Chapter 3: Pair of Linear Equations in Two Variables

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Pair of linear equations in two variables CBSE NCERT Solutions

Question:

Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i)  x – 3y – 3 = 0

     3x – 9y – 2 = 0

(ii) 2x + y = 5

     3x + 2y = 8

(iii) 3x − 5y = 20

      6x − 10y = 40

(iv) x − 3y – 7 = 0

      3x − 3y – 15 = 0

Answer:

(i) − 3y – 3 = 0

   3x − 9y – 2 = 0

We compare equation − 3– 3 = 0 with a1x +b1y + c1 = 0 and 3− 9– 2 = 0 with a2x +b2y + c2 = 0.

We get a1 = 1, b1= -3,  c1 = -3 and a2 = 3, b2 = -9, c2 = -2.

Here  \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}    this means that the two lines are parallel.

Therefore, there is no solution for the given equations i.e. it is inconsistent.

(ii) 2x + y = 5

    3x + 2y = 8

Comparing equation 2= 5 with a1x + b1y + c1 = 0 and 3+ 2= 8 with a2x + b2y + c2 = 0.

We get a1 = 2, b1= 1,  c1 = -5 and a2 = 3, b2 = 2, c2 = -8

Here   \frac{a_1}{a_2} \neq \frac{b_1}{b_2}  this means that there is a unique solution for the given equations.

\frac{x}{(-8)(1)-(2)(-5)} = \frac{y}{(-5)(3)-(-8)(2)} = \frac{1}{(2)(2)-(3)(1)}

\Rightarrow \frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}

\Rightarrow \frac{x}{2}=\frac{y}{1}=\frac{1}{1}

 = 2 and = 1

(iii) 3x − 5y = 20

     6x − 10y = 40

Comparing equation 3− 5= 20 with a1x + b1y + c1 = 0 and 6− 10= 40 with a2x + b2y + c2 = 0.

We get a1 = 3, b1= -5,  c1 = -20 and a2 = 6, b2 = -10, c2 = -40

Here,    \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

It means lines coincide with each other.

Hence, there are infinitely many solutions.

 

(iv) − 3y – 7 = 0

  3x − 3y – 15 = 0

Comparing equation − 3– 7 = 0 with a1x + b1y + c1 = 0  and 3− 3– 15 = 0 with a2x + b2y + c2 = 0.

We get a1 = 1, b1= -3,  c1 = -7 and a2 = 3, b2 = -3, c2 = - 15

Here   \frac{a_1}{a_2} \neq \frac{b_1}{b_2}   this means that we have a unique solution for these equations.

\frac{x}{(-3)(-15)-(-3)(-7)} = \frac{y}{(-7)(3)-(-15)(1)} = \frac{1}{(-3)(1)-(-3)(3)}

\Rightarrow \frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}

\Rightarrow \frac{x}{24}=\frac{y}{-6}=\frac{1}{6}

 = 4 and = –1

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