In the previous sections we completed the discussion on triangles. In this section, we will learn about equations.
Consider the following problem:
There are 100 beads in a box. Some of them are green, and the rest are red. There are 10 red beads more than the green beads. Then What is the number of red beads and green beads?
Solution: We can solve this using algebra.
1. Let the number of green beads = x
2. Then the number of red beads = x + 10
3. So we have written the ‘number of green beads’ and the ‘number of red beads’.
4. The total number of beads is given as 100.
5. So x + (x + 10) = 100 ⇒ 2x + 10 = 100
Bringing 10 to the right side we get: 2x = 100 – 10 ⇒2x = 90
∴ x = 90/2 = 45
■ That is., number of green beads = 45
■ Substitute this value of x in (2). We get: number of red beads = x+10 = 45 + 10 =55
■ Check: Number of green beads + number of red beads = x + y = 45 + 55 = 100
Another method:
1. Let the number of green beads = x
2. Let the number of red beads = y
3. We have, total number of beads = 100
4. So we can write x+y = 100
5. We have, number of red beads = 10 greater than that of the green beads
6. So we can write y = x +10
7. Substitute this value of y in (4)
8. So (4) becomes: x + (x +10) = 100 ⇒ 2x + 10 = 100
Bringing 10 to the right side we get: 2x = 100 – 10 ⇒2x = 90
∴ x = 90/2 = 45
■ That is., number of green beads = 45
■ Substitute this value of x in (4). We get: 45+y = 100 ⇒ y = 100 -45 =55
■ Check: Number of green beads + number of red beads = x + y = 45 + 55 = 100
We solved the above problem by two different methods.
• In the first method we used only one variable ‘x’.
• In the second method, we used two variables ‘x’ and ‘y’.
• Also in the second method, we used two equations (4) and (6)
• We used two conditions to form those two equations:
♦ 'The total number of beads = 100' was used to form equation (4)
♦ 'Red is 10 more than green' was used to form equation (6)
■ So we can say: in the second method, we used 'two equations' and 'two variables'
Another problem:
The price of a table and a chair together is ₹ 5000. The price of a table and 4 chairs is ₹8000. What is the price of each?
Solution: We will use 'two equations' and 'two variables'.
1. Let the price of a table = x
2. Let the price of a chair = y
3. We are given two conditions:
• Price of a table and chair = 5000
• Price of a table and 4 chairs = 8000
4. From the first condition, we can write: x+y = 5000
5. From the second condition, we can write: x+4y = 8000
6. From (4), we can isolate x, by bringing y to the right side. So we have: x = 5000 - y
7. We can substitute this value of x in (5). We get:
(5000-y) + 4y = 8000 ⇒ 5000-y+4y =8000 ⇒ 5000+3y =8000 ⇒ 3y = 8000 – 5000 ⇒ 3y = 3000
∴ y = 3000/3 = 1000
8. So the price of a chair = y = ₹1000
9. Substituting this in (6) we get price of a table = x = 5000 – 1000 = ₹4000
10. Check: Price of a table and chair together = x +y = 4000 +1000 = ₹5000
Another problem:
In a fraction, if 1 is added to the numerator, the value of the fraction becomes 1⁄2 . If 1 is added to the denominator instead, the value becomes 1⁄3. What is the fraction?
Solution:
1. Let the numerator be x, and the denominator be y. Then the fraction is x⁄y
2. When 1 is added to the numerator, the fraction becomes 1/2
3. So we can write: (x+1)⁄y = 1⁄2
4. When 1 is added to the denominator, the fraction becomes 1/3
5. So we can write: x⁄(y+1) = 1⁄3
6. • In (3), the quantities on either sides of the '=' sign are equivalent fractions.
• Similarly in (5), the quantities on either sides of the '=' sign are equivalent fractions.
• We can apply 'cross multiplication' for each of them
7. Applying cross multiplication to (3) we get:
2(x+1) =y
8. Applying cross multiplication to (5) we get:
3x = y+1
9. In (7), we already have y in the isolated form. We can substitute it in (8). So we get: 3x = [2(x+1)] + 1.
⇒ 3x = 2x + 2 + 1 ⇒ 3x = 2x + 3 ⇒ 3x – 2x = 3
∴ x = 3
10. Substituting this value of x in (7), we get: 2(3+1)=y ⇒ y = 2 × 4 = 8
11. Thus the fraction x/y = 3/8
Now we will see some solved examples. We can use either ‘one equation with one variable’ OR, ‘two equations with two variables’
Solved example 14.1:
The perimeter of a rectangle is 1 m. The length is 5 cm longer than the width. Find the actual length and width.
Solution:
1. Let the length of the rectangle = x
2. Let the width of the rectangle = y
It is given that one side is 5 cm longer than the other. We take the longer side as the length and shorter side as the width. So x is 5 cm greater than y. Thus we can write:
3. x = y+5
4. Now we consider the second condition that is given: Perimeter = 1 m = 100 cm. So we can write:
5. 2(x+y) = 100. In this we can substitute for x from (3). So we get
6. 2[(y+5)+y]=100 ⇒ 2[2y+5]=100 ⇒ 4y+10=100 ⇒4y = 90
7. ∴ y = 90/4 = 22.5 cm
8. Substitute this value of y in (3). We get: y = 22.5 +5 = 27.5 cm
9. check: perimeter = 2(22.5 +27.5) = 2 × 50 = 100 cm
Solved example 14.2
A class has 4 more girls than boys. On a day when only 8 boys were absent, number of girls was twice that of boys. How many boys and girls are there in the class?
Solution:
1. Let the number of boys = x
2. Let the number of girls = y
3. Use condition 1: Number of girls is 4 more than the number of boys. From this condition we get: y = x+4
4. Use condition 2: When 8 boys are absent, the number of boys on that day = (x-8). No girls were absent on that day. So number of girls on that day = y
Thus we can write: y = 2(x-8)
5. Substitute for y from (3). We get:
6. x+4 =2(x-8) ⇒ x+4 = 2x-16
7. Bring the x to the right side. We get: 4 =2x -x -16
Bring the 16 to the left side. We get:
4+16 = 2x-x ⇒ 20 = x. So the number of boys = x = 20
8. Substitute this x in (3). We get y = 20+4 = 24
9. Check: When 8 boys are absent, number of boys = 20-8 = 12. This is half of 24. So the number of girls on that day was indeed twice the number of boys present
In the next section we will see a more advanced case involving 'two equations' and 'two variables'.
Consider the following problem:
There are 100 beads in a box. Some of them are green, and the rest are red. There are 10 red beads more than the green beads. Then What is the number of red beads and green beads?
Solution: We can solve this using algebra.
1. Let the number of green beads = x
2. Then the number of red beads = x + 10
3. So we have written the ‘number of green beads’ and the ‘number of red beads’.
4. The total number of beads is given as 100.
5. So x + (x + 10) = 100 ⇒ 2x + 10 = 100
Bringing 10 to the right side we get: 2x = 100 – 10 ⇒2x = 90
∴ x = 90/2 = 45
■ That is., number of green beads = 45
■ Substitute this value of x in (2). We get: number of red beads = x+10 = 45 + 10 =55
■ Check: Number of green beads + number of red beads = x + y = 45 + 55 = 100
Another method:
1. Let the number of green beads = x
2. Let the number of red beads = y
3. We have, total number of beads = 100
4. So we can write x+y = 100
5. We have, number of red beads = 10 greater than that of the green beads
6. So we can write y = x +10
7. Substitute this value of y in (4)
8. So (4) becomes: x + (x +10) = 100 ⇒ 2x + 10 = 100
Bringing 10 to the right side we get: 2x = 100 – 10 ⇒2x = 90
∴ x = 90/2 = 45
■ That is., number of green beads = 45
■ Substitute this value of x in (4). We get: 45+y = 100 ⇒ y = 100 -45 =55
■ Check: Number of green beads + number of red beads = x + y = 45 + 55 = 100
We solved the above problem by two different methods.
• In the first method we used only one variable ‘x’.
• In the second method, we used two variables ‘x’ and ‘y’.
• Also in the second method, we used two equations (4) and (6)
• We used two conditions to form those two equations:
♦ 'The total number of beads = 100' was used to form equation (4)
♦ 'Red is 10 more than green' was used to form equation (6)
■ So we can say: in the second method, we used 'two equations' and 'two variables'
Another problem:
The price of a table and a chair together is ₹ 5000. The price of a table and 4 chairs is ₹8000. What is the price of each?
Solution: We will use 'two equations' and 'two variables'.
1. Let the price of a table = x
2. Let the price of a chair = y
3. We are given two conditions:
• Price of a table and chair = 5000
• Price of a table and 4 chairs = 8000
4. From the first condition, we can write: x+y = 5000
5. From the second condition, we can write: x+4y = 8000
6. From (4), we can isolate x, by bringing y to the right side. So we have: x = 5000 - y
7. We can substitute this value of x in (5). We get:
(5000-y) + 4y = 8000 ⇒ 5000-y+4y =8000 ⇒ 5000+3y =8000 ⇒ 3y = 8000 – 5000 ⇒ 3y = 3000
∴ y = 3000/3 = 1000
8. So the price of a chair = y = ₹1000
9. Substituting this in (6) we get price of a table = x = 5000 – 1000 = ₹4000
10. Check: Price of a table and chair together = x +y = 4000 +1000 = ₹5000
Another problem:
In a fraction, if 1 is added to the numerator, the value of the fraction becomes 1⁄2 . If 1 is added to the denominator instead, the value becomes 1⁄3. What is the fraction?
Solution:
1. Let the numerator be x, and the denominator be y. Then the fraction is x⁄y
2. When 1 is added to the numerator, the fraction becomes 1/2
3. So we can write: (x+1)⁄y = 1⁄2
4. When 1 is added to the denominator, the fraction becomes 1/3
5. So we can write: x⁄(y+1) = 1⁄3
6. • In (3), the quantities on either sides of the '=' sign are equivalent fractions.
• Similarly in (5), the quantities on either sides of the '=' sign are equivalent fractions.
• We can apply 'cross multiplication' for each of them
7. Applying cross multiplication to (3) we get:
2(x+1) =y
8. Applying cross multiplication to (5) we get:
3x = y+1
9. In (7), we already have y in the isolated form. We can substitute it in (8). So we get: 3x = [2(x+1)] + 1.
⇒ 3x = 2x + 2 + 1 ⇒ 3x = 2x + 3 ⇒ 3x – 2x = 3
∴ x = 3
10. Substituting this value of x in (7), we get: 2(3+1)=y ⇒ y = 2 × 4 = 8
11. Thus the fraction x/y = 3/8
Now we will see some solved examples. We can use either ‘one equation with one variable’ OR, ‘two equations with two variables’
Solved example 14.1:
The perimeter of a rectangle is 1 m. The length is 5 cm longer than the width. Find the actual length and width.
Solution:
1. Let the length of the rectangle = x
2. Let the width of the rectangle = y
It is given that one side is 5 cm longer than the other. We take the longer side as the length and shorter side as the width. So x is 5 cm greater than y. Thus we can write:
3. x = y+5
4. Now we consider the second condition that is given: Perimeter = 1 m = 100 cm. So we can write:
5. 2(x+y) = 100. In this we can substitute for x from (3). So we get
6. 2[(y+5)+y]=100 ⇒ 2[2y+5]=100 ⇒ 4y+10=100 ⇒4y = 90
7. ∴ y = 90/4 = 22.5 cm
8. Substitute this value of y in (3). We get: y = 22.5 +5 = 27.5 cm
9. check: perimeter = 2(22.5 +27.5) = 2 × 50 = 100 cm
Solved example 14.2
A class has 4 more girls than boys. On a day when only 8 boys were absent, number of girls was twice that of boys. How many boys and girls are there in the class?
Solution:
1. Let the number of boys = x
2. Let the number of girls = y
3. Use condition 1: Number of girls is 4 more than the number of boys. From this condition we get: y = x+4
4. Use condition 2: When 8 boys are absent, the number of boys on that day = (x-8). No girls were absent on that day. So number of girls on that day = y
Thus we can write: y = 2(x-8)
5. Substitute for y from (3). We get:
6. x+4 =2(x-8) ⇒ x+4 = 2x-16
7. Bring the x to the right side. We get: 4 =2x -x -16
Bring the 16 to the left side. We get:
4+16 = 2x-x ⇒ 20 = x. So the number of boys = x = 20
8. Substitute this x in (3). We get y = 20+4 = 24
9. Check: When 8 boys are absent, number of boys = 20-8 = 12. This is half of 24. So the number of girls on that day was indeed twice the number of boys present
In the next section we will see a more advanced case involving 'two equations' and 'two variables'.
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