SOLUTION: The length of a rectangle is 3 times it's width. Is both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?

Question 824982: The length of a rectangle is 3 times it's width. Is both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?
Found 2 solutions by jim_thompson5910, TimothyLamb:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 3 times it's width. If both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?

Width = x
Length = 3x

"If both dimensions are increased by 2 centimeters, the area is 32 square centimeters" --->

or

or

or

Toss the negative solution. So the only solution is

Width = x = 2

Length = 3x = 3*2 = 6

The width is 2 centimeters and the length is 6 centimeters
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
---
x = length
y = width
---
x = 3y
(x + 2)(y + 2) = 32
---
(x + 2)(y + 2) = 32
(3y + 2)(y + 2) = 32
3yy + 6y + 2y + 4 = 32
3yy + 8y - 28 = 0
---
the above quadratic equation is in standard form, with a=3, b=8, and c=-28
---
to solve the quadratic equation, by using the quadratic formula, copy and paste this:
3 8 -28
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
---
this quadratic has two real roots (the x-intercepts), which are:
y = 2
y = -4.66666667
---
the negative root does not fit the problem statement, so use the positive root
---
answer:
x = length = 6 cm
y = width = 2 cm
---
Solve and graph linear equations:
https://sooeet.com/math/linear-equation-solver.php
---
Solve quadratic equations, quadratic formula:
https://sooeet.com/math/quadratic-formula-solver.php
---
Solve systems of linear equations up to 6-equations 6-variables:
https://sooeet.com/math/system-of-linear-equations-solver.php
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