(i) By elimination method
x + y = 5 …………….(1)
2x – 3y = 4 ………….(2)
Multiplying equation (1) by 2, we obtain
2x + 2y = 10 .............(3)
Subtracting equation (2) from equation (3), we obtain
5y = 6
y = \(\frac{6}{5}\)
Substituting the value in equation (1), we obtain
x = 5 - \(\frac{6}{5}\) = \(\frac{19}{5}\)
∴ x = \(\frac{19}{5}\) , y =\(\frac{6}{5}\)
By substitution method
From equation (1), we obtain
x = 5-y ...............(4)
Putting this value in equation (2), we obtain
2(5-y) - 3y = 4
-5y = -6
y= \(\frac{6}{5}\)
Substituting the value in equation (4), we obtain
x = 5 -\(\frac{6}{5}\) = \(\frac{19}{5}\)
∴ x = \(\frac{19}{5}\) , y = \(\frac{6}{5}\)
(ii) By elimination method
3x + 4y = 10 .......(1)
2x - 2y + 2............(2)
Multiplying equation (2) by 2, we obtain
4x - 4y = 4 .............(3)
Adding equation (1) and (3), we obtain
7x = 14
x =2
Substituting in equation (1), we obtain
6+ 4y = 10
4y = 4
y=1
Hence, x = 2, y = 1
By substitution method
From equation (2), we obtain
x= 1+y .........(4)
Putting this value in equation (1), we obtain
3(1+y) +4y = 10
7y = 7
y=1
Substituting the value in equation (4), we obtain
x = 1 +1 = 2
∴ x = 2, y =1
(iii) By elimination method
3x - 5y -4 = 0 .............(1)
and 9x = 2y + 7
9x -2y -7 = 0...............(2)
Multiplying equation (1) by 3, we obtain
9x - 15y -12 =0............(3)
Subtracting equation (3) from equation (2), we obtain
13y = -5
y= \(-\frac{5}{13}\)
Substituting in equation (1), we obtain
3x +\(\frac{25}{13}\) -4 =0
3x = 27 /13
x= \(\frac{9}{13}\)
∴ x = \(\frac{9}{13}\), y = \(-\frac{5}{13}\)
By substitution method
From equation (1), we obtain
x = \(\frac{5y +4}{ 3}\).............(4)
Putting this value in equation (2), we obtain
9(\(\frac{5y +4}{ 3}\)) -2y -7 =0
13y = -5
y= \(-\frac{5}{13}\)
Substituting the value in equation (4), we obtain
x =\(\frac{5 (-\frac{5}{13}) + 4 }{3}\)
x = \(\frac{9}{13}\)
∴ x = \(\frac{9}{13}\), y = \(-\frac{5}{13}\)
(iv) By elimination method
\(\frac{x}{2} + \frac{2y} {3} = -1\)
\(3x + 4y = -6\).............(1)
and
\(x- \frac{y}{3} = 3\)
\(3x -y = 9\).............(2)
Subtracting equation (2) from equation (1), we obtain
5y = -15
y= -3 ............(3)
Substituting this value in equation (1), we obtain
3x - 12 = -6
3x= 6
x = 2
Hence, x = 2, y = −3
By substitution method
From equation (2), we obtain
\(x = \frac{y+9}{3}\) .............(4)
Putting this value in equation (1), we obtain
3(\(\frac{y+9}{3}\)) + 4y + -6
5y = −15
y = -3
Substituting the value in equation (4), we obtain
x =\(\frac{-3 + 9}{3}\) = 2
∴ x = 2, y = −3