High school Mathematics Lessons: Equations

Showing posts with label Equations. Show all posts

Friday, December 2, 2016

Chapter 20 - Polynomials

In the previous section we completed the discussion on Similar triangles. In this section we will learn about Polynomials.

Consider a rectangle shown in fig.20.1.1(a). It has a length (l) = 3 cm and width (b) = 2 cm.

Fig.20.1

• We can easily calculate it's perimeter.

p = 2(l+b) = 2(3+2) = 2 × 5 = 10 cm

1. Suppose we increase the length by 1 cm, and width also by the same 1 cm, the new length = 4 cm, and new width = 3 cm

• Then new perimeter = 2(4+3) = 14 cm. This is shown in fig.20.1.1(c)

2. Suppose we increase the original length by 2 cm, and original width also by the same 2 cm, the new length = 3 +2 =5 cm, and new width = 2 + 2 = 4 cm

• Then new perimeter = 2(5+4) = 18 cm. This is shown in fig.20.1.2(c)

3. Suppose we increase the original length by 1¹⁄₂cm, and original width also by the same 1¹⁄₂cm, the new length = 3 + 1¹⁄₂ = 4¹⁄₂ cm, and new width = 2 + 1¹⁄₂ = 3¹⁄₂ cm

• Then new perimeter = 2(4¹⁄₂ + 3¹⁄₂) = 16 cm

■ So we find that, when the dimensions of the rectangle changes, the perimeter also changes. We can write a general form (see fig.20.1.3):

• Let the increase in length be 'x'. Then the new length = (3+x)

• Let the increase in width also be 'x'. Then the new width = (2+x)

• Then, the new perimeter = 2 [(3+x) + (2+x)] = 2 [5+2x] = 10+4x = 4x +10

We can say this:

• We have a rectangle with length 3 cm and width 2 cm

• If the length and width are increased by the same amount 'x', the new perimeter will be given by 4x+10

• So this is an easy method to calculate new perimeter. Let us put some values for x. The new perimeters are tabulated below:

Value of perimeter for different values of 'x':

x (cm)	1	1¹⁄₂	2	2¹⁄₄	3
p (cm)	14	16	18	19	22

Now we will consider area.
The original area = l × b = 3 × 2 = 6 cm (see fig.20.2.1 below)

Fig.20.2

1. When the length and width are increased by 1 cm, new length = 4 cm, and new width = 3 cm. So new area = 4 × 3 = 12 cm²

• In the fig.20.2.1.c, the additional areas are marked by green rectangles. We can write:

• Original area = 6, additional area = 3 +1 +2 = 6. So total area = 6 +6 = 12

2. When the original length and width are increased by 2 cm, new length = 5 cm, and new width = 4 cm. So new area = 5 × 4 = 20. This is shown in fig.20.2.2.c

• In the fig.20.2.2.c, the additional areas are marked by green rectangles. We can write:

• Original area = 6, additional area = 6 +4 +4 = 14. So total area = 6 +14 = 20
■ So we find that, when the dimensions of the rectangle changes, the area also changes. We can write a general form (see fig.20.2.3):

• Let the increase in length be 'x'. Then the new length = (3+x)

• Let the increase in width also be 'x'. Then the new width = (2+x)

• Then new area = [(3+x)(2+x)]. We have learned how to multiply them, when we saw identities. We get: [(3+x)(2+x)] = 6 + 2x + 3x + x²= x²+ 5x + 6

[The value '6' in the above result is the area of the red rectangle in fig.c. The others 2x, 3x and x² are the areas of additional green rectangles]

• We can write: New area = x²+ 5x + 6

• So this is an easy method to calculate new areas. Let us put some values for x. The new areas are tabulated below:

Value of area for different values of 'x':

x (cm)	1	1¹⁄₂	2	3	4
a (cm)	12	15³⁄₄	20	30	42

So we have seen two equations:

■ New perimeter = 4x+10

■ New area = x²+ 5x + 6

Apart from making our calculations easier, such equations have wider applications.

• Instead of 'New perimeter', let us write 'p(x)'. Then the equation will become: p(x) = 10 + 4x

• 'p(x)' indicates that we must not put a 'permanent value' for perimeter.
• It is kind of a 'warning', which tells us that, in this particular problem, the perimeter, (which is denoted by 'p') will change based on the value of 'x'

• The manner in which the change takes place, will be given on the right side of the '=' sign.

• Similarly, in the second equation, instead of 'New area', let us write 'a(x)'. Then the equation will become: a(x) = x²+ 5x + 6

• 'a(x)' indicates that we must not put a 'permanent value' for area.
• It is kind of a 'warning', which tells us that, in this particular problem, the area, (which is denoted by 'a') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign.

We have seen perimeter and area. Now let us see volume.

1. In fig.20.3(a) below, we have a red box of length 3 cm, width 2 cm, and height 1 cm. So it's volume = 3 × 2 × 1 = 6 cm³

Fig.20.3

2. Its length and width are increased by 'x' cm as indicated by the green rectangles in fig.a. It's height is increased by the same 'x' cm, as indicated by the yellow rectangle in fig.a.

3. If we give a height of 1 cm for the green rectangles, they will reach the top level of the red box. This is shown in fig.b. Then, the additional volumes will be:
♦ 3 ×x ×1 = 2x cm³
♦ 2 ×x ×1 = 3x cm³
♦ x ×x ×1 = x² cm³

• So the total volume in fig.b = 6 + 2x + 3x + x² = x²+ 5x + 6

4. Now, in fig.c, the additional height x cm is also given volume. The volume of this yellow box = [(2+x)x(3+x)] = [(6+5x+x²)x] = x³+ 5x² + 6x

5. Thus, the total volume in fig.c = x²+ 5x + 6 + x³+ 5x² + 6x
= x³+ 6x² + 11x + 6

The same result can be obtained by multiplying the new length, width and height together. This is shown below:

(3+x)(2+x)(1+x) = (6+5x+x²)(1+x)
= 6+5x+x²+6x+5x²+x³ = x³+6x²+11x+6

We can write: v(x) = x³+6x²+11x+6

• 'v(x)' indicates that we must not put a 'permanent value' for volume.
• It is kind of a 'warning', which tells us that, in this particular problem, the volume, (which is denoted by 'v') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign

In the next section, we will see some solved examples demonstrating the creation of such expression in real life situations.

Monday, August 8, 2016

Chapter 15.3 - Equations in two variables - Identities

In the previous sections we saw some advanced cases in the solution of 'two equations with two variables'. We also saw some solved examples. In this section, we will see another type of equations.

We will be using some identities that we have learned before:
1. (x+y)² = x² + 2xy + y²
2. (x-y)² = x² - 2xy + y²
3. (x+y)² - (x-y)² = 4xy
4. (x+y)² = (x-y)²+ 4xy
5. x² - y² = (x+y)(x-y)

• (1) and (2) are basic identities
• (3) is simply (1) - (2)
• (4) is obtained by rearranging (3)
• (5) is easily obtained by expanding it's right side

Consider the problem: There are two squares of different sizes. The sides of the larger square is 5 cm greater than the sides of the smaller square. The area of the larger square is 55 sq.cm more than the smaller square. What is the length of the sides of each square?
Solution:
1. Let the length of the larger square = x. So area of the larger square = x²
2. Let the length of the smaller square = y. So area of the smaller square = y²
3. Use condition 1: Larger side is 5 cm greater: x = y+5
4. Use condition 2: Area is 55 greater: x² = y² + 55
5. From (3) we get x-y = 5
6. From (4) we get: x²-y² = 55
7. Now we use an identity that we have learned before: x² - y² = (x+y)(x-y)
8. From (7) we get x+y = ^(x²-y²⁾⁄_(x-y)
9. So we can write: x+y = 55/5 = 11
10. Isolate x from (9). We get: x = 11-y
11. Substitute this value of x in (5). We get: (11-y)-y = 5 ⇒ 11-2y = 5 ⇒ 2y = 11-5 ⇒ 2y = 6
∴ y = 6/2 = 3
12. Substitute this value of y in (3). We get: x = 3 + 5 = 8
13. So we can write: Side of the larger square = x = 8 cm, and, the side of the smaller square = 3 cm
14. Check: From (4) we get: 82 = 32 + 55 ⇒ 64 = 9+55 ⇒ 64 = 64

Another problem:
The perimeter of a rectangle is 10 m and it's area is 5¹⁄₄ sq.m. What are the lengths of the sides?
Solution:
1. Let the length of the rectangle = x
2. Let the breadth of the rectangle = y
3. Use condition 1: perimeter = 10: 2(x+y) = 10 ⇒ x+y = 10/2 ⇒ x+y = 5
4. Use condition 2: Area = 5¹⁄₄ : xy = 5¹⁄₄
5. Now we use an identity that we have learned before: (x+y)² -(x-y)² = 4xy
6. In (5), we have the value of (x+y). This we have from (3)
7. In (5), we have the value of xy. This we have from (4)
8. What we do not have in (5) is (x-y). So isolate it: (x-y)² = (x+y)²-4xy
9. Substitute the values of (x+y) and xy in (8). We get:
10. (x-y)² = 5²- 4 × 5¹⁄₄ ⇒ (x-y)² = 25 - 21 ⇒ (x-y)² = 4 ⇒ x-y = 2
11. Now, from (3), we have x+y, and from (10), we have x-y
12. In (10), isolate x. We get: x = 2+y.
13. Substitute this x in (3). We get: (2+y)+y = 5 ⇒ 2 +2y = 5 ⇒ 2y = 3
∴ y = 3/2 = 1¹⁄₂ m
14. Substitute this value of y in (3). We get: x+ 3/2 = 5 ⇒ x = 5 - 3/2 = 7/2 = 3¹⁄₂ m
15. So we can write: The length of the rectangle = x = 3¹⁄₂ m, and breadth = y = 1¹⁄₂ m
16. Check: use condition 2: Area = xy = 7/2 × 3/2 = 21/4 = 5¹⁄₄ sq.m

Now we will see some solved examples:
Solved example 14.11
A 10 m long rope is to be cut into two pieces, and a square is to be made using each. The difference in the areas enclosed must be 1¹⁄₄ sq.m. How should it be cut?
Solution:
1. Let the length of one part = x
2. Let length of the remaining = y
3. Use condition 1: Total length of rope = 10 m: x+y = 10
4. Use condition 2:
• x is made into a square. Side of that square will be ^x⁄₄. So area of that square will be ^x²⁄₁₆
• y is made into a square. Side of that square will be ^y⁄₄. So area of that square will be ^y²⁄₁₆
• Difference in areas enclosed = ^x²⁄₁₆ - ^y²⁄₁₆ = ^(x²-y²⁾⁄₁₆
• Condition 2 states that this difference must be 1¹⁄₄sq.mt. So we can write: ^(x²-y²⁾⁄₁₆ = 1¹⁄₄
⇒ ^(x²-y²⁾⁄₁₆ = ⁵⁄₄ ⇒ x² - y² = ^(16×5)⁄₄⇒ x² - y² = 20
5. Use the identity: x² - y² = (x+y)(x-y)
6. Rearranging (5) we get: x-y = ^(x²-y²⁾⁄_(x+y) ⇒ x-y = ²⁰⁄₁₀ ⇒ x-y = 2
7. Isolate x from (6). We get: x = y+2
8. Substitute this value of x in (3). We get: (y+2)+y = 10 ⇒ 2y+2 = 10 ⇒ 2y = 8
∴ y = 8/2 = 4
9. Substitute this value of y in (7). We get: x = 4+2 = 6
10. So we can write: The rope should be cut in such a way that, one piece has a length = x = 6m, and the remaining piece has a length = y = 4 m
11. Check: Use condition 2:
• Area of first square = ^x²⁄₁₆ = ^6²⁄₁₆ = ³⁶⁄₁₆ = ⁹⁄₄ = 2¹⁄₄
• Area of second square = ^y²⁄₁₆ = ^4²⁄₁₆ = ¹⁶⁄₁₆ = 1
• Difference in area = 2¹⁄₄ - 1 = 1¹⁄₄

Solved example 14.12
The length of a rectangle is 1 m more than it's breadth. It's area is 3³⁄₄ sq.m. What are it's length and breadth?
Solution:
1. Let the length of the rectangle = x
2. Let the breadth of the rectangle = y
3. Use condition 1: Length more than breadth by 1 m: x = y+1 ⇒ x-y=1
4. Use condition 2: Area = 3³⁄₄: xy = 3³⁄₄
5. Use the identity: (x+y)² = (x-y)²+ 4xy. We get: (x+y)² = 1²+ 4 × 3³⁄₄
⇒ (x+y)² = 1 + 15 ⇒ (x+y)² =16 ⇒ (x+y) = √16 ⇒ (x+y)= 4
6. Isolate x from (5). We get: x = 4-y
7. Substitute this value of x in (3). We get: (4-y)-y = 1 ⇒ 4-2y = 1 ⇒ 2y = 3 ⇒ y = 3/2 = 1¹⁄₂
8. Substitute this value of y in (6). We get: x = 4 - 3/2 = 5/2 = 2¹⁄₂
9. So we can write: Length of the rectangle = x = 2¹⁄₂ m, and breadth = y = 1¹⁄₂ m
10. check: Area = xy = ³⁄₂ × ⁵⁄₂ = ¹⁵⁄₄ = 3³⁄₄

Solved example 14.13
The hypotenuse of a right triangle is 6¹⁄₂ cm, and it's area is 7¹⁄₂ sq.cm. Calculate the length of it's perpendicular sides
Solution:
1. Let the length of one perpendicular side = x
2. Let the length of the other perpendicular side = y
3. Use condition 1: Hypotenuse = 6¹⁄₂
• Using Pythagoras theorem, x² + y² = [ 6¹⁄₂ ]²⇒ x² + y² = [ ¹³⁄₂ ]²  ⇒ x² + y² = ¹⁶⁹⁄₄
4. Use condition 2: Area = 7¹⁄₂
• In a right triangle, one perpendicular side can be taken as the base, and the other perpendicular side can be taken as the height. Let x be the base, and y, the height
• So area = ¹⁄₂× base × height = ¹⁄₂× xy = ^xy⁄₂
Thus we can write: ^xy⁄₂ = 7¹⁄₂ ⇒ ^xy⁄₂ = ¹⁵⁄₂ ⇒ xy = 15 ⇒ 2xy = 30
5. Use identity: (x+y)² = x² + 2xy + y² ⇒ (x+y)² = (x² + y²)+ 2xy
6. We have all values on the right side. Substituting those values we get: (x+y)² = (¹⁶⁹⁄₄) + 30 ⇒ (x+y)² = ^(169+120)⁄₄
⇒ (x+y)² = ²⁸⁹⁄₄⇒ x+y = √(²⁸⁹⁄₄) ⇒ x+y = ¹⁷⁄₂
7. Use identity: (x-y)² = x² - 2xy + y² ⇒ (x-y)² = (x² + y²) - 2xy
8. We have all values on the right side. Substituting those values we get: (x-y)² = (¹⁶⁹⁄₄) - 30 ⇒ (x-y)² = ^(169-120)⁄₄
⇒ (x-y)² = ⁴⁹⁄₄⇒ x-y = √(⁴⁹⁄₄) ⇒ x-y = ⁷⁄₂
9. Isolate x from (8). We get: x = ⁷⁄₂ + y
10. Substitute this value of x in (6). We get: ⁷⁄₂ + y + y = ¹⁷⁄₂  ⇒ ⁷⁄₂ + 2y = ¹⁷⁄₂
⇒ 2y = ¹⁷⁄₂ - ⁷⁄₂ = ¹⁰⁄₂ = 5 ⇒ 2y = 5
∴ y = ⁵⁄₂
11. Substitute this value of y in (6). We get: x + ⁵⁄₂ = ¹⁷⁄₂ ⇒ x = ¹⁷⁄₂ - ⁵⁄₂ = ¹²⁄₂ = 6
12. So we can write: One perpendicular side = x = 6 cm, and the other perpendicular side = y = ⁵⁄₂ cm
13. Check: Area = ¹⁄₂× 6 × ⁵⁄₂ = ³⁰⁄₄ = ¹⁵⁄₂ = 7¹⁄₂ sq.cm

We have completed the discussion on Equations in two variables. In the next section we will see irrational numbers.

Saturday, August 6, 2016

Chapter 15.2 - Equations in Two variables - Solved examples

In the previous sections we saw some advanced cases in the solution of two equations with two variables. We also saw some solved examples. In this section, we will see more solved examples.

Solved example 14.6
The distance travelled in t seconds by an object starting with a speed of u metres per second, and moving along a straight line, with speed increasing at the rate of a metres/second every second is given by d = ut + ¹⁄₂at² metres. An object moving in this manner travels 10 metres in 2 seconds and 28 metres in 4 seconds. With what speed did it start? At what rate does it's speed change?
Solution:
• The equation d = ut + ¹⁄₂at² gives us the distance d travelled by an object. But to use the equation, we must have three parameters: initial speed u, acceleration a, and time t.
• The object travelled with a certain initial speed. This speed is not given to us. But two readings were taken during it's travel
• According to the first reading, the object travelled 10 metres during a time interval of 2 seconds. That is., d= 10, and t = 2.
1. So we can write: 10 = (u × 2) + (¹⁄₂ × a × 2²) ⇒ 10 = 2u + 2a
• According to the second reading, it travelled 28 metres in 4 seconds. That is., d = 28, and t = 4
2. So we can write: 28 = (u × 4) + (¹⁄₂ × a × 4²) ⇒ 28 = 4u + 8a
3. Multiply (1) by 2 ⇒ (1) × 2 ⇒ 20 = 4u + 4a
4. Isolate 4u from (3): 4u = 20 - 4a
5. Substitute this 4u in (2): 28 = (20 - 4a) + 8a ⇒ 28 = 20 + 4a ⇒ 4a = 28 - 20 ⇒ 4a = 8
∴ a = 8/4 = 2
6. substitute this value of a in (1). We get: 10 = 2u + 2 × 2 ⇒ 10 = 2u + 4 ⇒ 2u = 6
∴ u = 6/2 = 3
7. So we can write: Initial speed = u = 3 meter/sec and, acceleration a = 2 metre/ sec²
8. Check: Use (2): 28 = 4u + 8a: 28 = 4 × 3 + 8 × 2 ⇒ 28 = 12 + 16 ⇒ 28 = 28

Solved example 14.7
Four times a number added to three times a second number gives 43. Two times the second number subtracted from three times the first number gives 11. What are the numbers?
Solution:
1. Let the first number = x
2. Let the second number = y
3. Use condition 1: 4 times first number + 3 times second number = 43: 4x + 3y = 43
4. Use condition 2: 3 times first number - 2 times second number = 11: 3x - 2y = 11
5. Multiply (3) by 3 (Here 3 is the coefficient of the x term in the 'other equation')
⇒  (3) × 3 ⇒ 12x + 9y = 129
6. Multiply (4) by 4 (Here 4 is the coefficient of the x term in the 'other equation')
⇒ (4) × 4 ⇒ 12x - 8y = 44
7. Isolate 12x from (6) ⇒ 12x = 44 + 8y
8. Substitute this 12x in (5)
⇒ (44 + 8y) + 9y = 129 ⇒ 44 + 17y = 129 ⇒ 17y = 129 - 44 ⇒ 17y = 85
∴ y = 85/17 = 5
9. Substitute this value of y in (3)
We get: 4x + 3 × 5 = 43 ⇒ 4x + 15 = 43 ⇒  4x = 43 -15 ⇒ 4x = 28
∴ x = 28/4 = 7
10. Thus we can write: First number = x = 7, and second number = y = 5
11. Check: Use condition 1: 3 times first number - 2 times second number = 11: (3 × 7) - (2 × 5) = 21 - 10 = 11

Solved example 14.8
The sum of the digits of a two digit number is 11. The number got by interchanging the digits is 27 more than the original number. What is the number ?
Solution:
• We have a two digit number. Let the digits be x and y
• Let x be in the tens place and y be in the ones place
• Then the value of the number = 10x +y
1. Use condition 1: Sum of the digits = 11: x + y = 11
2. Use condition 2: The number got by interchanging the digits is 27 more than the original number
• The value of the number got by interchanging the digits will be 10y + x
• This value is 27 more than the value of the original number. So we can write:
10y + x = (10x +y) + 27 ⇒ 10y - 10x + x - y = 27 ⇒ 9y - 9x = 27 ⇒ y - x = 3 (dividing both sides by 9)
3. Isolate y from (2): y = 3 + x
4. Substitute this value of y in (1): x + (3 + x) = 11 ⇒ 2x + 3 = 11 ⇒ 2x = 8
∴ x = 8/2 = 4
5. Substitute this value of x in (1). We get: 4 + y = 11 ⇒ y = 11 - 4 = 7
6. So we can write: the original number = 47, and the new number = 74
7. Check: second number - first number = 74 - 47 = 27

Solved example 14.9
Four years ago, Mr.A's age was three times the age of Mr.B. After two years, Mr. A's age will be two times the age of Mr.B. What are their ages now?
Solution:
1. Let the present age of Mr.A = x
2. Let the present age of Mr.B = y
3. Use condition 1: Age 4 years ago:
• Age of Mr.A 4 years ago = x-4
• Age of Mr.B 4 years ago = y-4
• Applying the condition, we get: x - 4 = 3(y-4) ⇒ x - 4 = 3y - 12 ⇒ 3y - x = 8
4. Use condition 2: Age after 2 years:
• Age of Mr.A after 2 years = x +2
• Age of Mr.B after 2 years = y +2
• Applying the condition, we get: x + 2 = 2(y +2) ⇒ x + 2 = 2y + 4 ⇒ x - 2y = 2
5. Isolate x from (4): x = 2 + 2y
6. Substitute this value of x in (3). We get: 3y -(2 + 2y) = 8 ⇒ 3y -2 - 2y = 8 ⇒ y -2 = 8 ⇒ y = 10
7. Substitute this value of y in (3). We get: 3 × 10 - x = 8 ⇒ 30 - x = 8 ⇒ x = 30 - 8 = 22
8. So we can write: Present age of Mr.A = x = 22, and present age of Mr.B = y = 10
9. Check: Use (4): 22 - 2 × 10 = 2 ⇒ 22 - 20 = 2 ⇒ 2 = 2

Solved example 14.10
If the length of a rectangle is increased by 5 m, and the breadth decreased by 3 m, the area will decrease by 5 sq.m. If the length is increased by 3 m, and breadth increased by 2 m, the area will increase by 50 sq.m. What is the original length and breadth?
Solution:
1. Let the original length = x
2. Let the original breadth = y
3. So original area = xy
4. Use condition 1: (x+5) (y-3) = xy-5 ⇒ xy+5y-3x-15 = xy-5 ⇒ 5y-3x = 10
5. Use condition 2: (x+3) (y+2) = xy+50 ⇒ xy+3y+2x+6 = xy+50 ⇒ 3y+2x = 44
6. Multiply (4) by 3 (Here 3 is the coefficient of the y term in the 'other equation')
⇒  (4) × 3 ⇒ 15y - 9x = 30
7. Multiply (5) by 5 (Here 5 is the coefficient of the y term in the 'other equation')
⇒ (5) × 5 ⇒ 15y + 10x = 220
8. Isolate 15y from (7). We get: 15y = 220-10x
9. Substitute this value of 15y in (6). We get: (220-10x)-9x = 30 ⇒ 220-19x = 30 ⇒ 190 = 19x
∴ x = 190/19 = 10
10. Substitute this value of x in (4). We get: 5y-3 10 = 10 ⇒ 5y-30 = 10 ⇒ 5y = 40
∴ y = 40/5 = 8
11. So we can write: length and breadth of the original rectangle are x = 10 m and y = 8 m respectively
12. Check: Use (5): 3 8 + 2 10 = 44 ⇒ 24 + 20 = 44 ⇒ 44 = 44

In the next section we will see another type of equation.

Friday, August 5, 2016

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

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:

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

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

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⁄₄cm
• 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

In the next section we will see more solved examples.