Question

Determine, by drawing graphs, whether the following system of linear equations has a unique solution or not:
(i) 2x – 3y = 6 and x + y = 1
(ii) 2y = 4x – 6 and 2x = y + 3

Answer 1

(1) unique solution

(2) Not unique solution

Step-by-step explanation:

$(i)2x-3y=6-(i)\\x+y=1-(ii)\\2x-3y-6=0-(iii)\\x+y-1=0(iv)$

$a_{1} =2,a_2=1,b_1=-3,b_2=1,c_1=-6,c_1=-1$

Condition for unique solution

$\frac{a_1}{a_2}\neq \frac{b_1}{b_2}\\=\frac{2}{1}\neq \frac{-3}{1}\\(ii)2y=4x-6-(i)\\2x=y+3-(ii)\\-4x+2y+6=0-(iii)\\2x-y-3=0-(iv)$

take common 2 in equation-(iii)

$-2x+y+3-(v)$

Now compare equation (iv) and (v)

$2x-y-3=0-(iv)\\-2x+y+3=0-(v)\\\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\\\frac{2}{-2}=\frac{-1}{1}=\frac{-3}{3}$

it is not unique answer,because

$\frac{a_1}{a_2} =\frac{b_1}{b_2}$