Question

linear inequalities with fractions
unique id: 0836
score: 4/5 penalty: none
question
solve the following inequality:
\\(-\frac{3}{5}x + \frac{1}{2} \leq x + \frac{1}{5}(-3x - 9)\\)
answer attempt 1 out of 5
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\\(\phi\\) or \\(\cup\\) \\(\infty\\) \\(<\\) \\(>\\) \\(\leq\\) \\(\geq\\) \\(\\:|\\) \\(\\{\\:\\}\\) \\((\\:|)\\) \\((\\:|)\\) all real numbers

Explanation:

Step1: Simplify right - hand side

First, simplify the right - hand side of the inequality \(-\frac{3}{5}x+\frac{1}{2}\leq x+\frac{1}{5}(-3x - 9)\).
Distribute \(\frac{1}{5}\) on the right - hand side: \(x+\frac{1}{5}(-3x-9)=x-\frac{3}{5}x-\frac{9}{5}\).
Combine like terms for \(x\) terms: \(x-\frac{3}{5}x=\frac{5}{5}x-\frac{3}{5}x=\frac{2}{5}x\). So the right - hand side becomes \(\frac{2}{5}x-\frac{9}{5}\).
The inequality is now \(-\frac{3}{5}x+\frac{1}{2}\leq\frac{2}{5}x-\frac{9}{5}\).

Step2: Eliminate fractions

Multiply every term in the inequality by the least common multiple of the denominators (5 and 2), which is 10.
\(10\times(-\frac{3}{5}x)+10\times\frac{1}{2}\leq10\times\frac{2}{5}x-10\times\frac{9}{5}\)
Simplify each term:
\(10\times(-\frac{3}{5}x)=- 6x\), \(10\times\frac{1}{2} = 5\), \(10\times\frac{2}{5}x = 4x\), \(10\times\frac{9}{5}=18\).
The inequality becomes \(-6x + 5\leq4x-18\).

Step3: Move variable terms to one side

Add \(6x\) to both sides of the inequality: \(-6x+6x + 5\leq4x+6x-18\).
Simplify: \(5\leq10x-18\).

Step4: Move constant terms to one side

Add 18 to both sides: \(5 + 18\leq10x-18+18\).
Simplify: \(23\leq10x\).

Step5: Solve for x

Divide both sides by 10: \(\frac{23}{10}\leq x\) or \(x\geq2.3\)

Wait, there is a mistake in the above steps. Let's re - do the simplification of the right - hand side correctly.

Starting over:

Step1: Simplify the right - hand side correctly

The original inequality is \(-\frac{3}{5}x+\frac{1}{2}\leq x+\frac{1}{5}(-3x - 9)\)
Expand the right - hand side: \(x+\frac{1}{5}\times(-3x)+\frac{1}{5}\times(-9)=x-\frac{3}{5}x-\frac{9}{5}\)
Combine like terms for \(x\): \(x-\frac{3}{5}x=\frac{5x - 3x}{5}=\frac{2x}{5}\)
So the right - hand side is \(\frac{2x}{5}-\frac{9}{5}\)

Now, the inequality is \(-\frac{3x}{5}+\frac{1}{2}\leq\frac{2x}{5}-\frac{9}{5}\)

Step2: Add \(\frac{3x}{5}\) to both sides

\(-\frac{3x}{5}+\frac{3x}{5}+\frac{1}{2}\leq\frac{2x}{5}+\frac{3x}{5}-\frac{9}{5}\)
Simplify: \(\frac{1}{2}\leq x-\frac{9}{5}\)

Step3: Add \(\frac{9}{5}\) to both sides

\(\frac{1}{2}+\frac{9}{5}\leq x-\frac{9}{5}+\frac{9}{5}\)
Find a common denominator for the left - hand side, which is 10. \(\frac{1\times5}{2\times5}+\frac{9\times2}{5\times2}=\frac{5 + 18}{10}=\frac{23}{10}\)
So \(\frac{23}{10}\leq x\) or \(x\geq2.3\)

Wait, but let's check the original simplification again. Wait, maybe I made a mistake in the sign when moving terms. Let's use another approach.

Let's bring all terms to the left - hand side:

\(-\frac{3}{5}x+\frac{1}{2}-x-\frac{1}{5}(-3x - 9)\leq0\)

Expand \(\frac{1}{5}(-3x - 9)\): \(-\frac{3}{5}x+\frac{1}{2}-x+\frac{3}{5}x+\frac{9}{5}\leq0\)

Combine like terms for \(x\): \((-\frac{3}{5}x - x+\frac{3}{5}x)+\frac{1}{2}+\frac{9}{5}\leq0\)

\(-x+\frac{5 + 18}{10}\leq0\) (since \(\frac{1}{2}+\frac{9}{5}=\frac{5 + 18}{10}=\frac{23}{10}\))

\(-x+\frac{23}{10}\leq0\)

Subtract \(\frac{23}{10}\) from both sides: \(-x\leq-\frac{23}{10}\)

Multiply both sides by - 1 (and reverse the inequality sign): \(x\geq\frac{23}{10}\) or \(x\geq2.3\)

Wait, but let's check the first simplification again. Let's substitute \(x = 3\) (which is greater than 2.3) into the original inequality:

Left - hand side: \(-\frac{3}{5}\times3+\frac{1}{2}=-\frac{9}{5}+\frac{1}{2}=-\frac{18 + 5}{10}=-\frac{13}{10}=-1.3\)

Right - hand side: \(3+\frac{1}{5}(-3\times3 - 9)=3+\frac{1}{5}(-9 - 9)=3+\frac{1}{5}\times(-18)=3-\frac{18}{5}=\frac{15 - 18}{5}=-\frac{3}{5}=-0.6\)

Since \(-1.3\leq - 0.6\), the inequality holds for \(x = 3\)

Substitute \(x = 2\) (which is less than 2.3) into the original inequality:

Left - hand side: \(-\frac{3}{5}\times2+\frac{1}{2}=-\frac{6}{5}+\frac{1}{2}=-\frac{12 + 5}{10}=-\frac{7}{10}=-0.7\)

Right - hand side: \(2+\frac{1}{5}(-3\times2 - 9)=2+\frac{1}{5}(-6 - 9)=2+\frac{1}{5}\times(-15)=2 - 3=-1\)

Since \(-0.7\leq - 1\) is false, so \(x\geq\frac{23}{10}\) is the solution.

But let's check the calculation of the right - hand side again.

Original right - hand side: \(x+\frac{1}{5}(-3x - 9)=x-\frac{3x}{5}-\frac{9}{5}=\frac{5x-3x}{5}-\frac{9}{5}=\frac{2x}{5}-\frac{9}{5}\)

Left - hand side: \(-\frac{3x}{5}+\frac{1}{2}\)

Subtract \(\frac{2x}{5}\) from both sides: \(-\frac{3x}{5}-\frac{2x}{5}+\frac{1}{2}\leq-\frac{9}{5}\)

\(-x+\frac{1}{2}\leq-\frac{9}{5}\)

Subtract \(\frac{1}{2}\) from both sides: \(-x\leq-\frac{9}{5}-\frac{1}{2}=-\frac{18 + 5}{10}=-\frac{23}{10}\)

Multiply both sides by - 1 (reverse inequality): \(x\geq\frac{23}{10}\)

Answer:

\(x\geq\frac{23}{10}\) or \(x\geq2.3\) (in boxed form \(\boxed{x\geq\frac{23}{10}}\))