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Algebra · real student question

Solve the inequality 4(x + 1) is greater than or equal to x + 2(x - 1).

Question

Solve the inequality

4(x+1)≥x+2(x−1).4(x+1)\ge x+2(x-1).

Step-by-step solution

  1. Expand each side independently. Distribute before comparing anything:

    4(x+1)=4x+4,x+2(x−1)=x+2x−2=3x−2.4(x+1)=4x+4,\qquad x+2(x-1)=x+2x-2=3x-2.

    On the right, the 22 multiplies only the bracket; the leading xx is a separate term.

  2. Rewrite the inequality with both sides simplified.

    4x+4≥3x−2.4x+4\ge 3x-2.

  3. Collect the x-terms on one side. Subtract 3x3x from both sides:

    x+4≥−2.x+4\ge -2.

    Adding or subtracting the same quantity never changes the direction of an inequality.

  4. Isolate x. Subtract 44:

    x≥−6.x\ge -6.

  5. Note why the sign did not flip. The direction reverses only when you multiply or divide by a negative number. Here every step was an addition or subtraction, so ≥\ge stayed ≥\ge throughout.

  6. Test one value inside and one outside. At x=0x=0: 4(1)=44(1)=4 and 0+2(−1)=−20+2(-1)=-2, and 4≥−24\ge -2 ✓. At x=−10x=-10: 4(−9)=−364(-9)=-36 and −10+2(−11)=−32-10+2(-11)=-32, and −36≥−32-36\ge -32 is false ✓. The boundary x=−6x=-6 gives −20≥−20-20\ge -20, true, so it is included.

Answer

x≥−6,that is x∈[−6,∞)x\ge -6,\qquad \text{that is }x\in[-6,\infty)

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