Let us consider a problem related to our daily life,
The total cost of 2 kg of tea and 3 kg of coffee is Rs. 725. If 3 kg of tea and 2 kg of coffee are bought, the cost becomes Rs. 75 less. What is the per kg price of each?
Here,
\begin{align*} i.\: \text{Combined cost of 2kg tea and 3 kg coffee} &= Rs. 725 \\ ii. \: \text{Total cost of 3 kg tea and 2 kg coffee} &= Rs. 725 - Rs. 75 \\ &= Rs. 650 \\ \end{align*}
Here are two prices which should be found.
Let the cost of 1 kg of tea = Rs. x
& the cost of 1 kg of coffee = Rs. y
Then, 2x + 3y = 725 ........... (i)
3x + 2y = 650 ...........(ii)
There are two variables x and y in the above pair of equations. The given pair of linear equations represents a pair of straight lines while plotting on the graph. The method of obtaining the value of the two variables (order pairs) which satisfy both the equation is known as 'solution of equations'. If the pair of values of two variables involved in a pair of equations satisfies both the equations, those equations are called linear simultaneous equation.
Let, the cost of the book = Rs x, the cost of the pen = Rs y
From 1^{st} condition , x + y = 475 --------------(i)
From 2^{nd} condition , x - y = 75 --------------(ii)
Adding (i) and (ii),
(x + y) + (x - y) = 475 + 75
or, 2x = 550
or x = \( \frac {550}{2}\)
∴ x = 275
Then, Putting the value of x in equation (i)
x + y = 475
or, 275 + y = 475
or, y = 475 - 275
∴ y = 200
∴ Cost of book = Rs 275 and cost of pen = Rs 200
\(
Let, the age of father = x years, the age of son = y years
From 1^{st} condition , x + y = 32 --------------(i)
From 2^{nd} condition , x - y = 24 --------------(ii)
Adding (i) and (ii),
(x + y) + (x - y) = 32 + 24
or, 2x = 56
or x = \( \frac {56}{2}\)
∴ x = 28
Then, Putting the value of x in equation (i)
x + y = 32
or, 28 + y = 32
or, y = 32 - 28
∴ y = 4
∴ Age of father = 28 years and age of son = 4 years
\)
Let, thetwo numbers are x and y.
From 1^{st} condition , x + y = 80--------------(i)
From 2^{nd} condition , x - y = 40--------------(ii)
Adding (i) and (ii),
(x + y) + (x - y) =80 + 40
or, 2x =120
or x = \( \frac {120}{2}\)
∴ x =60
Then, Putting the value of x in equation (i)
x + y =80
or, 60+ y = 80
or, y =80 - 60
∴ y =20
∴ The required numbers are 60 and 40.
\(
Let, thetwo numbers are xand y.
From 1^{st} condition , x+ y = 45^{o}--------------(i)
From 2^{nd} condition ,x- y = 15^{o}--------------(ii)
Adding (i) and (ii),
(x + y) + (x - y) =45^{o}+15^{o}
or, 2x =60^{o}
or x = \( \frac {60^o}{2}\)
∴ x =30^{o}
Then, Putting the value of x in equation (i)
x + y = 45^{o}
or, 30^{o}+ y =45^{o}
or, y = 15^{o}
∴ y =15^{o}
∴ The required anglesare 45^{o} and 15^{o.}
^{\)}
\(
Let, thetwo numbers are x and y.
From 1^{st} condition , x =3y--------------(i)
From 2^{nd} condition , x +y = 32--------------(ii)
Putting value of x in(ii),
3y + y=32
or, 4y=32
or y= \( \frac {32}{4}\)
∴ x =8
Then, Putting the value of yin equation (i)
x = 3y
or, x = 3× 8
or, y = 24
∴ y =20
∴ The required numbers are 8 and 24.
\)
Let the two numbers be x and y and x be the greater one .
From 1^{st} condition, x + y = 14 ---------(i)
From 2^{nd} condition, x= y + 4 ------------(ii)
Putting value of x in equation (i)
x + y = 14
or, y + 4 + y = 14
or, 2y = 14 - 4
or, 2y = 10
or y = \(\frac {10}{2}\)
∴y = 5
Now, putting value of y in equation (ii)
x = y + 4 = 5 + 4 = 9
Hence, the required numbers are 9 and 5.
Let, the number be x.
According to question,
x^{2} - 10 = 54
or, x^{2} = 54 + 10
or,x^{2} = 64
or,x^{2} = (±8)^{2}
∴ x = ±8
Let, breadth of the pond (b) = x m
Then, length of the pond = (x + 10) m
Perimeter of the pond (P) = 120m
or, 2 (l + b) = 120
or, l + b = \( \frac{120}{2}\)
or, l + b = 60
or, ( x + 10) + x = 60
or, 2x + 10 = 60
or, 2x = 60 - 10
or, 2x = 50
or, x = \( \frac{50}{2} \)
or, x = 25
∴Breadth = 25m
and Length = (x +10) = 35 m
Hence, the length and the breadth of the pond are 35m and 25m respectively.
Solve:
2x-3y = 7, 7x + 27 = 12
Solve the following equations:
4x -3y =0, 5x -4y = -1
The sum of two numbers is 59. If the smaller number is less than the bigger one by 7. Find the numbers.
The total cost of TV and Radio is Rs 500. If the watch is cheaper than Radio by 150, find their cost.
Solve: x+3y = 6, x +y = 4
The sum of the ages of a father and his son is 42 years. Six years ago, the father was 5 times as old as his son was. Find their present ages.
Ramesh is three times as old as his son. If the differences of their ages is 24 years. Find their present ages.
The sum of two numbers is 77 and their difference is 13. Find the numbers.
Solve:
x + y = 6, x + 2y = 8
The difference of two numbers is 12. If the greater number is three times the smaller one, find the numbers.
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Rijen
In two digit number, the product of two digit is 14 and if 45 are sutracted from the numbers, the number will be reversed what is the number find it
Mar 18, 2017
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Aasesh
If the age of father was 13 times than his son 1 year ago.Now the age of father is equal to cube of his son age.Find their present age.
Mar 18, 2017
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Resham
the present age of mother is equal to the squre of the age of her daughter after one year . If the age of daughter in 10 year heance will be 1 year less then the age of her mother in 10 years ago then find their present sges
Mar 17, 2017
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the present age of mother is equal to the square of the age of her daughter after one year. If theage of daughter in 10 year hence will be 1 year less than the age of her mother in 10 year ago then their present agess
Mar 17, 2017
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