KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 pdf, free KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 book pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Question 1.
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.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:
Let the present age of father be x’
Let the age of daughter be ‘y’.
Before seven years, age of father is x – 7
Before seven years, age of daughter is y – 7.
Father was 7 times as old as daughter.
∴ x – 7 = 7 (y – 7)
x – 7 = 7y – 49
x – 7y = -49 + 7
x – 7y = – 42
∴ x – 7y + 42 = 0 …………….. (i)
After three years, age of father is x + 3
After three years, age of daughter is y + 3
Three years from now will three times as old as father.
∴ x + 3 = 3(y + 3)
x + 3 = 3y + 9
x – 3y + 3 – 9 = 0
3 – 3y – 6 = 0 ………….. (ii)
∴ Algebraically linear equations :
x – 7y + 42 = 0
x – 3y – 6 =0
If we represent this linear equations through graph, we have
i) x – 7y + 42 = 0
-7y = -x – 42
7y = x + 42
y=x+427

x07
y=x+42767

ii) x – 3y – 6 = 0
-3y = -x + 6
3y = x – 6
y=x−63

x0+6
y=x−63-20
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 1

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebriacally and geometrically.
Solution:
Let the cost of each bat be Rs. ‘x’
Let the cost of each ball be Rs. ‘y’.
3x + 6y = 3900
∴ x + 2y = 1300 ………. (i)
Cost of another bat and ball be x + 3y = 1300 ………….. (ii)
∴Algebriacally
x + 2y = 1300
x + 3y = 1300
And to represent through graph :
i) x + 2y = 1300
2y = 1300 – x
x=1300−x2

x100200
y=1300−x2600500

ii) x + 3y = 1300
3y = 1300 – x
y=1300−x3

x100400
y=1300−x3400900
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 2

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.
Solution:
Algebraically,
If Cost of each kg. of apple be Rs. ‘x’,
Cost of grapes be Rs. ‘y’. Then
2x + y = 160
4x + 2y = 300
To represent geometrically,
i) 2x + y = 160
y = -2x + 160

x2080
y = 2x + 1601200

ii) 4x + 2y = 300
2x + y= 150
y = 150 – 2x

x4060
y = 150 – 2x7030
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 3

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