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Which of the following pairs of linear equations are consistent/inconsistent If consistent, obtain the solution graphically:
(i) $x + y = 5, 2x + 2y = 10$
(ii) $x – y = 8, 3x – 3y = 16$
(iii) $2x + y – 6 = 0, 4x – 2y – 4 = 0$
(iv) $2x – 2y – 2 = 0, 4x – 4y – 5 = 0$.
To do:
We have to find out whether the given pairs of linear equations are consistent or inconsistent and obtain the solution graphically.
Solution:
(i) $x+y-5=0$
$2x+2y-10=0$
$x+y=5\ \ ...( i)$
$2x+2y=10\ \ ...( ii)$
For equation $( i)$,
$x+y=5$
$\Rightarrow y=5-x$
$x$ | 0 | 5 |
$y$ | 5 | 0 |
Plot point $( 0,\ 5)$ and $( 5,\ 0)$ on a graph and join then to get equation
$x+y=5$
For equation $( ii)$,
$2x+2y=10$
$\Rightarrow y=\frac{10-2x}{2}$
$x$ | 5 | 5 |
$y$ | 0 | 0 |
Plot point $( 5,\ 0)$ and $( 0,\ 5)$ on a graph and join them to get equation $2x+2y=0$
From the above figure, we can observe that the lines are coincident
Therefore, the equations have infinite possible solutions
Here, $a_1=1,\ b_1=-1,\ c_1=8$ and $a_2=3,\ b_2=-3,\ c_2=16$.
$\frac{a_1}{a_2}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}$
$\frac{c_1}{c_2}=\frac{8}{16}=\frac{1}{2}$
Here, we find that $\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
Thus, given pair of linear equations has no solution.
Therefore, the given pair of linear equations is inconsistent.
(iii) $2x+y-6=0$
$4x-2y-4=0$
$2x+y=6\ \ ...( i)$
$4x-2y=4\ \ ...( ii)$
For equation $( i)$,
$2x+y=6$
$\Rightarrow y=6-2x$
x | 0 | 3 |
y | 6 | 0 |
x | 1 | 0 |
y | 0 | -2 |
![](/assets/questions/media/148618-40854-1615660855.png)
Here, $a_1=2,\ b_1=-2,\ c_1=-2$ and $a_2=4,\ b_2=-4,\ c_2=-5$.
$\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}$
Here, we find that $\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
Thus, given pair of linear equations has no solution.
Therefore, the given pair of linear equations is inconsistent.