#### Find the area bounded by the lines y = 4x + 5, y = 5 - x and 4y = x + 5

The lines mentioned are y = 4x + 5, y = 5 - x and 4y = x + 5

Below if the figure,

By solving these equations,

$\\y=4 x+5 \ldots \ldots(1)\\ y=5-x \ldots \ldots(2)\\ 4 y=x+5 \ldots . .(3)\\$

From (1) and (2)

$\\4 x+5=5-x \Rightarrow x=0 ; \\\therefore y=5-x=5\\$

From (2) and (3)

$\\4(5-x)=x+5\\ \Rightarrow x=3 ; \therefore y=5-x=2 \\$
\begin{aligned} &\text { From }(1) \text { and }(3)\\ &4(4 x+5)=x+5\\ &\Rightarrow x=-1 ; \therefore y=4 x+5=1 \end{aligned} \\

The point of interaction have found to be (0, 5), (3, 2) and (-1, 1)

Area of the region bounded by the curve y=f(x), the x -axis and the ordinates x=a and x=b, where f(x) is a continuous function defined on [a, b], is given by $A=\int_{a}^{b} f(x) d x or \int_{a}^{b} y d x$

Required area

$\\=\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(-x+5) d x-\int_{-1}^{3}\left(\frac{x}{4}+\frac{5}{4}\right) d x\\ =\left[2 x^{2}+5 x\right]_{-1}^{0}+\left[-\frac{x^{2}}{2}+5 x\right]_{0}^{3}-\left[\frac{x^{2}}{8}+\frac{5 x}{4}\right]_{-1}^{3} \\=((0+0)-(2-5))+\left(\left(-\frac{9}{2}+15\right)-(0+0)\right)-\left(\left(\frac{9}{8}+\frac{15}{4}\right)-\left(\frac{1}{8}-\frac{5}{4}\right)\right) \\$

$\\=3+\frac{21}{2}-\frac{39}{8}-\frac{9}{8} \\ =\frac{15}{2} \text { sq. units } \\$