Example-1 :- Akhila goes to a fair with Rs 20 and wants to have rides on the Giant Wheel and play Hoopla. Represent this situation algebraically and graphically (geometrically).
Solution :-The pair of equations formed is: y = x/2 x - 2y = 0 .......(i) 3x + 4y = 20 ......(ii) Lets represent these equations in graphically: x - 2y = 0 .......(i)
| x | 0 | 2 |
| y | 0 | 1 |
| x | 0 | 4 |
| y | 5 | 2 |
Now, Observe that the two lines representing the two equations are intersecting at the point (4, 2).
Example-2 :- Romila went to a stationery shop and purchased 2 pencils and 3 erasers for cost 9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought 4 pencils and 6 erasers of the same kind for cost 18. Represent this situation algebraically and graphically.
Solution :-Let us denote the cost of 1 pencil by x and one eraser by y. Then the algebraic representation is given by the following equations: 2x + 3y = 9 .......(i) 4x + 6y = 18 ......(ii) Lets represent these equations in graphically: 2x + 3y = 9 .......(i)
| x | 0 | 4.5 |
| y | 3 | 0 |
| x | 0 | 3 |
| y | 3 | 1 |
So, both the lines coincide in figure. This is so, because, both the equations are equivalent.
Example-3 :- Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.
Solution :-The algebraic representation is given by the following equations: x + 2y – 4 = 0 x + 2y = 4 ............(i) 2x + 4y – 12 = 0 2x + 4y = 12 x + 2y = 6 ...........(ii) Lets represent these equations in graphically: x + 2y = 4 ............(i)
| x | 0 | 4 |
| y | 2 | 0 |
| x | 0 | 6 |
| y | 3 | 0 |
So, both the lines parallels in figure.
Example-4 :- Check graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically.
Solution :-x + 3y = 6 .......(i) 2x – 3y = 12 ......(ii) Lets represent these equations in graphically: x + 3y = 6 .......(i)
| x | 0 | 6 |
| y | 2 | 0 |
| x | 0 | 6 |
| y | -4 | 0 |
Now, Observe that the two lines representing the two equations are intersecting at the point (6, 0).
So, it is consistent.
Example-5 :- Graphically, find whether the following pair of equations has no solution, unique solution or infinitely many solutions: 5x – 8y + 1 = 0 and 3x - 24y/5 +3/5 = 0.
Solution :-5x – 8y + 1 = 0 5x - 8y = -1 .......(i) 3x - 24y/5 +3/5 = 0 3x - 24y/5 = -3/5 15x - 24y = -3 5x - 8y = -1 ........(ii) Lets represent these equations in graphically: 5x - 8y = -1 .......(i)
| x | 0 | -1/5 |
| y | 1/8 | 0 |
| x | 0 | -1/5 |
| y | 1/8 | 0 |
But, this is the same as Equation (i).
Hence the lines represented by Equations (i) and (ii) are coincident.
Therefore, Equations (i) and (ii) have infinitely many solutions.
Example-6 :- Champa went to a ‘Sale’ to purchase some pants and skirts. When her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought.
Solution :-Let us denote the number of pants by x and the number of skirts by y. The algebraic representation is given by the following equations: y = 2x – 2 2x - y = 2 ...........(i) y = 4x – 4 4x - y = 4 ...........(ii) Lets represent these equations in graphically: 2x - y = 2 ............(i)
| x | 2 | 0 |
| y | 2 | -2 |
| x | 0 | 1 |
| y | -4 | 0 |
The two lines intersect at the point (1, 0).
So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e.,
the number of pants she purchased is 1 and she did not buy any skirt.
Example-7 :- Solve the following pair of equations by substitution method: 7x – 15y = 2 and x + 2y = 3.
Solution :- 7x – 15y = 2 .......(i)
x + 2y = 3 ......(ii)
Let us consider the Equation (ii)
x + 2y = 3
x = 3 - 2y ......(iii)
Put the value of x = 3 - 2y in Equation (i), we get
7(3 - 2y) - 15y = 2
21 - 14y - 15y = 2
-29y = 2 - 21
-29y = -19
y = 19/29
Now, put the value of y = 19/12 in Equation (iii)
x = 3 - 2 x 19/29
x = 3 - 38/29
x = (87 - 38)/29
x = 49/29, y = 19/29
Example-8 :- Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Solve by substitution method.
Solution :- Let us denote age of Aftab x and age of his daughter y.
Then the algebraic representation is given by the following equations:
x – 7 = 7 (y – 7)
x - 7 = 7y - 49
x - 7y = -49 + 7
x - 7y = -42 .......(i)
x + 3 = 3 (y + 3)
x + 3 = 3y + 9
x - 3y = 9 - 3
x - 3y = 6 ......(ii)
Let us consider the Equation (ii)
x - 3y = 6
x = 6 + 3y ......(iii)
Put the value of x = 6 + 3y in Equation (i)
6 + 3y - 7y = -42
-4y = -42 - 6
-4y = -48
y = 48/4
y = 12
Put the value of y = 12 in Equation (iii)
x = 6 + 3 x 12
x = 6 + 36
x = 42, y = 12
So, Age of Aftab is 42 and age of his daughter is 12.
Example-9 :- Romila went to a stationery shop and purchased 2 pencils and 3 erasers for cost 9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and she also bought 4 pencils and 6 erasers of the same kind for cost 18. Solve by the Substitution method.
Solution :- Let us denote the cost of 1 pencil by x and one eraser by y.
Then the algebraic representation is given by the following equations:
2x + 3y = 9 .......(i)
4x + 6y = 18 ......(ii)
Let us consider the Equation (ii)
4x + 6y = 18
2x + 3y = 9
2x = 9 - 3y
x = (9 - 3y)/2 .....(iii)
Put the value of x = (9 - 3y)/2 in Equation (i)
2 x (9 - 3y)/2 + 3y = 9
9 - 3y + 3y = 9
0 = 0
This statement is true for all values of y.
However, we do not get a specific value of y as a solution.
Therefore, we cannot obtain a specific value of x.
This situation has arisen bcause both the given equations are the same.
Therefore, Equations (i) and (ii) have infinitely many solutions.
Example-10 :- Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Solve by the Substitution Method.
Solution :- The algebraic representation is given by the following equations:
x + 2y – 4 = 0
x + 2y = 4 ............(i)
2x + 4y – 12 = 0
2x + 4y = 12
x + 2y = 6 ...........(ii)
Let us consider the Equation (ii)
x + 2y = 6
x = 6 - 2y ......(iii)
Put the value of x = 6 - 2y in Equation (i), we get
6 - 2y + 2y = 4
-2y + 2y = 4 - 6
0 = -2
which is a false statement.
Therefore, the equations do not have a common solution.
So, the two rails will not cross each other.
Example-11 :- The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes. Solve by the Elimination Method.
Solution :- Let us denote the incomes of the two person by 9x and 7x and their expenditures by 4y and 3y respectively.
Then the algebraic representation is given by the following equations:
9x - 4y = 2000 .......(i)
7x - 3y = 2000 ......(ii)
For equal of any one variable(x or y), mutiplied by their coefficients following below:
9x - 4y = 2000 ....X 7
7x - 3y = 2000 ....X 9
Now,
63x - 28y = 14000 ....(iii)
63x - 27y = 18000 ....(iv)
By substracting eq (iv) and (iii)
63x - 27y - 63x + 28y = 18000 - 14000
y = 4000
Put the value of y = 4000 in Equation (i)
9x - 4 x 4000 = 2000
9x - 16000 = 2000
9x = 2000 + 16000
9x = 18000
x = 18000/2
x = 9000
So, the solution of the equations is x = 2000, y = 4000.
Therefore, the monthly incomes of the persons are Rs. 18,000 and RS. 14,000, respectively.
Example-12 :- Use elimination method to find all possible solutions of the following pair of linear equations : 2x + 3y = 8 and 4x + 6y = 7.
Solution :- 2x + 3y = 8 .......(i)
4x + 6y = 7 ......(ii)
For equal of any one variable(x or y), mutiplied by their coefficients following below:
2x + 3y = 8 ....X 4
4x + 6y = 7 ....X 2
Now,
8x + 12y = 32 ....(iii)
8x + 12y = 14 ....(iv)
By substracting eq (iv) and (iii)
8x + 12y - 8x - 12y = 32 - 14
0 = 18
which is a false statement.
Therefore, the pair of equations has no solution.
Example-13 :- The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?
Solution :- Let the ten’s and the unit’s digits in the first number be x and y, respectively.
So, the first number may be written as 10 x + y in the expanded form (for example, 23 = 10(2) + 3).
When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s
digit. This number, in the expanded notation is 10y + x (for example, when 23 is
reversed, we get 32 = 10(3) + 2).
According to condition:
(10x + y) + (10y + x) = 66
11x + 11y = 66
x + y = 6 ............(i)
We are also given that the digits differ by 2, therefore,
x - y = 2 .......(ii) or y - x = 2 ......(iii)
Now, adding eq(i) and eq(ii)
x + y + x - y = 6 + 2
2x = 8
x = 8/2
x = 4
Put the value of x = 4 in eq(i)
4 + y = 6
y = 6 - 4
y = 2
x = 4 and y = 2
Now again, adding eq(i) and eq(iii)
x + y + y - x = 6 - 2
2y = 8
y = 8/2
y = 4
Put the value of y = 4 in eq(i)
x + 4 = 6
x = 6 - 4
x = 2
x = 2 and y = 4
If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.
In this case, we get the number 42.
If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.
In this case, we get the number 24.
Thus, there are two such numbers 42 and 24.
Example-14 :- From a bus stand in Bangalore , if we buy 2 tickets to Malleswaram and 3 tickets to Yeshwanthpur, the total cost is Rs. 46; but if we buy 3 tickets to Malleswaram and 5 tickets to Yeshwanthpur the total cost is Rs. 74. Find the fares from the bus stand to Malleswaram, and to Yeshwanthpur.
Solution :- Let x be the fare from the bus stand in Bangalore to Malleswaram, and y to Yeshwanthpur.
Then the algebraic representation is given by the following equations:
2x + 3y = 46
2x + 3y – 46 = 0 .......(i)
3x + 5y = 74
3x + 5y – 74 = 0 ......(ii)
By using of Cross Multipication method:
a₁ = 2, b₁ = 3, c₁ = -46
a₂ = 3, b₂ = 5, c₂ = -74
x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
x/(-222 + 230) = y/(-138 + 148) = 1/(10 - 9)
x/8 = y/10 = 1/1
Now,
x/8 = 1/1; y/10 = 1/1
x = 8; y = 10
Hence, the fare from the bus stand in Bangalore to Malleswaram is Rs. 8 and the fare to
Yeshwanthpur is Rs. 10.
Example-15 :- For which values of p does the pair of equations given below has unique solution? 4x + py + 8 = 0 and 2x + 2y + 2 = 0
Solution :- 4x + py + 8 = 0 .......(i)
2x + 2y + 2 = 0 ......(ii)
By using of Cross Multipication method:
a₁ = 4, b₁ = p, c₁ = 8
a₂ = 2, b₂ = 2, c₂ = 2
Here, a₁/a₂ = b₁/b₂
4/2 ≠ p/2
p ≠ 4
Therefore, for all values of p, except 4, the given pair of equations will have a unique solution.
Example-16 :- For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y – (k – 3) = 0 and 12x + ky – k = 0.
Solution :- kx + 3y – (k – 3) = 0
12x + ky – k = 0
By using of Cross Multipication method:
a₁ = k, b₁ = 3, c₁ = -(k - 3)
a₂ = 12, b₂ = k, c₂ = -k
For a pair of linear equations to have infinitely many solutions :
a₁/a₂ = b₁/b₂ = c₁/c₂
Now,
k/12 = 3/k
k = 36
k² = 6²
k = 6
Therefore, the value of k, that satisfies both the conditions, is k = 6.
For this value, the pair of linear equations has infinitely many solutions.
Example-17 :- Solve the pair of equations: 2/x + 3/y = 13 and 5/x - 4/y = -2.
Solution :- 2/x + 3/y = 13 .....(i)
5/x - 4/y = -2 .....(ii)
Now, Let us consider 1/x = u and 1/y = v
2u + 3v = 13 .......(iii)
5u - 4v = -2 .......(iv)
For equal of any one variable(u or v), mutiplied by their coefficients following below:
2u + 3v = 13 ----X 5
5u - 4v = -2 ----X 2
10u + 15v = 65 ......(v)
10u - 8v = -4 .......(vi)
Substracting of both eq (vi) and eq (v)
10u - 8v - 10u - 15v = -4 - 65
-23v = -69
v = 69/23
v = 3
Put the value of v in eq (iii)
2u + 3 x 3 = 13
2u + 9 = 13
2u = 13 - 9
2u = 4
u = 4/2
u = 2
Now, u = 1/x and v = 1/y.
so, x = 1/2 and y = 1/3
Example-18 :- Solve the following pair of equations by reducing them to a pair of linear equations : 5/(x - 1) + 1/(y - 2) = 2 and 6/(x - 1) - 3/(y - 2) = 1.
Solution :- 5/(x - 1) + 1/(y - 2) = 2 .....(i)
6/(x - 1) - 3/(y - 2) = 1 .....(ii)
Now, Let us consider 1/(x - 1) = u and 1/(y - 2) = v
5u + v = 2 .......(iii)
6u - 3v = 1 .......(iv)
For equal of any one variable(u or v), mutiplied by their coefficients following below:
5u + v = 2 ----X 6
6u - 3v = 1 ----X 5
30u + 6v = 12 ......(v)
30u - 15v = 5 .......(vi)
Substracting of both eq (vi) and eq (v)
30u - 15v - 30u - 6v = 5 - 12
-21v = -7
v = 7/21
v = 1/3
Put the value of v in eq (iii)
5u + 1/3 = 2
5u = 2 - 1/3
5u = 5/3
u = 1/3
Now, u = 1/(x - 1) and v = 1/(y - 2).
so, x - 1 = 3 and y - 2 = 3
x = 3 + 1 and y = 3 + 2
x = 4 and y = 5
Example-19 :- A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km down-stream. Determine the speed of the stream and that of the boat in still water.
Solution :- Let the speed of the boat in still water be x km/h and speed of the stream be y km/h.
Then the speed of the boat downstream = (x + y) km/h,
and the speed of the boat upstream = (x – y) km/h.
Also, time = distance/speed
In the first case, when the boat goes 30 km upstream, let the time taken, in hour,
be t₁. Then, t₁ = 30/(x - y)
Let t₂ be the time, in hours, taken by the boat to go 44 km downstream.
Then, t₂ = 44/(x + y)
The total time taken, t₁ + t₂, is 10 hours. Then Equation is
30/(x - y) + 44/(x + y) = 10 .......(i)
In the second case, in 13 hours it can go 40 km upstream and 55 km downstream. We
get the equation
40/(x - y) + 55/(x + y) = 13 .......(ii)
Now, Let us consider 1/(x - y) = u and 1/(x + y) = v
30u + 44v = 10
30u + 44v - 10 = 0 .......(iii)
40u + 55v = 13
40u + 55v - 13 = 0 .......(iv)
Using Cross multiplication Method:
a₁ = 30, b₁ = 44, c₁ = -10
a₂ = 40, b₂ = 55, c₂ = -13
u/(b₁c₂ - b₂c₁) = v/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)
u/(-572 + 550) = v/(-400 + 390) = 1/(1650 - 1760)
u/(-22) = v/(-10) = 1/(-110)
So, u/22 = 1/110 and v/10 = 1/110
u = 22/110 and v = 10/110
u = 1/5 and v = 1/11
Now, 1/(x - y) = u and 1/(x + y) = v
x - y = 5 ......(v)
x + y = 11 .....(vi)
Solving eq (v) and eq (vi)
x + y + x - y = 11 + 5
2x = 16
x = 16/2
x = 8
Put the value of x = 8 in eq (v)
8 - y = 5
-y = 5 - 8
-y = -3
y = 3
x = 8, y = 3
Hence, the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.
