Question

solve each system of inequalities.
\

$$\begin{cases} 8x + 3(2 - x) > 4 \\\\ 9x - 4(5 + x) < 8 \\end{cases}$$

Explanation:

Step1: Solve the first inequality \(8x + 3(2 - x) > 4\)

First, expand the brackets: \(8x + 6 - 3x > 4\)
Combine like terms: \(5x + 6 > 4\)
Subtract 6 from both sides: \(5x > 4 - 6\) which simplifies to \(5x > -2\)
Divide both sides by 5: \(x > -\frac{2}{5}\)

Step2: Solve the second inequality \(9x - 4(5 + x) < 8\)

Expand the brackets: \(9x - 20 - 4x < 8\)
Combine like terms: \(5x - 20 < 8\)
Add 20 to both sides: \(5x < 8 + 20\) which simplifies to \(5x < 28\)
Divide both sides by 5: \(x < \frac{28}{5}\) or \(x < 5.6\)

Step3: Find the intersection of the two solutions

We have \(x > -\frac{2}{5}\) and \(x < \frac{28}{5}\), so the solution to the system is \(-\frac{2}{5} < x < \frac{28}{5}\)

Answer:

\(-\frac{2}{5} < x < \frac{28}{5}\) (or \(-0.4 < x < 5.6\) in decimal form)