Solve the following equation. 6x−15=4x+55 6 x − 15 = 4 x + 55

x = 35

Step-by-step explanation:

Solve for x:

6 x - 15 = 4 x + 55

Hint: | Move terms with x to the left hand side.

Subtract 4 x from both sides:

(6 x - 4 x) - 15 = (4 x - 4 x) + 55

Hint: | Combine like terms in 6 x - 4 x.

6 x - 4 x = 2 x:

2 x - 15 = (4 x - 4 x) + 55

Hint: | Look for the difference of two identical terms.

4 x - 4 x = 0:

2 x - 15 = 55

Hint: | Isolate terms with x to the left hand side.

Add 15 to both sides:

2 x + (15 - 15) = 15 + 55

Hint: | Look for the difference of two identical terms.

15 - 15 = 0:

2 x = 55 + 15

Hint: | Evaluate 55 + 15.

55 + 15 = 70:

2 x = 70

Hint: | Divide both sides by a constant to simplify the equation.

Divide both sides of 2 x = 70 by 2:

(2 x)/2 = 70/2

Hint: | Any nonzero number divided by itself is one.

2/2 = 1:

x = 70/2

Hint: | Reduce 70/2 to lowest terms. Start by finding the GCD of 70 and 2.

The gcd of 70 and 2 is 2, so 70/2 = (2×35)/(2×1) = 2/2×35 = 35:

x = 35

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.