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The path of the train A is given by the equation $3x\ +\ 4y\ -\ 12\ =\ 0$ and the path of another train B is given by the equation $6x\ +\ 8y\ -\ 48\ =\ 0$. Represent this situation graphically.
Given:
The path of the train A is given by the equation $3x\ +\ 4y\ -\ 12\ =\ 0$ and the path of another train B is given by the equation $6x\ +\ 8y\ -\ 48\ =\ 0$.
To do:
We have to represent the above situation graphically.
Solution:
The given pair of equations are:
$3x\ +\ 4y\ -\ 12\ =\ 0$....(i)
$4y=12-3x$
$y=\frac{12-3x}{4}$
$6x\ +\ 8y\ -\ 48\ =\ 0$....(ii)
$8y=48-6x$
$y=\frac{48-6x}{8}$
To represent the above equations graphically we need at least two solutions for each of the equations.
For equation (i),
If $x=0$ then $y=\frac{12-3(0)}{4}=\frac{12}{4}=3$
If $y=0$ then $0=\frac{12-3x}{4}$
$12=3x$
$x=\frac{12}{3}=4$
$x$ | $0$ | $4$ |
$y=\frac{12-3x}{4}$ | $3$ | $0$ |
For equation (ii),
If $x=0$ then $y=\frac{48-6(0)}{8}=\frac{48}{8}=6$
If $y=0$ then $0=\frac{48-6x}{8}$
$\Rightarrow 6x=48$
$\Rightarrow x=\frac{48}{6}=8$
$x$ | $0$ | $8$ |
$y=\frac{48-6x}{8}$ | $6$ | $0$ |
The above situation can be plotted graphically as below:
The line AB represents the equation $6x+8y-48=0$ and the line PQ represents the equation $3x+4y-12=0$.
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