Someone help me please with this algebra problem

7 in.

Step-by-step explanation:

2L + 2W = 16

Try L = 7

2(7) + 2W = 16

14 + 2W = 16

2W = 2

W = 1

If the length is 7 in., then the width is 1 in. That is perfectly acceptable, so L = 7 in. is a good value.

Try L = 8

2(8) + 2W = 16

16 + 2W = 16

2W = 0

W = 0

A length of 8 would make a width of 0. You can't have a rectangle with 0 width, so L = 8 does not work.

When L = 9 or L = 10, the width would be negative. The width of a rectangle cannot be a negative number, so these values doe not work.

7 in.

Since we want to solve for the variable x, we want to isolate x

a²x + (a - 1) = (a + 1)x ⇒ Distribute x to (a+1). Also, remove parentheses

a²x + a - 1 = ax + x ⇒ Subtract a from both sides

a²x - 1 = ax + x - a ⇒ Add 1 to both sides

a²x = ax + x - a + 1 ⇒ Subtract (ax + x) from both sides

a²x - (ax + x)= ax + x - a + 1 - (ax+x) ⇒ Simplify. Remember that multiplying positive by negative = negative

a²x - ax - x = ax + x - a + 1 - ax - x ⇒ Simplify

a²x - ax - x = -a + 1 ⇒ Factor out the x from a²x - ax - x

x(a² - a - 1) = -a + 1 ⇒ Divide both sides by (a² - a - 1)

x = (-a + 1) / (a² - a - 1)

However, we need to make sure that the denominator does not equal 0. Therefore, you set the denominator = 0 (just use the quadratic formula for this), and it gives that the denominator =0 when a = (1+√5)/2 AND (1-√5)/2

Therefore, the final answer is

x = (-a + 1) / (a² - a - 1) given that a ≠ (1+√5)/2, a ≠ (1-√5)/2