Chapter 20.2 - Checking whether Polynomial or not

In the previous section we saw the conditions for a polynomial. In this section we will see a few expression in which we have to apply those conditions in order to decide whether they are polynomials or not.

Solved example 20.3
In a rectangle, the length of the longer side is 1 cm greater than the shorter side. Write the expression for calculating the length of diagonal of all such triangles. Check whether it is a polynomial or not.
Solution:
The given rectangle has length 1 cm greater than the width. We have to find an expression to calculate it's diagonal. Once we find that expression, we will be able to quickly calculate the diagonal of all such rectangles.
Fig.20.5(a) below shows such a rectangle

Fig.20.5

1. Let width = x cm. Then length = (x+1) cm
2. The diagonal is the hypotenuse of a right triangle with legs x and (x+1) cm
Applying Pythagoras theorem, we get: Diagonal = √[(x+1)² + x²]
⇒ Diagonal = √[x²+2x+1+x²] = √[2x²+2x+1]
3. Using this expression, we can calculate the diagonal of any rectangle whose length is 1 unit greater than the width. 
4. But it is not a polynomial. Because, the relation between the terms is not just addition. Square root is also coming.

Solved example 20.4
In fig.20.5(b), ABC is an isosceles triangle. It is right angled at C. A rectangle of 1 cm width is attached to the base of ΔABC. Write an expression for the total area of the triangle and the rectangle? Check whether it is a polynomial or not.
Solution:
1. ΔABC has two properties: 
• It is right angled at C
• It is isosceles
2. In an isosceles triangle, two sides will be equal. 
• Since it is also right angled, the two equal sides are the 'legs'. (∵ there can be only one hypotenuse, which is the longest side)
• The side AB opposite the 90^o vertex C will be the hypotenuse. So the legs are CA and CB.
• These legs are equal. Let their lengths be 'x'.
■ Knowing the value of x, the ABC can be easily constructed. And then a 1 cm wide rectangle can be attached to it's base. But in this problem, we do not have to make an actual construction. We can derive the expression using a rough sketch.
3. Applying Pythagoras theorem to ⊿ABC we get hypotenuse AB = √[x²+x²] = √[2x²] = x√2
4. We want the area of ⊿ABC
• Area = ¹⁄₂ × base × height 
• Base is the hypotenuse AB.
• Now we want the height. It is the length of the perpendicular CF dropped from C to AB. This is shown in fig.20.6 below:

Fig.20.6

• Perpendicular CF will split ⊿ABC into two right triangles: ⊿ACF and ⊿BCF
• Consider any one. Let us take ⊿BCF. It's base BF = ¹⁄₂ × x√2 = (^x√2⁄₂)
• Applying Pythagoras theorem, we get: CF² = BC² - BF².
• So CF = √[x² - (^x√2⁄₂)²] = √[x² - (^2x2⁄₄)] 
= √[^{(4x2 - 2x2)}⁄₄] = √[^(2x2)⁄₄] = √[^x2⁄₂] =  ^x⁄_√2.
•Area = ¹⁄₂ × base × height = ¹⁄₂ × AB × CF = ¹⁄₂ × x√2 × ^x⁄_√2 = ^x2⁄₂.
5. Now we want the area of the rectangle ADEB
Area = lb = AB × BE = x√2 × 1 = x√2
6. So total area = ^x2⁄₂ + x√2
7. Let us check the conditions for a polynomial:
• There is only addition between terms
• The exponents in all terms are whole numbers
• So it is a polynomial. We can write: a(x) = ^x2⁄₂ + x√2

Solved example 20.5
Area of a rectangle is 25 cm. It's length is x cm. Write an expression for the perimeter of the rectangle. Check whether it is a polynomial or not.
Solution:

1. For perimeter, we need both length and width. Here, we are given only the length. But, since area is given, we can write width in terms of length and area. That is., width = area / length = 25/x

2. So perimeter = 2(l+b) = 2(x + ²⁵⁄_x) = 2x + ⁵⁰⁄_x

This can be written as: 2x + 50x^-1. In this expression, the exponent of x is -1. It is not a whole number. So 2x + ⁵⁰⁄_x is not a polynomial.

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So now we know how to decide whether a given algebraic expression is a polynomial or not. We will see a few more of their properties:
■ We have seen that a polynomial will consist of ‘terms’. Consider the term with the largest exponent. This term has a special name. It is called the leading term. The power of the leading term is called the degree of the whole polynomial. Let us see some examples:
• a(x) = x² + x. In this polynomial, the leading term is x². The power of this term is 2.
So this polynomial is of degree 2.
• p(x) = 4x+2. In this polynomial, the leading term is 4x. The power of this term is 1.
So this polynomial is of degree 1.
• p(y) = y⁴+y²-1. In this polynomial, the leading term is y⁴. The power of this term is 4.
So this polynomial is of degree 4.

■ Instead of saying ‘polynomial of degree 1’, we say: ‘First degree polynomial’

■ Instead of saying ‘polynomial of degree 2’, we say: ‘Second degree polynomial’

■ Instead of saying ‘polynomial of degree 3’, we say: ‘Third degree polynomial’

so on . . .

■ Based on the degree, we can write the general form of all polynomials:

• First degree polynomial: ax + b
    ♦ This is also called a linear polynomial

• Second degree polynomial: ax² + bx + c
    ♦ This is also called a quadratic polynomial

• Third degree polynomial: ax³ + bx² + cx + d
    ♦ This is also called a cubic polynomial

so on . . .

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Do we see any peculiarity in the above general forms?
Yes. we see the following peculiarity:
First degree polynomial has two terms in it's general form
Second degree polynomial has three terms in it's general form
Third degree polynomial has four terms in it's general form

Will we ever meet a first degree polynomial with more than two terms?
The answer is 'No'. Let us analyse:
■ A first degree polynomial has the general form: ax + b
1. If there are any more terms, they must all belong to either one of the two categories below:
• Terms with exponent 1
• Terms with exponent less than 1
(Note that terms with exponent greater than 1 are not accepted as first degree polynomials)
2. In the above two categories,
• All terms with exponent 1 will combine together to give the general form 'ax'
• All terms with exponent less than 1 will combine together to give the general form 'b'. [∵ 'less than 1' is zero. And x⁰ is 1. So (b × x⁰) = (b×1) = b]
3. So the maximum number of terms possible in a first degree polynomial is 2

■ A second degree polynomial has the general form: ax² + bx + c
1. If there are any more terms, they must all belong to either one of the three categories below:
• Terms with exponent 2
• Terms with exponent 1
• Terms with exponent less than 1
(Note that terms with exponent greater than 2 are not accepted as second degree polynomials)
2. In the above three categories,
• All terms with exponent 2 will combine together to give the general form 'ax²'
• All terms with exponent 1 will combine together to give the general form 'bx'
• All terms with exponent less than 1 will combine together to give the general form 'c'. [∵ 'less than 1' is zero. And x⁰ is 1. So (c × x⁰) = (c×1) = c]
3. So the maximum number of terms possible in a second degree polynomial is 3

We can continue like this and write the following:
• The maximum number of terms possible in a first degree polynomial is 2
• The maximum number of terms possible in a second degree polynomial is 3
• The maximum number of terms possible in a third degree polynomial is 4
So on . . . 

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We will now see a few more solved examples

Solved example 20.6

Write each of the relations below in algebra, and see if it gives a polynomial. Give reasons for your answer.

(i). A 1 m wide path goes all around a square ground. The relation between the following two:

• Length of the side of the ground

• The area of the path

(ii). A liquid contains 7 litres of water and 8 litres of acid. More acid is added to it. The relation between the following two:

• Amount of acid added 

• The change in percentage of acid in the liquid.

(iii). Two poles of heights 3 m and 4 m are erected upright on the ground. The distance between the poles is 5 m. A rope is to be stretched from the top of one pole to some point on the ground, and from there, to the top of the other pole. The relation between the following two:

• The distance of the point on the ground from the foot of any pole

• Total length of the rope

Solution:

(i) A rough sketch is shown in the fig.20.7(a) below:

Fig.20.7

1. Let the side of the square ground = x m

2. So side of the outer square = (x+2) m

3. Area of the outer square = (x+2)² m²

4. Area of the inner square = x² m² 

5. So area of the path alone = Area of outer square – Area of inner square 

= (x+2)² - x² 

= x² + 4x + 4 - x² 

= 4x+4

6. The above expression gives the relation between the side of the ground (x), and the area of the path. We can write: a(x) = 2x+4

7. It is a polynomial. Because (Details here):

• It consists of terms which are added

• Power of all terms are whole numbers

(ii) 1. Amount of water = 7 litres

• Amount of acid = 8 litres

• Total amount = 7 + 8 = 15 litres

2. Percentage of acid in the above liquid = (⁸⁄₁₅) × 100

3. Let the amount of acid newly added = x litres

• Amount of water after adding additional acid = 7 litres

• Amount of acid after adding additional acid = (8+x)

• Total amount after adding additional acid = 7 + (8+x) = (15+x)

4. Percentage of acid after adding additional acid = [^(8+x)⁄_(15+x)]×100

5. We can write: When x litres of acid is added to a liquid containing 7 litres of water and 8 litres of acid, the 'change in the percentage of acid' is given by = [^(7x)⁄_15(15+x)]×100
■ Using this expression, we can easily calculate such a 'percentage change'. But the expression is not a polynomial. Because an x term comes in the denominator.

(iii) • A rough sketch is shown in fig.20.7(b). AB and CD are the poles. They are erected upright. So they are perpendicular to the ground BC. The distance BC is 5 m. E is the point on the ground.

• If we take BE as x, CE will be (5-x) m. We have to find the length of the rope which is equal to (AE + ED)
1. ABE and DCE are right triangles. Applying Pythagoras theorem to ⊿ABE, we get:
AE = √[3²+x²] 
2. Applying Pythagoras theorem to ⊿DCE, we get:
ED = √[(5-x)²+4²] = √[25 -10x +x² +16] = √[41 -10x + x²]
3. So the length of rope = √[3²+x²] + √[41 -10x + x²]
4. This is not a polynomial because square root of terms are taken

Solved example 20.7

Write each of the operations below as an algebraic expression. Find out which are polynomials and explain why

(i) Sum of a number and it’s reciprocal

(ii) Sum of a number and it’s square root

(iii) Product of the following two:

• Sum of a number and it’s square root

• Difference of a number and it’s square root

Solution:

(i) 1. Let the number = x. Then it's reciprocal = ¹⁄_x.

2. Then sum = (x + ¹⁄_x) = (x + x^-1)

3. This is not a polynomial because The power of one term is -1, which is not a whole number.

(ii) 1. Let the number = x. Then it's square root = √x

2. Then sum = (x + √x) = (x + x^1/2)

3. This is not a polynomial because The power of one term is ¹⁄₂, which is not a whole number.

(iii) 1. Let the number = x. Then it's square root = √x

2. Then sum = (x + √x) 

3. Difference = (x - √x)

4. Product of the above two: (x + √x)(x - √x) = x² - x [∵ (a+b)(a-b) = a² - b²]

5. It is a polynomial. Because (Details here):

• It consists of terms, in which one term is subtracted from the other.

• Power of all terms are whole numbers

Solved example 20.8

Find polynomials p(x) satifying each of the set of conditions given below:

(i) First degree polynomial with p(1) = 1 and p(2) = 3

(ii) First degree polynomial with p(1) = -1 and p(-2) = 3

(iii) Second degree polynomial with p(0) = 0, p(1) = 2 and p(2) = 6

(iv) Three different second degree polynomials with p(0) = 0 and p(1) = 2
Solution:
(i) 1. General form of a first degree polynomial is: p(x) = ax + b 
2. Given that p(1) = 1. So we get:
p(1) = (a×1) + b = 1 ⇒ a+b = 1
3. Given that p(2) = 3. So we get:
p(2) = (a×2) + b = 3 ⇒ 2a+b = 3
4. From (2) we get: b = (1-a). Substituting this value of a in (3) we get: 2a+(1-a) =3
⇒ a+1 = 3. So we get a = 3-1 = 2
Substituting this value of a in (2) we get: 2+b = 1 ⇒ b = 1-2 = -1
5. So the first degree polynomial p(x) satisfying  p(1) = 1 and p(2) = 3 is: p(x) = 2x - 1

(ii) 1. General form of a first degree polynomial is: p(x) = ax + b 
2. Given that p(1) = -1. So we get:
p(1) = (a×1) + b = -1 ⇒ a+b = -1
3. Given that p(-2) = 3. So we get:
p(-2) = (a×-2) + b = 3 ⇒ -2a+b = 3
4. From (2) we get: b = (-1-a). Substituting this value of a in (3) we get: -2a+(-1-a) =3
⇒ -1-3a = 3 same as 3a = -4. So we get a = -⁴⁄₃
Substituting this value of a in (2) we get: -⁴⁄₃+b = -1 ⇒ b = -1 + ⁴⁄₃ = ¹⁄₃
5. So the first degree polynomial p(x) satisfying  p(1) = -1 and p(-2) = 3 is: p(x) = (-⁴⁄₃)x + ¹⁄₃

(iii) 1. General form of a second degree polynomial is: p(x) = ax² + bx + c
2. Given that p(0) = 0. So we get: p(0) = a×0² + b×0 + c = 0
⇒ 0 + 0 + c = 0 ⇒ c = 0
3. Given that p(1) = 2. So we get:
p(1) = a×1² + b×1 + c = 2 ⇒ a+b+0 = 2 ⇒ a+b = 2
4. Given that p(2) = 6. So we get:
p(2) = a×2² + b×2 + c = 6 ⇒ 4a+2b+0 = 6 ⇒ 2a+b = 3
5. From (3) we get b = (2-a) 
Substituting this value of b in (4) we get: 2a+2-a = 3
a+2 = 3 ⇒ a = 1
Substituting this value of a in (3) we get: 1+b =2 ⇒ b = 1
6. So the second degree polynomial p(x) satisfying p(0) =0,  p(1) = 2,  and p(2) = 6 is:
p(x) = x² + x 

(iv) 1. General form of a second degree polynomial is: p(x) = ax² + bx + c
2. Given that p(0) = 0. So we get: p(0) = a×0² + b×0 + c = 0
⇒ 0 + 0 + c = 0 ⇒ c = 0
3. Given that p(1) = 2. So we get:
p(1) = a×1² + b×1 + c = 2 ⇒ a+b+0 = 2 ⇒ a+b = 2
4. There are two unknowns 'a' and 'b'. But there is only one equation: a+b =2
5. So it is not possible to find the exact value of either 'a' or 'b'. 
But there are many combinations which will satisfy a+b =2
Three possible combinations are:
• a = 1, b = 1. 
• a = ¹⁄₂, b = ³⁄₂
• a = 3, b = -1
So three polynomials that satisfy p(0) = 0 and p(1) = 2 are:
• p(x) = x² + x 
• p(x) = (¹⁄₂)x² +(³⁄₂) x 
• p(x) = 3x² - x 

Solved example 20.9
What is the value of b in p(x) = 3x² - bx +1 
Given that p(1) = 2
Solution:
Given that p(1) = 2. So we get: p(1) =  3×1² - b×1 + 1 = 2
⇒ 3 -b +1 = 2 ⇒ 4 -b = 2 ⇒ b = 2

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In the next section we will see polynomial operation.

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Chapter 20.1 - Conditions for a Polynomial

In the previous section we saw some examples for the formation of algebraic expressions. In this section we will see a few more examples.

Solved example 20.1
In rectangles with one side 1 cm shorter than the other, take the length of the shorter side as 'x' cm.
(i) Taking perimeters as p(x), write the relation between p(x) and (x) as an equation
(ii) Taking their areas as a(x), write the relation between a(x) and (x) as an equation
(iii) Calculate p(1), p(2), p(3), p(4) and p(5)
(iv) Calculate a(1), a(2), a(3), a(4) and a(5)
Solution:
1. Length of the shorter side (b) = x cm
2. So length of the longer side (l) = (x+1)
3. Perimeter (p) = 2(l+b) = 2(x+1+x) = 2(2x+1) = 4x+2
4. Area (a) = l×b = x(x+1) = x² + x 
5. So the algebraic expression showing the 'relation between perimeter and x' for the given rectangle is: p(x) = 4x+2. This is solution of part (i)
6. Also, the algebraic expression showing the 'relation between area and x' for the given rectangle is:
a(x) = x² + x. This is solution of part (ii) 
7. To find p(1), put x = 1 in p(x)
• From (5), we have p(x) = 4x+2. 
• So p(1) = 4×1 + 2 = 6
• Similarly, p(2) = 4×2 + 2 = 10
• p(3) = 4×3 + 2 = 14
• p(4) = 4×4 + 2 = 18
• p(5) = 4×5 + 2 = 22
This is solution of part (iii)
8. To find a(1), put x = 1 in a(x)
From (6), we have a(x) = x² + x
So a(1) = 1²+ 1 = 1 + 1 = 2
a(2) = 2²+ 2 = 4 + 2 = 6
a(3) = 3²+ 3 = 9 + 3 = 12
a(4) = 4²+ 4 = 16 + 4 = 20
a(5) = 5²+ 5 = 25 + 5 = 30
This is solution of part (iv)

Solved example 20.2
From the four corners of a rectangle 7 × 5 cm (fig.20.4.a), equal squares are cut off (fig.20.4.b), and the edges are folded up to make a box (fig.20.4.c).

Fig.20.4

(i) Taking sides of the squares as 'x' cm, write the dimensions of the box in terms of 'x'
(ii) Taking volume of the box as v(x) cm³, write the relation between v(x) and x as an equation
(iii) Calculate v(¹⁄₂), v(1) and v(1¹⁄₂)
Solution:
1. When squares of side x cm are cut off, 
• the remaining length becomes (7-2x) cm
• the remaining width becomes (5-2x) cm
• When these remaining edges are folded to make the box, the heigth of the box is x cm. So
    ♦ length of box (l) = (7-2x) cm
    ♦ width of box (b) = (5-2x) cm
    ♦ height of box (h) = x cm
This is solution of part (i)
2. Thus volume of box (v) = l×b×h = (7-2x)(5-2x)x 
= (35-10x-14x+4x²)x = (35-24x+4x²)x = 35x-24x²+4x³
3. So the algebraic expression showing the 'relation between volume and x' for the given box is:
v(x) = 35x-24x²+4x³. This is solution of part (ii)
4. For calculating v(¹⁄₂), it is more convenient to use the initial expression in (2)
• We have v(x) = (7-2x)(5-2x)x.
• Put x = ¹⁄₂ . We get: v(¹⁄₂) = [(7-(2×¹⁄₂)][(5-(2×¹⁄₂)]×¹⁄₂
⇒ v(¹⁄₂) = [(7-1)][(5-1)]×¹⁄₂ = 6 × 4 × ¹⁄₂ = 12
■ For calculating v(1), we can use the final expression in (2)
• We have v(x) = 35x-24x²+4x³
• Put x = 1. We get: v(1) = (35 × 1) - (24 × 1²) + (4 × 1³) 
⇒ v(1) = 35 - 24 + 4 = 15
■ For calculating v(1¹⁄₂), it is more convenient to use the initial expression in (2)
We have v(x) = (7-2x)(5-2x)x.
Put x = ³⁄₂ . We get: v(³⁄₂) =  [(7-(2׳⁄₂)][(5-(2׳⁄₂)]׳⁄₂.
⇒ v(³⁄₂) = [(7-3)][(5-3)]׳⁄₂ = 4 × 2 × ³⁄₂ = 12
This is solution of part (iii)

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So far we have seen a number of algebraic expressions.

Algebraic expressions like:

p(x) = 10 + 4x

a(x) = 6 + 5x + x²

v(x) = x³ + 6x² + 11x + 6

are called polynomials. But they must satisfy some conditions. Then only they will be called as polynomials. Let us analyse the various properties first:

• In p(x) = 10 + 4x, there are two components on the right side of the '=' sign

• Each component is called a term. So one term is '10', and the other term is '4x'

• Every term can be split into two parts: A coefficient and a variable part. In '4x', 4 is the coefficient, and x is the variable part.

• '10' does not seem to have a variable part. But in reality, it also has a variable part. This can be explained as follows:

    ♦ 10 can be written as 10 × 1 

    ♦ Now, this 1 is same as x⁰ 

    ♦ So 10 = 10x⁰

    ♦ Thus, in the term '10', 10 is the coefficient, and x⁰ is the variable part.

• Variables are usually denoted by letters x, y or z

• In a polynomial, the variables can take different values. But there are some terms that will not change values

• For example, in the polynomial 'p(x) = 10 + 4x' , what ever be the value of x, values of 10 and 4 does not change. Such values are called constants.

• Constants are usually denoted by letters a, b, c etc.,

■ Let us take the next polynomial:

• a(x) = 6 + 5x + x²

• Terms: 6, 5x and x²

• Splitting the terms: 

    ♦ 6: coefficient is 6, variable part is x⁰

    ♦ 5x: coefficient is 5, variable part is x

    ♦ x²: coefficient is 1 (∵ x² = 1x²), variable part is x²

• Constants: 6 and 5 are the constants

• This polynomial has only one variable 'x'

■ Let us take the next polynomial:

• v(x) = x³ + 6x² + 11x + 6

• Terms: x³, 6x², 11x and 6

• Splitting the terms: 

    ♦ x³: coefficient is 1, variable part is x³

    ♦ 6x²: coefficient is 6, variable part is x²

    ♦ 11x: coefficient is 11, variable part is x

• Constants: 6, 11 and 6 are the constants

• This polynomial has only one variable 'x'

■ Let us consider one more example:

• r(y) = 2y² – 5y – 3

• Terms: 2y², 5y, and 3

• Splitting the terms: 

    ♦ 2y²: coefficient is 2, variable part is y²

    ♦ 5y: coefficient is 5, variable part is y

    ♦ 3: coefficient is 3, variable part is y0

• Constants: 2, 5 and 3 are the constants

• This polynomial has only one variable 'y'

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Number of terms in a polynomial

■ Polynomials having only one term are called monomials ('mono' means one)

Examples:  

• a(x) = 5x  

• t(u) = u² 

• p(x) = x³

■ Polynomials having only two terms are called binomials ('bi' means two)

Examples: 

• p(y) = y²+2  

• a(x) = 5x²+2x  

• r(x) = x¹² - 1

■ Polynomials having only three terms are called trinomials ('tri' means three)

Examples:

• r(x) = x²-x+2

• p(y) = y⁴+y²-1

• t(y) = y³²-2y+7

■ A polynomial can have a large number of terms. An example with 151 terms is given below:

t(x) = x¹⁵⁰  + x¹⁴⁹  + x¹⁴⁸  + ..... + x² + x + 1

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Conditions for an algebraic expression to be called as a polynomial

Condition 1:

• We have seen so many polynomials above. All of them have a definite number of terms. Look at the sign between those terms. We find that the sign will either be '+' or '-'. 

• That means, there are only two possible operations between the terms of a polynomial: addition or subtraction. 

• If any other operation is involved, it is not a polynomial. 

Condition 2:

• We have seen that every term in a polynomial can be split into two parts. A coefficient and a variable part.

• Consider the variable part. Every variable will be having an exponent. (If there is no exponent written, it is understood that, the exponent is '1').

• For an algebraic expression to be a polynomial, the exponent in every term should be a whole number. We have seen whole numbers here. So the exponent should belong to the list: 0, 1, 2, 3. . .

    ♦ The exponent cannot be a negative number.

    ♦ The exponent cannot be a fraction

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Based on the above two conditions, we will see some algebraic expressions that fall outside the category of Polynomials in the next section.

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Chapter 15.1 - Equations in Two variables

In the previous sections we saw the solution of two equations with two variables. In this section, we will see a more advanced case.

Consider the following problem:
The price of 2 pens and 3 notebooks is ₹40. And the price of the same 2 pens and 5 notebooks is ₹60. What is the price of one notebook? What is the price of one pen?
Solution:
1. Let the price of one pen = x
2. Let the price of one notebook = y
3. Use condition 1: 2 pens and 3 notebooks cost 40: 2x + 3y = 40
4. Use condition 2: 2 pens and 5 notebooks cost 60:  2x + 5y = 60
5. From (4) we get 2x = 60 – 5y. Substitute for 2x in (3). We get:
6. (60 – 5y) + 3y = 40 ⇒ 60 – 2y = 40 ⇒ 20 = 2y
∴ y = 20/2 = 10
7. Substitute this value of y in (3). We get: 2x + 3 × 10 = 40 ⇒ 2x +30 =40 ⇒ 2x = 40 – 30 = 10
∴ x = 10/2 = 5
8. So the price of a pen = x = 5, and the price of a notebook = y = 10

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In this problem, we isolated '2x', from (4) and substituted it in (3). So the calculations were easy. Let us see if this method will work in all situations:

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Another problem:
The price of 3 pencils and 4 pens is ₹26. The price for 6 pencils and 3 pens is ₹27. What is the price of each?
Solution:
1. Let the price of one pencil = x
2. Let the price of one pen = y
3. Use condition 1: 3 pencils and 4 pens cost 26: 3x + 4y = 26
4. Use condition 2: 6 pencils and 3 pens cost 27: 6x + 3y = 27
• From (4) we can isolate '6x' as: 6x = 27-3y. But it does not have much use. Because, (3) does not have '6x'. Let us try an alternate way:
• From (3), we can isolate '3x' as: 3x = 26-4y. But this also does not have much use. Because, (4) does not have '3x'. So we need a new method to proceed
• Consider (3): 3x + 4y = 26. It is an equation. 
• If we multiply both sides of this equation by '2', we will get: 6x + 8y = 52. (Note that, if we multiply or divide both sides of the '=' sign of an equation by the same number, the equation will not change)
• Now we have '6x' in both equations. They can be easily solved. We can write the steps as follows:
5. Multiply (3) by 2 ⇒ (3) × 2 ⇒ 6x+8y =52
6. Isolate 6x from (5): 6x = 52 – 8y. Substitute this value of 6x in (4)
7. (52-8y)+3y =27 ⇒ 52-5y=27 ⇒ 5y = 52-27 ⇒ 5y =25
∴ y = 25/5 = 5
8. Substitute this value of y in (3). 
We get: 3x + 4 × 5 = 26 ⇒ 3x +20 =26 ⇒ 3x = 26 -20 ⇒ 3x = 6
∴ x = 6/3 = 2
9. Thus we can write: Price of one pencil = x = 2, and price of one pen = y = 5
10. Check: Use condition (1): 3 pencils and 4 pens cost 26: 3 × 2 + 4 × 5 = 6 + 20 = 26

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Now we will see a still more advanced case. Consider the following problem:
5 small buckets and 2 large buckets make 20 litres. 2 small buckets and 3 large buckets make 19 litres. What is the capacity of a small bucket? What is the capacity of the large bucket? 
Solution:
1. Let the capacity of the small bucket = x
2. Let the capacity of the large bucket = y
3. Use condition 1: 5 small buckets and 2 large buckets make 20 litres: 5x + 2y = 20
4. Use condition 2: 2 small buckets and 3 large buckets make 19 litres: 2x + 3y = 19
• From (4) we can isolate '2x'as: 2x = 19-3y. But it does not have much use. Because, (3) does not have '2x'
• From (3), we can isolate '5x' as: 5x = 20-2y. But it does not have much use. Because, (4) does not have '5x'
• Can we proceed as in the previous problem? That is., can we make '5x', common?
    ♦ The answer is 'No'. Because, there is no natural number which when multiplied with 2, will give 5. So we have to use a modified method:
• In the modified method, we multiply each equation by the 'coefficient of x in the other equation'
• So (3) is to be multiplied by 2, and, (4) is to be multiplied by 5. This will give a common term in x. We can write the steps as follows:
5. Multiply (3) by 2 ⇒ (3) × 2 ⇒ 10x+4y =40
6. Multiply (4) by 5 ⇒ (4) × 5 ⇒ 10x+15y =95
7. Isolate 10x from (6): 10x = 95 – 15y. Substitute this value of 10x in (5)
8. (95-15y)+4y =40 ⇒ 95-11y=40 ⇒ 11y = 95-40 ⇒ 11y =55
∴ y = 55/11 = 5
9. Substitute this value of y in (3). 
We get: 5x+2 × 5 = 20 ⇒ 5x +10 =20 ⇒ 5x = 20 -10 ⇒ 5x = 10
∴ x = 10/5 = 2
10. Thus we can write: Capacity of small bucket = x = 2, and capacity of large bucket = y = 5
11. Check: Use condition (1): 5 small buckets and 2 large buckets make 20 litres: 5 × 2 + 2 × 5 = 10 + 10 = 20

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Now we will see some solved examples:
Solved example 14.3:
Student A bought 7 notebooks of 200 pages and 5 notebooks of 100 pages for ₹107. Student B bought 5 notebooks of 200 pages and 7 notebooks of 100 pages for ₹97. What is the price of each type of note book?
Solution:
1. Let the price of one 200 page notebook = x
2. Let the price of one 100 page notebook = y
3. Use condition 1: seven 200 pages and five 100 pages notebooks cost 107: 7x + 5y =107
4. Use condition 2: five 200 pages and seven 100 pages notebooks cost 97: 5x + 7y =97
5. Multiply (3) by 5 (Here 5 is the coefficient of the x term in the 'other equation') 
⇒  (3) × 5 ⇒ 35x + 25y =535
6. Multiply (4) by 7 (Here 7 is the coefficient of the x term in the 'other equation') 
⇒ (4) × 7 ⇒ 35x + 49y =679
7. Isolate 35x from (6) ⇒ 35x = 679 – 49y
8. Substitute this 35x in (5) 
⇒ (679 -49y) + 25y =535 ⇒ 679 – 24y =535 ⇒ 24y = 679 -535 ⇒ 24y = 144
∴ y = 144/24 = 6
9. Substitute this value of y in (3)
We get: 7x + 5 × 6 =107
⇒ 7x + 30 =107 ⇒  7x = 107 -30 ⇒ 7x = 77
∴ x = 77/7 = 11
10. Thus we can write: Cost of 200 page notebook  = x = ₹11, and cost of 100 page notebook  = y = ₹6
11. Check: Use condition 1: seven 200 pages and five 100 pages notebooks cost ₹107: 7 × 11 + 5 × 6 = 77 + 30 = ₹107

Solved example 14.4
A man split an amount of ₹10000 into two parts. He invested one part in a bank which gives an annual interest of 8%. He invested the other part in another bank which gives an annual interest of 9%. After one year, he got a total interest of ₹875. How much did he invest in each bank?
Solution:
1. Let the amount invested in the first bank = x
2. Let the amount invested in the second bank = y
3. Use condition 1: Total amount = 10000: x + y =10000
4. Use condition 2: 
• Interest obtained from first bank = 8% of x = x × ⁸⁄₁₀₀ = 0.08x
• Interest obtained from second bank = 9% of y = y × ⁹⁄₁₀₀ = 0.09y
• Given that total interest amount = ₹875. So we can write:
• 0.08x + 0.09y = 875
5. Multiply (3) by 0.08
⇒ (3) × 0.08 ⇒ 0.08x + 0.08y = 10000 × 0.08  ⇒ 0.08x + 0.08y = 800
6. Isolate 0.08x from (5): 0.08x = 800 – 0.08y. 
Substitute this value of 0.08x in (4): (800 - 0.08y) + 0.09y = 875
⇒ 800 + 0.01y = 875 ⇒ 0.01y = 875 - 800 ⇒ 0.01y = 75 ⇒ ^y⁄₁₀₀ = 75 
⇒ y = 75 × 100 = 7500
7. Substitute this value of y in (3):
x + 7500 = 10000 ⇒ x = 10000 - 7500 = 2500
8. So we can write: Amount invested in the first bank = x = 2500 and, amount invested in the second bank = 7500
9. Check: Use condition 2: 0.08x + 0.09y = 875: 
⇒ 0.08 × 2500 + 0.09 × 7500 = 200 + 675 = ₹875

Solved example 14.5
A three and a half metres long rod is to be cut into two pieces. One piece is to be bent into a square, and the other into an equilateral triangle. The length of the side of both must be the same. How should it be cut?
Solution:
1. Let length of first piece = x
2. Let length of second piece = y
3. Use condition 1: Total length = three and a half metre = 3.5 m = 350 cm. So we get x+y =350
4. Use condition 2: 
• We have to bend the x cm into a square
    ♦ Length of all sides of a square are equal   
    ♦ So each side of the square will be ^x⁄₄ cm
• We have to bend the y cm into an equilateral triangle
    ♦ Length of all sides of an equilateral triangle are equal
    ♦ So each side of the equilateral triangle will be ^y⁄₃ 
The condition states that, the sides of the square and equilateral triangle are to be the same. So we can write: ^x⁄₄ = ^y⁄₃  ⇒ 3x = 4y
5. Multiply (3) by 3 ⇒ (3) × 3 ⇒ 3x + 3y =  1050
6. Substitute 3x = 4y [obtained from (4)] in (5). We get:
4y + 3y = 1050 ⇒ 7y = 1050
∴ y = 1050/7 = 150
7. Substitute this value of y in (3). We get: x + 150 = 350 ⇒ x = 350 - 150 ⇒ x = 200
8. So we can write: The 350 cm long rod should be cut in such a way that, length of one piece = x = 200 cm, and the length of the other piece = y = 150 cm
9. Check: Use condition 2: 3x = 4y ⇒ 3 × 200 = 4 ×150 ⇒ 600 = 600

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In the next section we will see more solved examples.

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