Question

solve for x.
38x² - 42x + 35 ≥ 39x² - 43x + 33
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3.

Explanation:

Step1: Move all terms to left side

Subtract \(39x^{2}\), add \(43x\), and subtract \(33\) from both sides:
\(38x^{2}-42x + 35-(39x^{2})+43x - 33\geq0\)
Simplify: \(-x^{2}+x + 2\geq0\)

Step2: Multiply by -1 (reverse inequality)

Multiply both sides by \(- 1\), so the inequality becomes \(x^{2}-x - 2\leq0\)

Step3: Factor the quadratic

Factor \(x^{2}-x - 2\) as \((x - 2)(x+1)\leq0\)

Step4: Find critical points

Set \((x - 2)(x + 1)=0\), critical points are \(x=-1\) and \(x = 2\)

Step5: Test intervals

• For \(x<-1\) (e.g., \(x=-2\)): \((-2 - 2)(-2 + 1)=(-4)(-1)=4>0\) (does not satisfy \(\leq0\))
• For \(-1
• For \(x>2\) (e.g., \(x = 3\)): \((3 - 2)(3 + 1)=(1)(4)=4>0\) (does not satisfy)
• At \(x=-1\) and \(x = 2\), the product is \(0\), which satisfies \(\leq0\)

Answer:

\(-1\leq x\leq2\)