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.
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.
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