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Algebra · real student question

Solve the inequality 1 - x/3 <= 1 - 2x/7 and show the solution set on a number line.

Question

Solve the inequality

1−x3≤1−2x71-\frac{x}{3}\le 1-\frac{2x}{7}

and represent the solution set on a number line.

Step-by-step solution

  1. Cancel the constant that appears on both sides. Subtracting 11 from each side is legal for an inequality (adding or subtracting never changes the direction) and immediately kills the constants: −x3≤−2x7.-\frac{x}{3}\le-\frac{2x}{7}. Clearing denominators first would work too, but cancelling the 11 keeps the numbers smaller.

  2. Gather the xx terms on one side. Add 2x7\dfrac{2x}{7} to both sides: −x3+2x7≤0.-\frac{x}{3}+\frac{2x}{7}\le 0. Collecting on the left rather than the right avoids a needless sign change later.

  3. Combine over the common denominator 2121. Since lcm⁡(3,7)=21\operatorname{lcm}(3,7)=21, −7x21+6x21≤0⟹−x21≤0.-\frac{7x}{21}+\frac{6x}{21}\le 0\quad\Longrightarrow\quad -\frac{x}{21}\le 0. The two fractions nearly cancel, leaving a single coefficient of −121-\tfrac{1}{21}.

  4. Divide by the negative coefficient and flip the sign. Multiplying both sides by −21-21 reverses the inequality: −x21≤0⟹x≥0.-\frac{x}{21}\le 0\quad\Longrightarrow\quad x\ge 0. This reversal is the only step where the direction changes, and forgetting it is the classic error - it would give the wrong answer x≤0x\le 0.

  5. Test values to confirm. At x=7x=7: the left side is 1−73=−1.33‾1-\tfrac73=-1.3\overline{3} and the right side is 1−2=−11-2=-1, and −1.333≤−1-1.333\le-1 holds. At x=−7x=-7: the left side is 1+73=3.33‾1+\tfrac73=3.3\overline{3} and the right side is 1+2=31+2=3, and 3.333≤33.333\le 3 is false. So non-negative xx works and negative xx does not, as claimed. At x=0x=0 both sides equal 11, so the endpoint is included.

  6. Describe the number-line picture. Mark a solid (filled) dot at 00 to show that 00 itself is a solution, then shade the ray running to the right, towards +∞+\infty. In interval notation the solution set is [0,+∞)[0,+\infty).

Answer

x≥0,i.e. x∈[0,+∞)x\ge 0,\qquad\text{i.e. } x\in[0,+\infty)

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