What is the solution set for the linear inequality \frac{1 - 4x}{3} \le \frac{x + 7}{2}? — JAMB UTME

What is the solution set for the linear inequality $\frac{1 - 4x}{3} \le \frac{x + 7}{2}$ ?

x \ge -\frac{19}{11}

Multiplying the inequality by 6 clears denominators to give

2(1 - 4x) \le 3(x + 7)

. Expanding gives

2 - 8x \le 3x + 21

, which simplifies to

-11x \le 19

. Dividing both sides by

-11

flips the inequality sign, yielding

x \ge -\frac{19}{11}

Clear the denominators by multiplying both sides of the inequality by the lowest common multiple of 3 and 2, which is 6.

6 \cdot \left(\frac{1 - 4x}{3}\right) \le 6 \cdot \left(\frac{x + 7}{2}\right) \implies 2(1 - 4x) \le 3(x + 7)

Eliminating fractions simplifies the algebraic manipulation.

Expand both sides by distributing the factors.

2 - 8x \le 3x + 21

Removing parentheses allows grouping of like terms.

Rearrange terms by collecting terms containing

x

on the left and constants on the right.

-8x - 3x \le 21 - 2 \implies -11x \le 19

Isolating the variable term prepares for the final division step.

Divide both sides by

-11

and reverse the direction of the inequality sign.

x \ge -\frac{19}{11}

Dividing or multiplying an inequality by a negative real number requires flipping the inequality sign.

Solving linear inequalities involving fractions and reversing the inequality sign when dividing by a negative number.

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