Chapter 3: Pair of Linear Equations in Two Variables

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Q
Pair of linear equations in two variables CBSE NCERT Solutions

Question:

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2+ 3= 7

(a − b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no

solution?

3= 1

(2k − 1) x + (k − 1) y = 2k + 1

Answer:

Solutions:

(i) We compare equation 2x + 3y – 7 = 0 with a1x + b1y + c1 = 0  and (a − b) x + (a + b) y − 3a – b + 2 = 0 with a2x + b2y + c2 = 0.

We get a1 = 2, b1= 3,  c1 = -7 and a2 = (a-b), b2 = (a+b), c2 = 2- b-3a.

Linear equations have infinitely many solutions if     

\Rightarrow \frac{2}{a-b} = \frac{3}{a+b}=\frac{-7}{2-b-3a}

\Rightarrow \frac{2}{a-b} = \frac{3}{a+b}   and   \Rightarrow \frac{3}{a+b} = \frac{-7}{2-b-3a}

2a + 2b = 3a − 3b and 6 − 3b − 9a = −7a − 7b

 = 5b… (1) and −2a = −4b – 6… (2)

Putting (1) in (2), we get

−2 (5b) = −4b – 6

10b + 4b = −6

6b = –6  = 1

Putting the value of b in (1), we get

= 5= 5 (1) = 5

Therefore, = 5 and = 1.

(ii) We compare (3x + y – 1 = 0) with a1x + b1y + c1 = 0 and (2k − 1)x + (k − 1)y −2k – 1 = 0) with a2x + b2y + c2 = 0.

We get a1 = 3, b1= 1,  c1 = -1 and a2 = (2k-1), b2 = (k-1), c2 = −2k – 1

Linear equations have no solution if   \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}    .

\Rightarrow \frac{3}{2k-1} = \frac{1}{k-1} \neq \frac{-1}{-2k-1}

\Rightarrow \frac{3}{2k-1} = \frac{1}{k-1}

3 (k − 1) = 2k – 1

3k – 3 = 2k − 1

 = 2

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