Linear Equations in Two Variables – Exercise 4.2 – (MATHEMATICS) – 9th Class

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Exercise 4.2

1. Which on of the following option is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii)  only two solutions,
(iii) infinitely many solutions.

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Sol. – (iii) As each linear equation in two variables has infinitely many solutions. Further, for every x there is a corresponding value of y and vice – versa.

2. Write four solutions for each of the following equations:
(i) 2x + y = 7

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Sol. –
Consider equation:   2x + y = 7    ⇒   y  = 7 – 2x
Let x = 0, then y = 7. Solution is x = 0, y = 7
Let x = 1, then y = 5. Solution is x = 1, y = 5
Let x = 2, then y = 3. Solution is x = 2, y = 3
Let x = 3, then y = 1.  Solution is x = 3,  y = 1.

(ii) πx + y = 9

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Sol. –
Consider equation: πx + y = 9   ⇒   y = 9  –  πx
Let x = 0, then y = 9.  Solution is x = 0, y = 9
Let x = 1, then y = 9  – π .Solution is x = 1, y =9 – π
Let x = 2, then y = 9 – 2π. Solution is x = 2, y = 9 – 2π.
Let x = 3, then y = 9  – 3π Solution is x = 3, y =9 – 3π.

(iii) x = 4y,

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Sol. –
Consider equation: x = 4y.
Let y = 0,   then x = 0.         Solution is x =  0, y  = 0
Let y = 1, then x = 4.  Solution is x = 4,  y = 1
Let y = – 1, then x = – 4.  Solution is x = – 4, y = – 1
Let y = 2, then x = 8.  Solution is x = 8, y = 2.

3. Check which of the following are solutions of the equation it x – 2y = 4 and which are not:
(i) (0, 2)

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Sol. – For (0,2) substituting x = 0, y = 2 in (A), we get 0 – 4 = 4  ⇒   – 4  = 4 , not true.
Hence, (0,2) is not a solution.

(ii) (2, 0)

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Sol. – For (2,0)  substituting x = 2, y = 0 in (A), we get
2 – 0 = 4  ⇒  2 = 4, not true.
Hence, (2,0) is not a solution.

(iii) (4, 0)

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Sol. –
For (4,0), substituting x = 4,   y = 0 in (A), we get
4 –  0 = 4   ⇒   4 = 4, true.
Hence, (4,0) is a solution.

(iv)  (\sqrt 2 ,4\sqrt {2)}

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Sol. –
For  (\sqrt 2 ,4\sqrt {2)} , substituting x = \sqrt 2 , y = 4 \sqrt 2  in (A), we get
\sqrt 2 –  8 \sqrt 2  = 4 ⇒  – 7 \sqrt 2 = 4, not true.
Hence, (\sqrt 2 ,4\sqrt {2)}  is not a solution.

(v) (1, 1)

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Sol. –
For (1, 1), substituting x = 1,  y = 1  in (A), we get
1 – 2 = 4 ⇒  – 1 = 4, not true.
Hence, (1,1) is not a solution.

4. Find the value of ℜ, if x = 2, y = 1 is a solution. of the equation 2x + 3y ,= ℜ.

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Sol. If x = 2, y = 1 is a solution of the equation 2x + 3y = ℜ  then
4 + 3 = ℜ    ⇒  ℜ  = 7

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