Draw the graphs of the equation $3 x + 4 y = 7$ and $3 x - 2 y = 1$ and find their point of intersection of lines representing the equation.

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Solution

$3 x + 4 y = 7 \Rightarrow \frac{x}{7 / 3} + \frac{y}{(7 / 4)} = 1$
$3 x - 2 y = 1 \Rightarrow \frac{x}{(1 / 3)} - \frac{y}{(1 / 2)} = 1$

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Draw the graphs of the equation 3x+4y=7 and 3x−2y=1 and find their point of intersection of lines representing the equation.

3x+4y=7⇒x7/3+y(7/4)=13x−2y=1⇒x(1/3)−y(1/2)=1

Draw the graphs of the equation $3 x + 4 y = 7$ and $3 x - 2 y = 1$ and find their point of intersection of lines representing the equation.

$3 x + 4 y = 7 \Rightarrow \frac{x}{7 / 3} + \frac{y}{(7 / 4)} = 1$
$3 x - 2 y = 1 \Rightarrow \frac{x}{(1 / 3)} - \frac{y}{(1 / 2)} = 1$