Question

Draw the graphs of the equations x – y = 1 and 2x + y = 8. Shade the area bounded by these two lines and y-axis. Also, determine this area.

Answer 1

Answer:

Using Geometry to answer the question would be the simplest:

Step-by-step explanation:

Remembering the formula for the area of a triangle which is $A=\frac12bh$. One can then tackle the question by doing the following:

Step 1 Find the y-intercepts

The y-intercepts are found by substituting in $x=0$.

Which gives you this when you plug it into both equations:

$-y=1\\y=-1\\y=8$

So the y-intercepts for the graphs are $(0,-1)$, and $(0,8)$ respectively.

Now one has to use elimination to solve the problems by adding up the equations we get:

$x-y=1\\2x+y=8\\3x=9\\x=3$

Now to solve for the y component substitute:

$2(3)+y=8\\y=2$

Therefore, the graphs intersect at the following:

$(3,2)$

Now we have our triangle which is accompanied by the graph.

now to solve it we must figure out how long the base is:

$b=8-(-1)\\b=9$

The height must also be accounted for which is the following:

$h=3$

Now the formula can be used:

$A=\frac12bh=\frac12(9)(3)=\frac{27}2\ \text{units}^2$

Answer 2

Answer: 13.5 units²

Step-by-step explanation:

Geometry Solution:

The base is along the y-axis from -1 to 8 = 9 units

The height is the largest x-value = 3

$Area=\dfrac{base\times height}{2}\quad =\dfrac{9\times 3}{2}\quad =\dfrac{27}{2}\quad =\large\boxed{13.5}$

Calculus Solution:

$\int^3_0[(-2x+8)-(x-1)]dx\\\\\\=\int^3_0(-3x+9)dx\\\\\\=\bigg(\dfrac{-3x^2}{2}+9x\bigg)\bigg|^3_0\\\\\\=\bigg(\dfrac{-3(3)^2}{2}+9(3)\bigg)-\bigg(\dfrac{-3(0)^2}{2}+9(0)\bigg)\\\\\\=\dfrac{-27}{2}+27-0-0\\\\\\=\dfrac{27}{2}\quad =\large\boxed{13.5}$