 1

⇒ \(8 x y\left(x^2+y^2\right)\)

= \(4 x y \cdot 2\left(x^2+y^2\right)\)4xy.2(x²+y²)

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

= \(\left\{(5)^2-(1)^2\right\}\left\{(5)^2+(1)^2\right\}\)

= (25 – 1) (25 + 1) = 24 x 26 = 624

∴ 624.

∴ \(8 x y\left(x^2+y^2\right)\) = 624

Example 21. If x= \(\frac{a}{b}\) – \(\frac{b}{a}\) and y= \(\frac{a}{b}\) – \(\frac{b}{a}\) then show that, \(x^4+y^4-2 x^2 y^2=16\).

Solution:

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

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

= \(\left\{\left(2 \cdot \frac{a}{b}\right)\left(2 \cdot \frac{b}{a}\right)\right\}^2=\left(2 \cdot \frac{a}{b} \cdot 2 \cdot \frac{b}{a} \cdot\right)^2\)

= \((4)^2\) = 16 = R.H.S.

Example 22. If m – \(\frac{1}{m-3}\) = 5, then what is the value of (m – 3)² + 1/(m-3)²?

Solution:

m – \(\frac{1}{m-3}\) = 5

⇒ \((m-3)^2+\left(\frac{1}{m-3}\right)^2-2 \cdot(m-3) \frac{1}{(m-3)}=4\)

or, \((m-3)^2+\frac{1}{(m-3)^2}-2=4\)

or, \((m-3)^2+\frac{1}{(m-3)^2}=6\)

∴ (m – 3)² + 1/(m-3)² = \((m-3)^2+\frac{1}{(m-3)^2}=6\)

Algebra Chapter 5 formulae And Their Applications Exercise 5 Some Examples On a² – b² = (a + b)(a – b).

Example 1. Find the product of (2a – 5b) and (2a + 5b).

Solution:

The required product = (2a – 5b) (2a + 5b)

= (2a)² – (5b)²

= 4a² – 25b² 

(2a – 5b) and (2a + 5b) = 4a² – 25b² 

Example 2.  Find the product of (1 + 5pq) and (1 – 5pq).

Solution:

The required product = (1 + 5pq) (1 – 5pq)

= (1)²-(5pq)² =1

∴ 1 – 25p²q².

(1 + 5pq) X (1 – 5pq) = 1 – 25p²q².

Example 3. Find the value of 255² – 254².

Solution:

255² – 254² = (255 + 254)(255 – 254)

= 509 x 1 = 509

∴ 509.

255² – 254² = 509.

Example 4. Find the continued product of \((a+b),(a-b),\left(a^2+b^2\right),\left(a^4+b^4\right)\).

Solution:

The required product = \((a+b)(a-b)\left(a^2+b^2\right)\left(a^4+b^4\right)\)

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

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

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

∴ \(a^8-b^8\).

∴ \((a+b),(a-b),\left(a^2+b^2\right),\left(a^4+b^4\right)\) = \(a^8-b^8\).

Math Solution Of Class 7 Wbbse

Example 5. Find the product of (x -y + a) and (x +y-z).

Solution:

The required product = (x-y + z) (x+y-z)

= {x- (y – z)} {x + (y – z)}

= \((x)^2-(y-z)^2=x^2-\left(y^2-2 y z+z^2\right)\)

= \(x^2-y^2-z^2+2 y z\)

∴ \(x^2-y^2-z^2+2 y z\).

⇒ (x -y + a) a X (x +y-z) = \(x^2-y^2-z^2+2 y z\).

Example 6. Find the product of (x + 2y + 3z) and (x + 2y – 3z).

Solution:

The required product = (x + 2y + 3 z) (x + 2y – 3 z)

= {(x + 2y) + 3z} {(x + 2y) – 3z}

= \((x+2 y)^2-(3 z)^2=(x)^2+2 \cdot x \cdot 2 y+(2 y)^2-(3 z)^2\)

= \(x^2+4 x y+4 y^2-9 z^2\)

= \(x^2+4 y^2-9 z^2+4 x y\)

∴ \(x^2+4 y^2-9 z^2+4 x y\).

⇒ (x + 2y + 3z) X (x + 2y – 3z) = \(x^2+4 y^2-9 z^2+4 x y\).

Step-by-Step Solutions for Class 7 Algebra Problems

Example 7. Multiply with the help of the formula 1001 x 999.

Solution:

1001 x 999 = (1000 + 1) X (1000 – 1)

= (1000)² – (1)² = 1000000 – 1 = 999999

∴ 999999.

1001 x 999 = 999999.

Math Solution Of Class 7 Wbbse

Example 8. Simplify with the help of the formula : \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\).

Solution:

⇒ \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\)

=\({(2 x+3 y+4 z)+(3 x-2 y+4 z)} x{(2 x+3 y+4 z)-(3 x-2 y+4 z)}\)

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

= (5x + y + 8z) (- x + 5y)

= \(-5 x^2-x y-8 z x+25 x y+5 y^2+40 y z\)

= \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\)

∴ \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\).

∴ \((2 x+3 y+4 z)^2-(3 x-2 y+4 z)^2\) = \(-5 x^2+5 y^2+24 x y+40 y z-8 z x\).

Example 9. Express as the product of two expressions  \(a^2-4 a b+4 b^2-4\)

Solution:

⇒ \(a^2-4 a b+4 b^2-4\)a² – 4ab + 4b² – 4

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

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

∴ \(a^2-4 a b+4 b^2-4\) = \((a-2 b)^2-(2)^2=(a-2 b+2)(a-2 b-2)\).