AlgebraDifficulty 3.9AMC 10/12Find the answerUnited States
Problem: If x,y,z are real numbers such that xy=6, x−z=2, and x+y+z=9, compute yx−xz−xyz2.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
Solution: Let k=yx−xz−xyz2=xyx2−yz−z2. We have k+1=xyx2+xy−yz−z2=xyx2−xz+xy−yz+zx−z2=xy(x+y+z)(x−z)=69⋅2=3, so k=2.
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Source: MathNet,
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