Chapter 2: Q 2.1 (page 99)

Determine whether the values are solutions to the equation: $9 y + 2 = 6 y + 3$ .
(a) $y = \frac{4}{3}$ (b) $y = \frac{1}{3}$ .

Short Answer

Expert verified

Part (a) $y = \frac{4}{3}$ is not a solution to the equation.

Part (b) $y = \frac{1}{3}$ is a solution to the equation.

Step by step solution

Part (a) Step 1. Given information.

Determine whether the values are solutions to the equation: $9 y + 2 = 6 y + 3$ .

(a) $y = \frac{4}{3}$ (b) $y = \frac{1}{3}$

Part (a) Step 2. Substitute the value y=43in the equation.

$9 y + 2 = 6 y + 3 9 (\frac{4}{3}) + 2 = 6 (\frac{4}{3}) + 3 3 \cdot 4 + 2 = 2 \cdot 4 + 3 14 \neq 11$

Thus $y = \frac{4}{3}$ does not result in a true equation.

Hence, $y = \frac{4}{3}$ is not a solution of the equation.

Part (b) Step 3. Substitute the value y=13in the equation.

$9 y + 2 = 6 y + 3 9 (\frac{1}{3}) + 2 = 6 (\frac{1}{3}) + 3 3 \cdot 1 + 2 = 2 \cdot 1 + 3 5 = 5$

Thus $y = \frac{1}{3}$ result in a true equation.

Hence, $y = \frac{1}{3}$ is a solution of the equation.

Step 4. Conclusion.

Hence, (a) $y = \frac{4}{3}$ is not a solution of the equation and (b) $y = \frac{1}{3}$ is a solution of the equation.

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Determine whether the values are solutions to the equation: $9 y + 2 = 6 y + 3$ .
(a) $y = \frac{4}{3}$ (b) $y = \frac{1}{3}$ .

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Substitute the value y=43in the equation.

Part (b) Step 3. Substitute the value y=13in the equation.

Step 4. Conclusion.

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Determine whether the values are solutions to the equation: 9y+2=6y+3.(a) y=43(b) y=13.

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Substitute the value y=43in the equation.

Part (b) Step 3. Substitute the value y=13in the equation.

Step 4. Conclusion.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Geometry

Pure Maths

Probability and Statistics

Mechanics Maths

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Determine whether the values are solutions to the equation: $9 y + 2 = 6 y + 3$ .
(a) $y = \frac{4}{3}$ (b) $y = \frac{1}{3}$ .