by using variables x and y form correct linear equations in two variables x+y+z=9, x-y=p, x=y-4, y=z
Answers
Step-by-step explanation:
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.