Chapter 3: Pair of Linear Equations in Two Variables

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Pair of linear equations in two variables CBSE NCERT Solutions

Question:

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14;   x – y = 4

(ii) s – t = 3;     frac{s}{3} + frac{t}{2} = 6 

(iii) 3x – y = 3;  9x − 3y = 9

(iv) 0.2x + 0.3y = 1.3;  0.4x + 0.5y = 2.3

(v) sqrt{2}x+ sqrt{3}y = 0 sqrt{3}x - sqrt{8}y = 0

(vi)  frac{3x}{2} - frac{5y}{3} = -2 ;  frac{x}{3} + frac{y}{2} = frac{13}{6}

Answer:

(i) x + y = 14 …(1)

     x – y = 4  … (2)

From equation (2),   x – y = 4 x = 4 + y 

Putting  x = 4 + y  in equation (1), we get

     4 + y + y = 14

2y = 10  = 5

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

   x + 5 = 14

 = 14 – 5 = 9

Therefore, x = 9 and y = 5

(ii) s – t = 3 … (1)

       frac{s}{3} + frac{t}{2} = 6 …(2)

Using equation (1), we can say that s = 3 + t

Putting s = 3 + t in equation (2), we get

\frac{3+t}{3} + \frac{t}{2} = 6

\Rightarrow \frac {6 + 2t + 3t }{6} = 6

6 + 2t + 3t = 36

5t + 6 = 36

5t = 30  = 6

Putting value of t = 6 in equation (1), we get

– 6 = 3  = 3 + 6 = 9

Therefore, t = 6 and s = 9

(iii) 3x – y = 3 … (1)

     9x − 3y = 9 … (2)

Comparing equation 3x – y = 3 with a_1x + b_1y + c_1 = 0  and equation 9x − 3y = 9 with a_2x + b_2y + c_2 = 0 .

We get  a_1 = 3 , b_1 = -1 , c_1 = -3,a_2 = 9 , b_2 = -3 and c_2 = -9

Here,

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

Therefore, we have infinite many solutions for x and y

(iv) 0.2x + 0.3y = 1.3 … (1)

       0.4x + 0.5y = 2.3 … (2)

Using equation (1), we can say that

0.2x = 1.3 − 0.3y

 =  \frac{1.3-0.3y}{0.2}

Putting this in equation (2), we get

0.4 (\frac{1.3-0.3y}{0.2} )+ 0.5y = 2.3

2.6 0.6y + 0.5y = 2.3

0.1y = −0.3  = 3

Putting value of y in (1), we get

0.2x + 0.3 (3) = 1.3

0.2x + 0.9 = 1.3

0.2x = 0.4  = 2

Therefore, x = 2 and y = 3

(v)   sqrt{2}x+ sqrt{3}y = 0    ….(1)

      sqrt{3}x - sqrt{8}y = 0  ….(2)

Using equation (1), we can say that

x = \frac{-\sqrt{3}y}{\sqrt2}

Putting this in equation (2), we get

\sqrt{3} \left (\frac{-\sqrt{3}y}{\sqrt2} \right ) - \sqrt{8}y = 0

 \frac{-{3}y}{\sqrt2} - \sqrt{8}y = 0

 y\left ( \frac{-{3}}{\sqrt2} - \sqrt{8} = 0 \right )

 = 0

Putting the value of y in (1), we get x = 0

Therefore, x = 0 and y = 0

(vi)  frac{3x}{2} - frac{5y}{3} = -2 … (1)

         frac{x}{3} + frac{y}{2} = frac{13}{6} … (2)

Using equation (2), we can say that

x = \left ( \frac{13}{6} - \frac{y}{2}\right ) \times 3
\Rightarrow x = \frac{13}{2} - \frac{3y}{2}

Putting this in equation (1), we get

\frac{3}{2}\left ( \frac{13}{2} - \frac{3y}{2} \right ) - \frac{5y}{3} = \frac{-2}{1}

 \frac{39}{4} - \frac{9y}{4} - \frac{5y}{3} = -2

 \frac{-27y-20y}{12} = -2 - \frac{39}{4}

 \frac{-47y}{12} = - \frac{-8-39}{4}

\frac{-47y}{12} = \frac{-47}{4}

 = 3

Putting the value of y in equation (2), we get

    \frac{x}{3} + \frac{3}{2}= \frac{13}{6}

\Rightarrow \frac{x}{3} = \frac{13}{6} - \frac{3}{2} = \frac{13-9}{6} = \frac{4}{6} = \frac{2}{3}

\Rightarrow \frac{x}{3} = \frac{2}{3}

x = 2

Therefore, = 2 and = 3

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