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⁄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
In the next section we will see more solved examples.
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⁄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
In the next section we will see more solved examples.
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