Find x so that b∥e.
∠3=(6x+4)∘ and ∠6=(8x−8)∘
x=?

x = 6

Step-by-step explanation:

b∥e

so <3 = <6

Given

∠3=(6x+4)∘ and ∠6=(8x−8)∘

So now

6x+4 = 8x−8

6x - 8x = - 8 - 4

-2x = - 12

  x = 6

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