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Elementary Algebra

Lynn Marecek, MaryAnne Anthony-Smith

$Math Studyset 91影视 Explanations$ Math

2nd Edition 路 1240 pages

ISBN: 9780998625713

Chapter 9: Q. 423 (page 1108)

In the following exercises, solve.
$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

Short Answer

Expert verified

Part (a) : $\frac{2401}{100}$

Part (b) : 6

Step by step solution

01

Step 1. Given Information

The given expressions are

$(a) \sqrt{x} + 10 = \sqrt{x + 2} (b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4}$

02

Step 2. Squaring on both sides

$(a) \sqrt{x} + 10 = \sqrt{x + 2} {(\sqrt{x} + 10)}^{2} = {(\sqrt{x + 2})}^{2} 98 + 20 \sqrt{x} = 0 20 \sqrt{x} = - 98 S q u a r i n g o n b o t h s i d e s {(20 \sqrt{x})}^{2} = 9604 x = \frac{2401}{100}$

The solution is $\frac{2401}{100}$

$(b) \sqrt{y - 2} + 2 = \sqrt{2 y + 4} {(\sqrt{y - 2} + 2)}^{2} = {(\sqrt{2 y + 4})}^{2} - 2 + 4 \sqrt{y - 2} = y 4 \sqrt{y - 2} = y + 2 s q u a r i n g o n b o t h s i d e s 16 (y - 2) = y^{2} + 4 + 4 y 16 y - 32 = y^{2} + 4 + 4 y y^{2} - 12 y + 36 = 0 y = 6$

The solution is 6

03

Step 3. The solution

The solution for are $\frac{2401}{100} a n d 6$

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