Elementary Algebra

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

Appendix - Extra Word Problems - Page 524: 37

Answer

3,6

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.