Question

Solve the compound inequality.
6 - 2(4x + 1) /geq -10 and 5-(x-9) < 2x + 7

Answer 1

To solve the compound inequality

1. \(6 - 2(4x + 1) \geq -10\)
2. \(5 - (x-9) < 2x + 7\)

we will solve each inequality separately.

### Solving the first inequality:

Given:
\[
6 - 2(4x + 1) \geq -10
\]

First, distribute the \(-2\):
\[
6 - 8x - 2 \geq -10
\]

Combine like terms:
\[
4 - 8x \geq -10
\]

Now, subtract \(4\) from both sides:
\[
-8x \geq -10 - 4
\]
\[
-8x \geq -14
\]

Now, divide by \(-8\) and remember to reverse the inequality when dividing by a negative number:
\[
x \leq \frac{-14}{-8}
\]
\[
x \leq \frac{14}{8}
\]
\[
x \leq \frac{7}{4}
\]

### Solving the second inequality:

Given:
\[
5 - (x-9) < 2x + 7
\]

Simplifying \(5 - (x - 9)\):
\[
5 - x + 9 < 2x + 7
\]
\[
14 - x < 2x + 7
\]

Now, add \(x\) to both sides:
\[
14 < 3x + 7
\]

Subtract \(7\) from both sides:
\[
14 - 7 < 3x
\]
\[
7 < 3x
\]

Now divide by \(3\):
\[
\frac{7}{3} < x
\]

Which can be rewritten as:
\[
x > \frac{7}{3}
\]

### Combining the results:

Now we combine both results:
1. \(x \leq \frac{7}{4}\)
2. \(x > \frac{7}{3}\)

To express this as a compound inequality, we can observe that \(\frac{7}{3} \approx 2.33\) and \(\frac{7}{4} = 1.75\). Therefore, there is no overlap between the inequalities since \(\frac{7}{3} > \frac{7}{4}\).

Thus, the solution is:
\[
\text{There is no solution.}
\]