look at the diagram. which equation can be used to solve for x? 3x + 90 = 180, 3x + 96 = 111, 3x + 6 = 111, 3x + 96 = 180 solve for x.

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Accepted Answer

### Part 1: Find the equation to solve for $ x $
# Explanation:
## Step 1: Identify right angle and angle sum
We know that $ \angle AB C = 90^\circ $ (right angle), and the sum of angles on a straight line is $ 180^\circ $. The angles $ (3x + 6)^\circ $, $ 90^\circ $, and the angle adjacent to $ 111^\circ $ (which is $ 180^\circ - 111^\circ = 69^\circ $, but wait, actually, looking at the vertical line $ DC $ and horizontal line $ AB $, the angle between $ BE $ and $ BC $ is $ (3x + 6)^\circ $, and the angle between $ BF $ and $ BC $ is $ 180^\circ - 111^\circ = 69^\circ $? Wait, no, let's re - examine.

Wait, the line $ DC $ is a straight line, and $ AB $ is perpendicular to $ DC $, so $ \angle ABC=90^\circ $. The angle $ \angle EBC=(3x + 6)^\circ $, and the angle $ \angle FBC = 180^\circ- 111^\circ=69^\circ $? No, actually, the angle between $ BE $ and $ BF $: Wait, the sum of $ (3x + 6)^\circ $, $ 90^\circ $, and the angle supplementary to $ 111^\circ $? Wait, no, let's use the fact that the angle between $ BE $ and $ DC $ is $ (3x + 6)^\circ $, and the angle between $ BF $ and $ DC $ is $ 180^\circ - 111^\circ = 69^\circ $? No, maybe a better approach: Since $ AB\perp DC $, $ \angle ABD = 90^\circ $. The angle $ \angle EBF $: Wait, the angle $ (3x + 6)^\circ $, the right angle $ 90^\circ $, and the angle $ 111^\circ $ related? Wait, no, let's look at the straight line. The sum of $ (3x + 6)^\circ $, $ 90^\circ $, and $ (180 - 111)^\circ=69^\circ $? No, that's not right. Wait, actually, the angle $ (3x + 6)^\circ $ and the angle $ 111^\circ $ are related through the right angle. Wait, the angle between $ BE $ and $ AB $ is $ (3x + 6)^\circ $, and $ AB\perp DC $, so the angle between $ BE $ and $ DC $ is $ 90^\circ-(3x + 6)^\circ $? No, maybe I made a mistake. Wait, let's look at the vertical line $ DC $ and the line $ EF $ intersecting at $ B $. The angle $ \angle FBC = 180^\circ - 111^\circ=69^\circ $, and since $ AB\perp DC $, $ \angle ABC = 90^\circ $, so the angle $ \angle ABE=(3x + 6)^\circ $, and $ \angle ABE+\angle ABC+\angle FBC = 180^\circ $? No, $ \angle ABE+(3x + 6)^\circ $? Wait, no, the angle $ (3x + 6)^\circ $ is $ \angle EBC $, and $ \angle FBC = 180 - 111 = 69^\circ $, and $ \angle EBC+\angle ABC+\angle FBC=180^\circ $? No, $ \angle ABC = 90^\circ $, so $ (3x + 6)+90+(180 - 111)=180 $? No, that's not. Wait, another approach: The angle $ (3x + 6)^\circ $ and the angle $ 111^\circ $ are such that $ (3x + 6)+90 = 111 + \text{something} $? No, wait, the correct relationship: Since $ AB\perp DC $, $ \angle ABD = 90^\circ $. The angle between $ BE $ and $ BD $ is $ 90^\circ-(3x + 6)^\circ $, and the angle between $ BF $ and $ BD $ is $ 180^\circ - 111^\circ = 69^\circ $. But since $ BE $ and $ BF $ are a straight line? No, $ EF $ is a straight line. So $ \angle EBC+(3x + 6)^\circ $? Wait, I think I messed up. Let's start over.

We know that $ AB $ is perpendicular to $ DC $, so $ \angle ABC = 90^\circ $. The line $ EF $ intersects $ DC $ at $ B $. The angle $ \angle FBC = 180^\circ - 111^\circ = 69^\circ $. The sum of $ \angle EBC $, $ \angle ABC $, and $ \angle FBC $ should be $ 180^\circ $ (since they are on a straight line $ EF $? No, $ EF $ is a straight line, so $ \angle EBC+\angle FBC = 180^\circ $? No, $ DC $ is a straight line. Wait, $ DC $ is a straight line, so the sum of angles around $ B $ on line $ DC $ is $ 180^\circ $. The angles are $ \angle ABD = 90^\circ $, $ \angle ABE=(3x + 6)^\circ $, and $ \angle FBC = 180 - 111 = 69^\circ $? No, I think the correct equation comes from the fact that $ (3x + 6)^\circ+90^\circ = 111^\circ+\text{supplementary angle} $? No, wait, the angle $ (3x + 6)^\circ $ and the angle $ 111^\circ $ are related by the right angle. Let's see: The angle between $ BE $ and $ DC $ is $ (3x + 6)^\circ $, and the angle between $ BF $ and $ DC $ is $ 180 - 111 = 69^\circ $. Since $ AB\perp DC $, the angle between $ BE $ and $ AB $ is $ 90-(3x + 6) $, and the angle between $ BF $ and $ AB $ is $ 111 - 90 = 21^\circ $. But since $ BE $ and $ BF $ are a straight line, the sum of angles on $ AB $ side: Wait, maybe the correct equation is $ 3x + 6+90 = 111 + (180 - 111) $? No, this is getting confusing. Wait, let's look at the answer options. The options are $ 3x + 90 = 180 $, $ 3x + 96 = 111 $, $ 3x + 6 = 111 $, $ 3x + 96 = 180 $.

Wait, another way: The angle $ (3x + 6)^\circ $ and the angle $ 111^\circ $ are such that $ (3x + 6)+90 = 111 + (180 - 111) $? No, let's use the fact that the sum of $ (3x + 6)^\circ $, $ 90^\circ $, and the angle supplementary to $ 111^\circ $ is $ 180^\circ $. The supplementary angle to $ 111^\circ $ is $ 180 - 111 = 69^\circ $. So $ (3x + 6)+90 + 69=180 $, which simplifies to $ 3x+165 = 180 $, which is not an option. Wait, maybe I made a mistake in identifying the angles. Let's look at the diagram again: $ AB $ is perpendicular to $ DC $, so $ \angle ABD = 90^\circ $. The line $ EF $ passes through $ B $, with $ \angle FBC = 111^\circ $? No, the diagram shows $ 111^\circ $ between $ BF $ and $ BC $. Wait, $ DC $ is a straight line, so $ \angle DBC = 180^\circ $? No, $ DC $ is a vertical line, $ AB $ is horizontal, so $ \angle ABC = 90^\circ $. The angle $ \angle EBC=(3x + 6)^\circ $, and $ \angle FBC = 111^\circ $. Since $ E $, $ B $, $ F $ are colinear, $ \angle EBC+\angle FBC = 180^\circ $? No, that would mean $ (3x + 6)+111 = 180 $, which is $ 3x+117 = 180 $, $ 3x = 63 $, $ x = 21 $, but that's not related to the right angle. Wait, no, $ AB $ is perpendicular to $ DC $, so $ \angle ABE + \angle EBC=90^\circ $? No, $ AB $ is horizontal, $ DC $ is vertical, so $ \angle ABC = 90^\circ $, so $ \angle ABE+(3x + 6)=90 $, and $ \angle ABF = 111 - 90 = 21^\circ $, and $ \angle ABE+\angle ABF = 180^\circ $? No, $ E $, $ B $, $ F $ are colinear, so $ \angle ABE+\angle ABF = 180^\circ $. $ \angle ABF = 111^\circ - 90^\circ=21^\circ $? No, $ \angle FBC = 111^\circ $, $ \angle ABC = 90^\circ $, so $ \angle ABF = \angle FBC-\angle ABC=111 - 90 = 21^\circ $. And $ \angle ABE=(3x + 6)^\circ $, so $ (3x + 6)+21 = 180 $? No, that's $ 3x+27 = 180 $, not an option.

Wait, maybe the correct equation is $ 3x + 6+90 = 111 $? No, that would be $ 3x+96 = 111 $, which is one of the options. Let's check: If $ 3x + 96 = 111 $, then $ 3x=15 $, $ x = 5 $. Let's see if that makes sense. If $ x = 5 $, then $ 3x + 6=21 $, and $ 21+90 = 111 $, which matches the $ 111^\circ $ angle. Ah! So the angle $ (3x + 6)^\circ $ plus the right angle $ 90^\circ $ equals $ 111^\circ $? Wait, no, $ (3x + 6)+90 = 111 $, so $ 3x+96 = 111 $. That must be the equation.

## Step 2: Verify the equation
We know that $ AB \perp DC $, so $ \angle ABC = 90^\circ $. The angle $ \angle EBC=(3x + 6)^\circ $, and the angle $ \angle FBC = 111^\circ $. Since $ \angle EBC+\angle ABC=\angle FBC $ (because $ \angle EBC $ and $ \angle ABC $ add up to $ \angle FBC $), we have $ (3x + 6)+90 = 111 $, which simplifies to $ 3x + 96 = 111 $.

# Answer (for the equation):
$ 3x + 96 = 111 $

### Part 2: Solve for $ x $
# Explanation:
## Step 1: Start with the equation
We have the equation $ 3x + 96 = 111 $.

## Step 2: Subtract 96 from both sides
Subtract 96 from each side of the equation: $ 3x+96 - 96=111 - 96 $.
Simplifying, we get $ 3x = 15 $.

## Step 3: Divide by 3
Divide both sides by 3: $ \frac{3x}{3}=\frac{15}{3} $.
Simplifying, we get $ x = 5 $.

# Answer (for $ x $):
$ x = 5 $

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QUESTION IMAGE

Question

Explanation:

Part 1: Find the equation to solve for \( x \)

Step 1: Identify right angle and angle sum

Step 1: Start with the equation

Step 2: Subtract 96 from both sides

Step 3: Divide by 3

Answer:

Part 2: Solve for \( x \)