Chapter 12: Q 36. (page 745)

Solve each equation. State any extraneous solutions.
$\frac{t - 3}{t + 3} - \frac{2 t}{t - 3} = - 1$

Short Answer

Expert verified

The solution of the equation is $0$ .

Step by step solution

Step 1. State the concept for observational studies.

The linear equations in one variable is an equation which is expressed in the form of $a x + b = 0$ , where a and b are two integers, andx is a variable and has only one solution.

Step 2. Solve the equation.

The equation can be solved as follows:

$\begin{matrix} \frac{t - 3}{t + 3} - \frac{2 t}{t - 3} = - 1 \\ \frac{(t - 3) (t - 3) - 2 t (t + 3)}{(t - 3) (t + 3)} = - 1 \\ (t - 3) (t - 3) - 2 t (t + 3) = - 1 (t - 3) (t + 3) \\ t^{2} - 3 t - 3 t + 9 - 2 t^{2} - 6 t = - t^{2} + 3 t - 3 t + 9 \\ - t^{2} - 12 t + 9 = - t^{2} + 9 \\ - 12 t = 0 \\ t = - \frac{0}{12} \\ t = 0 \end{matrix}$

Step 3. Check the solution.

To check the solution, substitute $t = 0$ for in the original equation,

$\begin{matrix} \frac{t - 3}{t + 3} - \frac{2 t}{t - 3} = - 1 \\ \frac{0 - 3}{0 + 3} - \frac{2 (0)}{0 - 3} = - 1 \\ \frac{- 3}{3} - \frac{0}{- 3} = - 1 \\ - 1 - 0 = - 1 \\ - 1 = - 1 \\ L H S = R H S \end{matrix}$

Since both sides are equal, the equation is correct.

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Solve each equation. State any extraneous solutions.
$\frac{t - 3}{t + 3} - \frac{2 t}{t - 3} = - 1$

Short Answer

Step by step solution

Step 1. State the concept for observational studies.

Step 2. Solve the equation.

Step 3. Check the solution.

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Solve each equation. State any extraneous solutions.t−3t+3−2tt−3=−1

Short Answer

Step by step solution

Step 1. State the concept for observational studies.

Step 2. Solve the equation.

Step 3. Check the solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Logic and Functions

Mechanics Maths

Decision Maths

Applied Mathematics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Solve each equation. State any extraneous solutions.
$\frac{t - 3}{t + 3} - \frac{2 t}{t - 3} = - 1$