Question

Part 1: Explain how you calculate the restricted value(s) of x for the equation and what those value(s) represent.
Part 2: Explain which method (cross-multiplication or LCD) you would use to solve the equation and why.
Part 3: Using complete sentences, explain the steps to solve the equation using the method you chose in step 2. Provide the solution to the equation. (6 points)

Answer 1

This is the review I was given.

Review

When an algebraic equation includes a fraction on one side of the equal sign, you eliminate the fraction by multiplying both sides of the equation by the denominator.

When solving an equation with a variable in the denominator, you need to determine any values for x that would make the equation undefined. These values will be restrictions on x.

When the denominators on both sides of the equation are the same, you can solve the equation by setting the numerators equal to one another and solving for the variable.

You can solve rational equations by setting the cross-products equal to one another, or by multiplying both sides of the equation by the LCD.

Cross-Multiplication Method:

Step 1: Identify any restrictions on the variable.

Step 2: Cross-multiply the fractions by multiplying the opposite numerators and denominators.

Step 3: Solve for x.

Step 4: Check your solution by making sure it is not a restricted value, and by substituting it back into the original equation.

LCD Method:

Step 1: Identify any restrictions on the variable.

Step 2: Determine the LCD of the fractions.

Step 3: Multiply both sides of the equation by the LCD and simplify.

Step 4: Solve for x.

Step 5: Check your solution by making sure it is not a restricted value, and by substituting it back into the original equation.

Answer 2

I dont understand what part 1 is asking I get everything else

Answer 3

am I just solving for x @ParthKohli ?

Answer 4

Yes, you get the restricted value of a variable when the variable is found in the denominator. This is because then the variable may also include a value that may bring the denominator to 0.

Answer 5

For example see this expression:

\( \color{Black}{\Rightarrow \large {3 \over x} }\)

Here x \(\ne\) 0 because a denominator of 0 will bring the value to undefined.

Answer 6

https://www.connexus.com/content/media/527936-7112011-15817-PM-1275663069.jpg this is the equation

Answer 7

ok

Answer 8

first step multiply both sides by x+2?

Answer 9

So you just equate the denominators which have the denominator as x and 0.

Answer 10

wait what?

Answer 11

I'm talking about the restricted value

Answer 12

x \(\ne \) -2, 0

Answer 13

ohh!!! So would the actual value be 14?

Answer 14

If you want to check, you may use wolfram for this, I think

Answer 15

ok

Answer 16

http://www.wolframalpha.com/input/?i=3%2F%28x%2B2%29+%2B+%281%2F2x%29+%3D+4%2F%28x+%2B+2%29

Answer 17

oh wow thats really helpful! Thank you so much!

Answer 18

:)