Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Appendix - Extra Word Problems - Page 524: 37



Work Step by Step

Let one number be $x$ and the other number be $y$. If the sum of the two numbers is 9, this means that $x+y=9$. Rearranging this equation to make $y$ the subject of the equation, $x+y=9$ becomes $y=9-x$ This is the first of the two simultaneous equations. Next, we know that the sum of the reciprocals of the two numbers is $\frac{1}{2}$. This means that: $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ Now we simplify the above equation, $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ $\frac{y+x}{xy}=\frac{1}{2}$ $xy(1)=2(y+x)$ $xy=2y+2x$ This is the second of the two simultaneous equations. Therefore, we now have a pair of simultaneous equations to solve: Substituting the first equation in the second equation, we obtain: $x(9-x)=2(9-x)+2x$ $9x-x^{2}=18-2x+2x$ $9x-x^{2}=18$ $9x-x^{2}-18=0$ $x^{2}-9x+18=0$ We now solve this equation using the rules of factoring trinomials: $x^{2}-3x-6x+18=0$ $x(x-3)-6(x-3)=0$ $(x-3)(x-6)=0$ $(x-3)=0$ or $(x-6)=0$ $x=3$ or $x=6$ Since the sum of two numbers is $9$, $x=3$ means that $y=6$ or $x=6$ means that $y=3$. Therefore, the two numbers are 3 and 6.
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