Q.1 Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution
(ii) only two solution
(iii) infinitely many solutions
Sol.
Given equation is y = 3 x + 5
For y = 0 , 3x + 5 = 0
Therefore is one of solution.
For x = 0, y = 0 + 5 = 5
Therefore, (0, 5) is another solution.
For x = 1, y = 3 × 1 + 5 = 8
Therefore (1, 8) is another solution.
Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with
this value of y constitutes another solution of the given equation. So, there is no end to different
solutions of a linear equation in two variables.
Therefore, A linear equation in two variables has infinitely many solutions.
Q.2 Write four solutions for each of the following equations :
(i) 2x + y = 7
(ii)
(iii) x = 4y
Sol.
(i) The given equation can be written as y = 7 – 2x.
For x = 0 , y = 7 – 2 × 0 = 7 – 0 = 7
For x = 1, y = 7 – 2 × 1 = 7 – 2 = 5
For x = 2, y = 7 – 2 × 2 = 7 – 4 = 3
For x = 3, y = 7 – 2 × 3 = 7 – 6 = 1
Therefore the four solutions of the given equation are (0, 7), (1, 5) , (2, 3) and (3, 1) .
(ii) The given equation can be written as
For x = 0, y = 9 – 0 = 9
For x = 1,
For x = 2 ,
For x = 3,
Therefore the four solutions of the given equation are (0, 9),
(1,9) , (2,9 ) and (3, 9 )
(iii) The given equation can be written as x = 4y
For x = 0, y = 0
For x = 1 ;
For x = 2 ;
For x = 3 ;
Therefore the four solutions of the given equation are
Q.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) Putting x = 0 , y = 2 in L.H.S. of x – 2y = 4, we have
L.H.S. = 0 – 2 × 2 = – 4 R.H.S.
Therefore x = 0 , y = 2 is not its solution.
(ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have
L.H.S. = 2 – 2 × 0 = 2 – 0 = 2 R.H.S.
Therefore x = 2, y = 0 is not its solution.
(iii) Putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 4 – 0 = 4 = R.H.S.
Therefore x = 4, y = 0 is its solution.
(iv) Putting in the L.H.S. of x – 2y = 4, we have
L.H.S.
R.H.S.
Therefore is not its solution.
(v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 1 – 2 × 1 = 1 – 2 = – 1 R.H.S.
Therefore x = 1, y = 1 is not its solution.
Q.4 Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Sol.
If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.
Therefore 2 × 2 + 3 × 1 = k k = 4 + 3 = 7.
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