Solve graphically the system of linear equations:
$4 x - 3 y + 4 = 0$
$4 x + 3 y - 20 = 0$
Find the area bounded by these lines and $x$ -axis.

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Solution

$\frac{x}{7} + \frac{y}{(4 / 3)} = 1 ⟶ (a)$
$\frac{x}{5} + \frac{y}{(20 / 3)} = 1 ⟶ (b)$
Intersection $\to$ $x = 2$ , $y = 3$
Area $= \frac{1}{2} \times 6 \times 3 = 9$ sq. units.

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Solve graphically the system of linear equations:4x−3y+4=04x+3y−20=0Find the area bounded by these lines and x-axis.

x7+y(4/3)=1⟶(a)x5+y(20/3)=1⟶(b)Intersection → x=2, y=3Area =12×6×3=9 sq. units.

Solve graphically the system of linear equations:
$4 x - 3 y + 4 = 0$
$4 x + 3 y - 20 = 0$
Find the area bounded by these lines and $x$ -axis.

$\frac{x}{7} + \frac{y}{(4 / 3)} = 1 ⟶ (a)$
$\frac{x}{5} + \frac{y}{(20 / 3)} = 1 ⟶ (b)$
Intersection $\to$ $x = 2$ , $y = 3$
Area $= \frac{1}{2} \times 6 \times 3 = 9$ sq. units.