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Zorluk: Çok zorSolving One-Variable Inequalities

Evaluate whether the following statement is true or false:

The only positive integer solutions to the linear inequality −35(103x−15)+4(x−2)≥−52(−2x+4)+2-\frac{3}{5}\left(\frac{10}{3}x - 15\right) + 4(x - 2) \geq -\frac{5}{2}(-2x + 4) + 2 are 1, 2, and 3.

Cevap: Cevap

Cevap

The statement is True.
By successfully distributing the fractional negative coefficients and combining like terms, the inequality simplifies to −3x≥−9-3x \geq -9. Dividing by −3-3 requires reversing the inequality to x≤3x \leq 3. The positive integers in this solution set are exactly 1, 2, and 3, meaning the statement is factually correct.

Adım Adım Çözüm

1
Distribute the negative fractions and constants on the left side of the inequality.
(−2x+9)+(4x−8)(-2x + 9) + (4x - 8)
Multiplying −35-\frac{3}{5} by 103x\frac{10}{3}x gives −2x-2x, and −35-\frac{3}{5} by −15-15 gives +9+9. Multiplying 44 by (x−2)(x-2) gives 4x−84x - 8.
2
Distribute the negative fraction on the right side of the inequality.
(5x−10)+2(5x - 10) + 2
Multiplying −52-\frac{5}{2} by −2x-2x gives 5x5x, and −52-\frac{5}{2} by 44 gives −10-10.
3
Combine the like terms on both sides.
2x+1≥5x−82x + 1 \geq 5x - 8
On the left side, −2x+4x=2x-2x + 4x = 2x and 9−8=19 - 8 = 1. On the right side, −10+2=−8-10 + 2 = -8.
4
Isolate the variable xx by subtracting 5x5x and 11 from both sides.
−3x≥−9-3x \geq -9
Grouping all variables on one side and constants on the other prepares the equation for the final division.
5
Divide both sides by −3-3 and reverse the inequality symbol.
x≤3x \leq 3
Dividing an inequality by a negative number requires flipping the inequality direction. Since x≤3x \leq 3, the only integers greater than zero that satisfy this are 1, 2, and 3.

Anahtar Kavram

Solving complex linear inequalities by distributing negative fractions, combining like terms, and properly applying the inequality sign flip rule.

Alternatif Yöntem

You can verify the statement by plugging the boundary values into the original inequality. Substitute x=3x=3 (which evaluates to 7≥77 \geq 7, a true statement), x=4x=4 (which evaluates to 9≥129 \geq 12, a false statement), and recognize that since it is a linear relation, solutions are valid downwards. Zero is not a positive integer, leaving 1, 2, and 3 as the only valid positive integers.
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Solving One-Variable Inequalities

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