All you need to do is plug -5 into the second equation and you see it is near (-5, -8). When plugged into the top, you get (-5, -27/4) which comes out to ABOUT -6.75 for the Y value. The closest is actually a tie. The first option is .8 from the first and .45 from the second leading in a total distance of 1.25. The second, which is the fellow answer, is 1.2 from the first and .05 from the second, leading to 1.25 away.
The third, which is next closest is 1.8 from the first and .55 from the second leading to a distance of over 2 from the optimal, so only the first two are answers.
Answer: The dimensions of the rectangle are 6ft and 16ft.
Step-by-step explanation:
Let the width of the rectangle be y
Let the length of the rectangle be y+10
Area= 96ft
Since area of a rectangle is length multiplied by width
y(y+10) = 96
y^2 +10y =96
y^2 +10y -96= 0
Solving algebraically,
y^2 +16y - 6y - 96= 0
y(y+16) - 6(y+16)= 0
(y-6)(y+16)= 0
y-6= 0
y =6ft
Recall that the width was denoted as y. That means the width is 6ft.
Since length is y+10= 6+10 = 16ft
Length is 16ft, width is 6ft.
Sorry, I didn't read the question fully. I don't know how to do this. Please disregard my previous answer.
