Question

If \( \frac {2-x}{4}-\frac{4+x}{6}\geq10 \) then :

• A. \(x\geq-\frac{122}{5}\)
• B. \(x\geq-\frac{5}{122}\)
• C. \(x\leq-\frac{122}{5}\)
• D. \(x\leq-\frac{5}{122}\)

Answer 1

The Correct Option is C

To solve the inequality \( \frac{2-x}{4}-\frac{4+x}{6}\geq10 \), follow these steps:

First, find a common denominator for the fractions. The least common multiple of 4 and 6 is 12. Rewrite each fraction with this common denominator:

\( \frac{2-x}{4} = \frac{3(2-x)}{12} = \frac{6-3x}{12} \)

\( \frac{4+x}{6} = \frac{2(4+x)}{12} = \frac{8+2x}{12} \)

Substitute back into the inequality:

\( \frac{6-3x}{12} - \frac{8+2x}{12} \geq 10 \)

Combine the fractions:

\( \frac{6-3x-(8+2x)}{12} \geq 10 \)

Simplify the expression inside the fraction:

\( \frac{6-3x-8-2x}{12} = \frac{-2-5x}{12} \)

The inequality now becomes:

\( \frac{-2-5x}{12} \geq 10 \)

To clear the fraction, multiply both sides by 12:

\( -2-5x \geq 120 \)

Next, isolate \( x \) by adding 2 to both sides:

\( -5x \geq 122 \)

Divide both sides by -5, remembering to reverse the inequality sign because we are dividing by a negative number:

\( x \leq -\frac{122}{5} \)

Thus, the solution to the inequality is:

\( x \leq -\frac{122}{5} \)