# Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 52

6 and 4.

#### Work Step by Step

$x+y=10$ $\implies y=10-x$ Substituting the value of y in $xy=24$, we have $x(10-x)=24$ or $10x-x^{2}=24$ or $x^{2}-10x+24=0$ x=$\frac{-b±\sqrt {b^{2}-4ac}}{2a}=\frac{10±\sqrt {100-4\times1\times24}}{2}$ $x=6$ or $x=4$ If x is 6 y is (10-6)=4. If x is 4, y is 6.

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