Find the domain of the Equation: (24:3t+4)/(24y+48) = 7z

24:3t+424y+212+6=7z \frac{24:3t+4}{24y+21\cdot2+6}=7z

What is the domain of the equation?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:10 Let's find the domain of the function.
00:13 Remember, dividing by zero is not possible, so we need to avoid it.
00:18 The denominator has a variable. It shouldn't be zero.
00:22 This is the first case. Make sure the denominator isn't zero.
00:27 So, the first condition for variable T is checked.
00:38 Now, look at the main denominator of the function. It has a variable too.
00:45 It must also not be zero. Let's handle this one.
00:49 We'll go through the right steps to solve it properly.
01:01 Let's focus and isolate variable Y now.
01:14 And that's how we solve the problem. Great job!

Step-by-step written solution

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1

Understand the problem

24:3t+424y+212+6=7z \frac{24:3t+4}{24y+21\cdot2+6}=7z

What is the domain of the equation?

2

Step-by-step solution

To find the domain of definition for this equation, we need to identify all values of the variables that would make the expression undefined. An expression becomes undefined when we have division by zero.

Let me first clarify the notation and rewrite the equation properly:
243t+424y+212+6=7z \frac{\frac{24}{3t}+4}{24y+21\cdot2+6}=7z

Now, let's identify where division by zero could occur:

Step 1: Analyze the numerator
The numerator is 243t+4 \frac{24}{3t}+4 . For this to be defined, we need:
3t0 3t \neq 0
Therefore: t0 t \neq 0

Step 2: Analyze the denominator
The denominator is 24y+212+6 24y+21\cdot2+6 . Let's simplify this:
24y+42+6=24y+48 24y + 42 + 6 = 24y + 48

For the entire fraction to be defined, we need:
24y+480 24y + 48 \neq 0
24y48 24y \neq -48
y2 y \neq -2

Step 3: State the domain restrictions
For the equation to be defined, both conditions must be satisfied simultaneously:

  • t0 t \neq 0 (to avoid division by zero in the numerator)
  • y2 y \neq -2 (to avoid division by zero in the main denominator)

Therefore, the domain of definition is: y2,t0 y \neq -2, t \neq 0

3

Final Answer

y2,t0 y\operatorname{\ne}-2 , t\operatorname{\ne}0

Practice Quiz

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Solve for X:

\( x - 3 + 5 = 8 - 2 \)

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