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

Using Geometry to answer the question would be the simplest:

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

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=(base* height)/(2)\quad =(9* 3)/(2)\quad =(27)/(2)\quad =\large\boxed{13.5}`

Calculus Solution:

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