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

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

(i)  x + y = 5  and   2x – 3y = 4

(ii)  3x + 4y = 10  and  2x – 2y = 2

(iii)  3x – 5y – 4 = 0  and  9x = 2y + 7

(iv)  \(x\over 2\) + \(2y\over 3\) = -1  and  x – \(y\over 3\) = 3

Solution :

(i)  By Elimination Method :

The given equations are

x + y = 5          ……(1)

and   2x – 3y = 4          …….(2)

Multiply equation (1) by 3 and adding with equation (2), we get

5x = 19   \(\implies\)  x = \(19\over 5\)

Now, Put the  value of x in equation (1), we get

\(19\over 5\) + y = 5    \(\implies\)  y = \(6\over 5\)

Hence, x = \(19\over 5\) and y = \(6\over 5\)

By Substitution Method : 

The given equations are

x + y = 5          ……(1)

and   2x – 3y = 4          …….(2)

From equation (1),  y = 5 – x

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

2x – 15 + 3x = 4     \(\implies\)   5x = 4 + 19

\(\implies\)  5x = 19       \(\implies\)    x = \(19\over 5\)

Now, Put the value of x in equation (1), we get

\(19\over 5\) + y = 5    \(\implies\)  y = \(6\over 5\)

Hence, x = \(19\over 5\) and y = \(6\over 5\)

(i)  By Elimination Method :

The given equations are

3x + 4y = 10          ……(1)

and   2x – 2y = 2         …….(2)

Multiply equation (2) by 2 and adding with equation (1), we get

7x = 14   \(\implies\)  x = 2

Now, Put the  value of x in equation (1), we get

3(2) + 4y = 10    \(\implies\)  4y = 10 – 6 = 4   \(\implies\)  y = 1

Hence, x = 2 and y = 1

By Substitution Method : 

The given equations are

3x + 4y = 10          ……(1)

and   2x – 2y = 2          …….(2)

From equation (2),  y = x – 1

Substituting the value of y in equation (1), we get

3x + 4(x – 1) = 10

\(\implies\)  7x = 14       \(\implies\)    x = 2

Now, Put the value of x in equation (1), we get

3(2) + 4y = 10   \(\implies\)    y = 1

Hence,  x = 2 and y = 1

(iii)  By Elimination Method :

The given equations are

3x – 5y – 4 = 0    \(\implies\)   3x – 5y = 4          …..(1)

and  9x = 2y + 7  \(\implies\)  9x – 2y = 7         …….(2)

Multiplying equation (1) by 3 and subtracting equation (2) from equation (3), we get

-13y = 5        \(\implies\)        y = \(-5\over 13\)

Now, Put the value of y in equation (1), we get

3x – 5(\(-5\over 13\)) = 4

\(\implies\)   3x = \(52 – 25\over 13\)

\(\implies\)  3x = \(27\over 13\)

\(\implies\)  x = \(9\over 13\)

Hence, x = \(9\over 13\)  and  y = \(-5\over 13\)

By Substitution Method :

The given equations are

3x – 5y – 4 = 0    \(\implies\)   3x – 5y = 4          …..(1)

and  9x = 2y + 7  \(\implies\)  9x – 2y = 7         …….(2)

From equation (2),    y = \(9x – 7\over 2\)

Substituting the value of y in equation (1), we get

3x – 5(\(9x – 7\over 2\)) = 4     \(\implies\)    6x – 45x + 35 = 8

\(\implies\)  -39x = 8 – 35    \(\implies\)   -39x = -27

\(\implies\)  x = \(9\over 13\)

Now, Put the value of x in equation (2), we get

3 \(\times\) \(9\over 13\) – 5y = 4

\(\implies\)  5y = \(-25\over 13\)

\(\implies\)  y = \(-5\over 13\)

Hence, x = \(9\over 13\)  and  y = \(-5\over 13\)

(iv)  By Elimination Method :

The given equations are

\(x\over 2\) + \(2y\over 3\) = -1      \(\implies\)   3x + 4y = -6       ………(i)

and  x – \(y\over 3\) = 3       \(\implies\)       3x – y = 9              ………..(ii)

Multiplying equation (2) by  4 and adding to equation (1), we get

15x = 30    \(\implies\)   x = 2

Now, put  x = 2 in equation (2), we get

3(2) – y  = 9    \(\implies\)   – y = 9 – 6 = 3

\(\implies\)  y = -3

Hence, x = 2 and y = -3

By Substitution Method :

The given equations are

\(x\over 2\) + \(2y\over 3\) = -1      \(\implies\)   3x + 4y = -6       ………(i)

and  x – \(y\over 3\) = 3       \(\implies\)       3x – y = 9              ………..(ii)

From equation (2),  y = 3x – 9

Putting the value of y in equation (1), we get  y = 3x – 9

3x + 4(3x – 9) = -6     \(\implies\)    3x + 12x – 36 = -6

\(\implies\)   15x = 30   \(\implies\)  x = 2

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

3(2) – y = 9  \(\implies\)   – y = 9 – 6 = 3

\(\implies\)  y = -3

Hence, x = 2 and y = -3

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