Question
Show graphically that the following system of equation is in-consistent (i.e. has no solution):
$3x - 5y = 20$
$6x - 10y = -40$
✓
Answer
We have,
$3x - 5y = 20$
$6x - 10y = -40$
Now, $3x - 5y = 20$
$\Rightarrow\text{x}=\frac{5\text{y}+20}{3}$
When $y = -1,$ we have,
$\text{x}=\frac{5(-1)+20}{3}=5$
When $y = -4,$ we have,
$\text{x}=\frac{5(-4)+20}{3}=0$
Thus we have the following table giving points on the line $3x - 5y = 20.$
Now, $6x - 10y = -40$
$\Rightarrow 6x = -40 + 10y$
$\Rightarrow\text{x}=\frac{-40+10\text{y}}{6}$
When $y = 4$, we have,
$\text{x}=\frac{-40+10\times4}{6}=0$
When $y = 1$, we have,
$\text{x}=\frac{-40+10\times1}{6}=-5$
Thus we have the following table giving points on the line $6x - 10y = -40$
Graph of the given equations:
Clearly, there is no common points between these two lines.
Hence, given system of equations is in-consistent.
$3x - 5y = 20$
$6x - 10y = -40$
Now, $3x - 5y = 20$
$\Rightarrow\text{x}=\frac{5\text{y}+20}{3}$
When $y = -1,$ we have,
$\text{x}=\frac{5(-1)+20}{3}=5$
When $y = -4,$ we have,
$\text{x}=\frac{5(-4)+20}{3}=0$
Thus we have the following table giving points on the line $3x - 5y = 20.$
|
$x$
|
$5$
|
$0$
|
|
$y$
|
$-1$
|
$-4$
|
$\Rightarrow 6x = -40 + 10y$
$\Rightarrow\text{x}=\frac{-40+10\text{y}}{6}$
When $y = 4$, we have,
$\text{x}=\frac{-40+10\times4}{6}=0$
When $y = 1$, we have,
$\text{x}=\frac{-40+10\times1}{6}=-5$
Thus we have the following table giving points on the line $6x - 10y = -40$
|
$x$
|
$0$
|
$-5$
|
|
$y$
|
$4$
|
$1$
|
Clearly, there is no common points between these two lines.
Hence, given system of equations is in-consistent.
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