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:

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5, 2x – 3y = 4

(ii) 3x + 4y = 10, 2x – 2y = 2

(iii) 3− 5– 4 = 0, 9= 2+ 7

(iv) frac{x}{2} + frac{2y}{3} = -1 , x - frac{y}{3} = 3

Answer:

(i) x + y = 5 … (1)

2x – 3y = 4 … (2)

Elimination method:

Multiplying equation (1) by 2, we get

2x + 2y = 10 … (3)

2x − 3y = 4 … (2)

Subtracting equation (2) from (3), we get

5y = 6  =  \frac{5}{6}

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

\frac{6}{5}  = 5

 = 5 − \frac{6}{5}  =  \frac{19}{5}

Therefore, x = \frac{19}{5}   and y =  \frac{6}{5} 

Substitution method:

+ y = 5 … (1)

2x − 3y = 4 … (2)

From equation (1), we get,

= 5 − y

Putting this in equation (2), we get

2 (5 − y) − 3y = 4

10 2y − 3y = 4

5y = 6  \frac{6}{5}

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

= 5 − \frac{6}{5}  =   \frac{19}{5}

Therefore, x = \frac{19}{5}   and y =   \frac{6}{5}

(ii) 3x + 4y = 10… (1)

      2x – 2y = 2… (2)

Elimination method:

Multiplying equation (2) by 2, we get

4x − 4y = 4 … (3)

3x + 4y = 10 … (1)

Adding (3) and (1), we get

7x = 14  = 2

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

3 (2) + 4y = 10

4y = 10 – 6 = 4

 = 1

Therefore, x = 2 and y = 1

Substitution method:

3x + 4y = 10… (1)

2x − 2y = 2… (2)

From equation (2), we get

2x = 2 + 2y

 = 1 + y … (3)

Putting this in equation (1), we get

3 (1 + y) + 4y = 10

3 + 3y + 4y = 10

7y = 7  = 1

Putting the value of y in (3), we get  

= 1 + 1 = 2

Therefore, x = 2 and y = 1

(iii) 3− 5– 4 = 0 … (1)

       9= 2+ 7… (2)

Elimination method:

Multiplying (1) by 3, we get (3)

9x − 15y – 12 = 0… (3)

9x − 2y – 7 = 0… (2)

Subtracting (2) from (3), we get

− 13y – 5 = 0

− 13y = 5

 = \frac {-5}{13}  

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

3x - 5 \left ( \frac {-5}{13} \right ) - 4 = 0

\Rightarrow 3x = 4 - \frac {-25}{13} = \frac{52 - 25}{13} = \frac {27}{13} 

\Rightarrow x = \frac {27}{13 \times 3} = \frac{9}{13}

Therefore, x = \frac{9}{13}   and  y = \frac{-5}{13} 

Substitution Method:

3x − 5y – 4 = 0 … (1)

9x = 2y + 7… (2)

From equation (1), we can say that

3x = 4 + 5y

\Rightarrow x = \frac{4 + 5y}{3}

Putting this in equation (2), we get

9 \left ( \frac{4 + 5y}{3} \right )  − 2y = 7

12 + 15y − 2y = 7

13y = −5  \frac{-5}{13}

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

3x - 5\left ( \frac{-5}{13} \right ) - 4 = 0

\Rightarrow 3x = 4 - \frac{-25}{13} = \frac{52 - 25}{13} = \frac{27}{13}   

\Rightarrow x = 4 - \frac{27 }{13\times 3} = \frac{9}{13}

\Rightarrow x = 4 - \frac{27 }{13\times 3} = \frac{9}{13}

Therefore , x = \frac{9}{13}  and   y = \frac{-5}{13}

(iv) 

\frac{x}{2} + \frac{2y}{3} = -1… (1)

x - \frac{y}{3} = 3… (2)

Elimination method:

Multiplying equation (2) by 2, we get

2x - \frac{2y}{3} = 6 … (3)

\frac{x}{2} + \frac{2y}{3} = -1 … (1)

Adding (3) and (1), we get

\frac{5x}{2} = 5

 = 2

Putting the value of x in (2), we get

2 - \frac{y}{3} = 3

 = − 3

Therefore, x = 2 and y = −3

Substitution method:

\frac{x}{2} + \frac{2y}{3} = -1 … (1)

x - \frac{y}{3} = 3 … (2)

From equation (2), we can say that 

x = 3 + \frac{y}{3} = \frac{9 + y}{3}

Putting this in equation (1), we get

\frac{9 + y}{6} + \frac{2y}{3} = -1

\frac{9 + y + 4y}{6} = -1 

5y + 9 = − 6

5y = − 15 = − 3

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

\frac{x}{2} + \frac{2}{3}(-3) = -1

 = 2

Therefore, x = 2 and y = −3

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