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
