if m/EBH=(6x+12) and m/HBC= (8x-10) find m/EBH

To solve for \( m/\angle EBH \), given the following information:

- \( m/\angle EBH = 6x + 12 \)
- \( m/\angle HBC = 8x - 10 \)

We need additional information about the relationship between these angles. If these two angles are somehow related (for example, they form a linear pair, are complementary, or supplementary), we can set up an equation to solve for \( x \).

Assuming \( m/\angle EBH \) and \( m/\angle HBC \) form a linear pair (meaning they are adjacent and sum to \( 180^\circ \)), we can set up the following equation:

\[
(6x + 12) + (8x - 10) = 180
\]

### Step-by-step solution:

1. Combine like terms:
\[
6x + 12 + 8x - 10 = 180
\]
\[
14x + 2 = 180
\]

2. Subtract 2 from both sides:
\[
14x = 178
\]

3. Divide both sides by 14:
\[
x = \frac{178}{14} = 12.71
\]

Now that we have \( x = 12.71 \), we can substitute it back into the expression for \( m/\angle EBH \):

\[
m/\angle EBH = 6x + 12 = 6(12.71) + 12 = 76.26 + 12 = 88.26^\circ
\]

Thus, \( m/\angle EBH \approx 88.26^\circ \).

### Do you need more details or have any questions?

### Related Questions:
1. What happens if \( m/\angle EBH \) and \( m/\angle HBC \) are complementary instead of forming a linear pair?
2. How would the solution change if the sum of these angles is given to be \( 90^\circ \)?
3. Can you verify the relationship between the angles from a diagram if provided?
4. What are other possible types of angle relationships in geometry?
5. How do you solve similar problems involving different kinds of angles, such as vertical angles?

### Tip:
Always look for key words that describe angle relationships, such as "linear pair," "complementary," or "supplementary." These words are crucial in setting up the correct equations for solving geometric problems.

if m/EBH = (6x + 12) and m/HBC = (8x - 10) find m/EBH

Learn how to find m/∠EBH when given m/∠EBH = (6x + 12) and m/∠HBC = (8x - 10). This problem involves solving a linear pair of angles using algebraic expressions, ideal for students in Grades 9-10.

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