Jen bowls, a 135 first game a 145 the second game.  Her average of four games is 135.  If she bowls a third game of x-8 and a fourth game of x, what is the value of x

Jen's bowling average over four games is 135. To find the value of \( x \), we need to set up an equation based on the sum of her four game scores divided by 4.

### Step 1: Sum of all scores
Jen’s scores are:
- 1st game: 135
- 2nd game: 145
- 3rd game: \( x - 8 \)
- 4th game: \( x \)

The average of these four games is 135, so we can express this as:

\[
\frac{135 + 145 + (x - 8) + x}{4} = 135
\]

### Step 2: Simplify the equation
First, add the constants:

\[
135 + 145 = 280
\]

So the equation becomes:

\[
\frac{280 + (x - 8) + x}{4} = 135
\]

Simplifying further:

\[
\frac{280 + x - 8 + x}{4} = 135
\]

\[
\frac{272 + 2x}{4} = 135
\]

### Step 3: Eliminate the denominator
Multiply both sides by 4 to eliminate the fraction:

\[
272 + 2x = 540
\]

### Step 4: Solve for \( x \)
Now subtract 272 from both sides:

\[
2x = 268
\]

Finally, divide by 2:

\[
x = 134
\]

### Conclusion:
The value of \( x \) is \( \boxed{134} \).

Would you like more details on any step?

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Here are 5 related questions to expand your understanding:
1. How would Jen's average change if her first game was 150 instead of 135?
2. What if Jen bowled a 160 in the fourth game? How would that affect the value of \( x \)?
3. How do averages change if the third and fourth games were both below 100?
4. How does the equation change if the average is 140 instead of 135?
5. Can we calculate the sum of the scores if the average over 5 games was given?

**Tip**: When working with averages, multiplying both sides of the equation by the number of items often helps eliminate fractions early in the process!

Jen bowls a 135 first game and a 145 second game. Her average of four games is 135. If she bowls a third game of x-8 and a fourth game of x, what is the value of x?

Discover how to calculate the value of x in Jen's bowling average problem. Jen bowls four games, and her scores follow an algebraic pattern. This detailed solution covers averages, linear equations, and step-by-step calculations. Ideal for students in Grades 6-8.

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