Question

Question 9 (i)
Solve the following pairs of equation
$x+y=3.3,\frac{0.6}{3x-2y}=-1,3x-2y\ne 0$

Answer 1

Given pair of linear equations are is
x + y = 33
$\frac{06}{2x-y}=-1\Rightarrow 06=-3x+2y\Rightarrow 3x-2y=-06$
Now, multiplying Eq. (i) by 2 and then adding with Eq. (ii), we get
2x + 2y= 6.6
3x – 2y = - 0.6
$5x=6\Rightarrow x=\frac{6}{5}=1.2$
Now, put the value of x in Eq. (i), we get
1.2 + y = 3.3
$\Rightarrow$ y = 33 – 1.2
$\Rightarrow$ y = 2.1
Hence, the required values of x and y are 1.2 and 2.1, respectively