Q.8. Which of these is a linear
equation in two variables *
a) 5x-2y=0
b) X+x2-2y+8
c)x-2y+10= x2 +y
d)3x-2y+1=0
Answers
Answer:
that is
Step-by-step explanation:
EXERCISE: 4.1
1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y).
Sol: Let the cost of a notebook = Rs x
The cost of a pen = y
According to the condition, we have
[Cost of a notebook] = 2 × [Cost of a pen]
i.e. [x] = 2 × [Y]
or x = 2y
or x – 2y = 0
Thus, the required linear equation is × – 2y = 0.
2. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) (ii) (iii) –2x + 3y = 6 (iv) x = 3y
(v) 2x = –5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x
Sol: (i) We have
Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and
(ii) We have
Comparing with ax + bx + c = 0, we get
Note: Above equation can also be compared by:
Multiplying throughout by 5,
or 5x – y – 50 = 0
or 5(x) + (–1)y + (–50) = 0
Comparing with ax + by + c = 0, we get a = 5, b = –1 and c = –50.
(iii) We have –2x + 3y = 6
⇒ –2x + 3y – 6 = 0
⇒ (–2)x + (3)y + (–6) = 0
Comparing with ax + bx + c = 0, we get a = –2, b = 3 and c = –6.
(iv) We have x = 3y
x – 3y = 0
(1)x + (–3)y + 0 = 0
Comparing with ax + bx + c = 0, we get a = 1, b = –3 and c = 0.
(v) We have 2x = –5y
⇒ 2x + 5y =0
⇒ (2)x + (5)y + 0 = 0
Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.
(vi) We have 3x + 2 = 0
⇒ 3x + 2 + 0y = 0
⇒ (3)x + (10)y + (2) = 0
Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y – 2 = 0
⇒ (0)x + (1)y + (–2) = 0
Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = –2.
(viii) We have 5 = 2x
⇒ 5 – 2x = 0
⇒ –2x + 0y + 5 = 0
⇒ (–2)x + (0)y + (5) = 0
Comparing with ax + by + c = 0, we get a = –2, b = 0 and c = 5.
EXERCISE: 4.2
1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions
Sol: Option (iii) is true because a linear equation has an infinitely many solutions.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y
Sol: (i) 2x + y = 7
When x = 0, 2(0) + y = 7
⇒ 0 + y = 7
⇒ y =7
∴ Solution is (0, 7).
When x = 1, 2(1) + y = 7
⇒ y = 7 – 2
⇒ y = 5
∴ Solution is (1, 5).
When x = 2, 2(2) + y = 7
⇒ y = 7 – 4
⇒ y = 3
∴ Solution is (2, 3).
When x = 3, 2(3) + y = 7
⇒ y = 7 – 6
⇒ y = 1
∴ Solution is (3, 1).
(ii) πx + y = 9
When x = 0 π(0) + y = 9
⇒ y = 9 – 0
⇒ y = 9
∴ Solution is (0, 9).
When × = 1, π(1) + y = 9
⇒ y = 9 – π
∴ Solution is {1, (9 – π)}
When x = 2, π(2) + y = 9
⇒ y = 9 – 2π
∴ Solution is {2, (9 – 2π)}
When × = –1, π(–1) + y = 9
⇒ – π + y = 9
⇒ y = 9 + π
∴ Solution is {–1, (9 + π)}
(iii) x = 4y
When x = 0, 4y = 0
⇒ y = 0
∴ Solution is (0, 0).
When x = 1, 4y = 1
⇒ y = 0
∴ Solution is (0, 0)
When x = 4, 4y = 4
⇒
3. Check which of the following are solutions of the equation x – 2y = 4 and which are not:
(i) (0, 2) (ii) (2, 0) (iii) (4, 0)
(iv) (v) (1, 1)
Sol: (i) (0, 2) means x = 0 and y = 2
Putting x = 0 and y = 2 in x – 2y = 4, we have
L.H.S. = 0 – 2(2) = –4
But R.H.S. = 4
L.H.S. ≠ R.H.S.
∴ x = 0, y = 0 is not a solution.
(ii) (2, 0) means x = 2 and y = 0
∴ Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H.S. = 2 – 2(0) = 2 – 0 = 2
But R.H.S. = 4
L.H.S. ≠ R.H.S.
∴ (2, 0) is not a solution.
(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = 0 in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4
But R.H.S. = 4
L.H.S. ≠ R.H.S.
∴ (4, 0) is a solution.
(v) (1, 1) means x = 1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
L.H.S. = 1 – 2(1) = 1 – 2 = –1
But R.H.S. = 4
⇒ L.H.S ≠ R.H.S.
∴ (1, 1) is not a solution.
4. Find the value of k, if x = 2, y = 1 is a solution fo the equation 2x + 3y = k.
Sol: We have 2x + 3y = k
Putting x = 2 and y = 1 in 2x + 3y = k, we get
2(2) + 3(1) = k
⇒ 4 + 3 = k
⇒ 7 = k
Thus, the required value of k = 7.
EXERCISE: 4.3
1. Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4 ii) x – y = 2 (iii) y = 3x (iv) 3 = 2x + y
Sol: (i) x + y = 4 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y = 4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:
Plot the ordered pairs (0, 4), (1, 3) and (2, 2) on the graph paper. Joining these points, we get a line AB as shown below.
Thus, the line AB is the required graph of x + y = 4.
(ii) x – y = 2
⇒ y = x – 2
If we have x = 0,then y = 0 – 2 = –2
x = 1,then y = 1 – 2 = –1
x = 2, then y = 2 – 2 = 0
∴ We have the following table:
Plot the ordered pairs (0, –2), (1, –1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown below:
Thus, the line PQ is required graph of x – y = 2.
(iii) y = 3x
If x = 0, then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) ⇒ y = 3
x = –1, then y = 3(–1) ⇒ y = –3
∴ We get the following table:
Plot the ordered pairs (0, 0), (1, 3) and (–1, –3) on the graph paper. Joining these points, we get the straight line LM.
Thus, LM is the required graph of y = 3x.
Note: The graph of the equation of the form y – kx is a straight line which always passes through the origin.
(iv) 3 = 2x + y ⇒ y – 3 – 2x
∴ If
Answer:
b because it is the linear equation In two variables