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.

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01-06-2020
Mathematics

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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.

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joseserra803 joseserra803 · Answer 1 · 2020-06-01T13:58:15+02:00

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 [tex]A=\frac12bh[/tex]. One can then tackle the question by doing the following:

Step 1 Find the y-intercepts

The y-intercepts are found by substituting in [tex]x=0[/tex].

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

[tex]-y=1\\y=-1\\y=8[/tex]

So the y-intercepts for the graphs are [tex](0,-1)\\[/tex], and [tex](0,8)[/tex] respectively.

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

[tex]x-y=1\\2x+y=8\\3x=9\\x=3[/tex]

Now to solve for the y component substitute:

[tex]2(3)+y=8\\y=2[/tex]

Therefore, the graphs intersect at the following:

[tex](3,2)[/tex]

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

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

[tex]b=8-(-1)\\b=9[/tex]

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

[tex]h=3[/tex]

Now the formula can be used:

[tex]A=\frac12bh=\frac12(9)(3)=\frac{27}2\ \text{units}^2[/tex]

tramserran tramserran · Answer 2 · 2020-06-01T15:01:19+02:00

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

[tex]Area=\dfrac{base\times height}{2}\quad =\dfrac{9\times 3}{2}\quad =\dfrac{27}{2}\quad =\large\boxed{13.5}[/tex]

Calculus Solution:

[tex]\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}[/tex]