First you had to add by seven from both sides of equation form.

Then simplify by equation.

Next, you multiply by two from both sides of equation form.

And finally, simplify by equation.





Final answer: 
Hope this helps!
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-Charlie
Slope is positive with a y intercept at (0,-5) hope this helps
The dimensions of the enclosure that is most economical to construct are; x = 14.22 ft and y = 22.5 ft
<h3>How to maximize area?</h3>
Let the length of the rectangular area be x feet
Let the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 320 square feet
y = 320/x -----(1)
Cost to fence the three sides = $6 per foot
Therefore cost to fence one length and two width of the rectangular area
= 6(x + 2y)
Similarly cost to fence the fourth side = $13 per foot
So, the cost of the remaining length = 13x
Total cost to fence = 6(x + 2y) + 13x
Cost (C) = 6(x + 2y) + 13x
C = 6x + 12y + 13x
C = 19x + 12y
From equation (1),
C = 19x + 12(320/x)
C' = 19 - 3840/x²
At C' = 0, we have;
19 - 3840/x² = 0
19 = 3840/x²
19x² = 3840
x² = 3840/19
x = √(3840/19)
x = 14.22 ft
y = 320/14.22
y = 22.5 ft
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<h2>
Explanation:</h2><h3>Part A.</h3>
The boundary lines of both inequalities will be dashed, because neither includes the "or equal to" case.
The first inequality solution area is bounded by a line with slope +5 and y-intercept +5. The solution area is above the line (y is greater than ...). Since the line rises steeply, the solution area looks to be to the left of the line. (It is shaded red on the attached graph.)
The second inequality solution area is bounded by a line with slope -1/2 and y-intercept +1. The solution area for this inequality is also above the line.
The solution area is where the two solution spaces overlap, in the quadrant to the upper left of the point where the lines intersect.
___
<h3>Part B.</h3>
The graph shows the point (-2, 5) to be in the solution space.
We can show this point satisfies both inequalities.
- 5 > 5(-2)+5 ⇒ 5 > -5 . . . true
- 5 > (-1/2)(-2) +1 ⇒ 5 > 2 . . . true