x is a whole number. If three-quarters of x is subtracted from 1, the result is always greater than 0. Find the four highest values of x.

1 answer

We start by setting up the given condition algebraically. Let \( x \) be the whole number in question.

The problem states that if three-quarters of \( x \) is subtracted from 1, the result is always greater than 0. Mathematically, this can be represented as:

\[
1 - \frac{3}{4}x > 0
\]

To solve this inequality, we first isolate \( x \). Subtract 1 from both sides:

\[
1 - \frac{3}{4}x > 0
\]

Rewriting the inequality for clarity:

\[
\frac{3}{4}x < 1
\]

Next, to remove the fraction, multiply both sides of the inequality by 4:

\[
3x < 4
\]

Now, divide both sides by 3:

\[
x < \frac{4}{3}
\]

Since \( x \) is a whole number, we need \( x \) to be less than \(\frac{4}{3}\), which is approximately 1.33. Therefore, the largest whole number satisfying this inequality is \( x = 1 \).

Now, we need to list the four highest whole number values of \( x \) that satisfy the inequality. Since the inequality requires \( x < \frac{4}{3} \), we can see that the only whole number that satisfies this is:

\[
x = 1
\]

Thus, there are not four different values of \( x \) that satisfy the condition; in fact, the only value is \( x = 1 \). Therefore, the four highest values would just be repeatedly \( 1 \).