Question

linear inequalities with fractions
question
solve the following inequality:
-2x - 5 < \frac{5}{4}(-4x - 1) + 2x

Explanation:

Step1: Simplify the right - hand side

First, distribute the \(\frac{5}{4}\) in \(\frac{5}{4}(-4x - 1)\):
\(\frac{5}{4}\times(-4x)=-5x\) and \(\frac{5}{4}\times(-1)=-\frac{5}{4}\)
So the right - hand side of the inequality \(\frac{5}{4}(-4x - 1)+2x\) becomes \(-5x-\frac{5}{4}+2x\)
Combine like terms: \(-5x + 2x=(-5 + 2)x=-3x\)
So the inequality \(-2x-5\lt\frac{5}{4}(-4x - 1)+2x\) simplifies to \(-2x-5\lt-3x-\frac{5}{4}+2x\)? Wait, no, wait. Wait, \(\frac{5}{4}(-4x - 1)+2x=-5x-\frac{5}{4}+2x\), and \(-5x + 2x=-3x\), so the inequality is \(-2x-5\lt-3x-\frac{5}{4}\)

Step2: Add \(3x\) to both sides

To get all the \(x\) terms on one side, add \(3x\) to both sides of the inequality \(-2x-5\lt-3x-\frac{5}{4}\)
\(-2x+3x - 5\lt-3x + 3x-\frac{5}{4}\)
Simplify: \(x-5\lt-\frac{5}{4}\)

Step3: Add 5 to both sides

Add 5 to both sides of the inequality \(x - 5\lt-\frac{5}{4}\)
\(x-5 + 5\lt-\frac{5}{4}+5\)
We know that \(5=\frac{20}{4}\), so \(-\frac{5}{4}+\frac{20}{4}=\frac{-5 + 20}{4}=\frac{15}{4}=3.75\)
So \(x\lt\frac{15}{4}\)

Wait, let's re - do the simplification of the right - hand side correctly.

Original inequality: \(-2x-5\lt\frac{5}{4}(-4x - 1)+2x\)

Distribute \(\frac{5}{4}\) over \((-4x - 1)\):

\(\frac{5}{4}\times(-4x)=-5x\), \(\frac{5}{4}\times(-1)=-\frac{5}{4}\)

So the right - hand side is \(-5x-\frac{5}{4}+2x\)

Combine like terms for the \(x\) terms: \(-5x+2x=-3x\)

So the inequality is \(-2x - 5\lt-3x-\frac{5}{4}\)

Now, add \(3x\) to both sides:

\(-2x+3x-5\lt-3x + 3x-\frac{5}{4}\)

Which simplifies to \(x - 5\lt-\frac{5}{4}\)

Now, add 5 to both sides. Since \(5=\frac{20}{4}\), we have:

\(x-5 + 5\lt-\frac{5}{4}+\frac{20}{4}\)

\(x\lt\frac{15}{4}\) (because \(-\frac{5}{4}+\frac{20}{4}=\frac{15}{4}\))

Answer:

\(x\lt\frac{15}{4}\) (or \(x\lt3.75\))