What Are the Variable Restrictions in the Equation 5/x = (10/3x) + 4?

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The equation 5/x = (10/3x) + 4 has variable restrictions that must be considered to avoid undefined expressions. Specifically, the variable x cannot equal zero, as this would make the denominators of both fractions undefined. The correct solution to the equation is x = 5/12, which adheres to the established restrictions. Therefore, the only restriction on x is that x ≠ 0.

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Write the restrictions on the variables for the following equation. Keeping in mind the restrictions, solve the equation

5/x=(10/3x)+4

Here is what I did, but I'm not sure if it's right.

(3x)(5/x)=(3x)[(10/3x)+4]

15=(3x)(10/3x)+(3x)4

15=10+12x

5=12x

x=5/12

But I don't know what to do about the restrictions. Can someone help? Thank you in advance.
 
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Hi rynesdad5,:)

Welcome to MHB!

You've solved the problem correctly but you have to keep in mind that the denominator of any fraction cannot be zero. If the denominator is zero, then the expression is not real because its overall value is undefined, since, e.g. $\dfrac{2}{0}=\infty$.

In your problem, you have two fractions there, one is $\dfrac{5}{x}$ and the other is $\dfrac{10}{3x}$. What can you say about the value(s) of $x$ that you couldn't take for these two fractions?
 
rynesdad5 said:
Write the restrictions on the variables for the following equation. Keeping in mind the restrictions, solve the equation

5/x=(10/3x)+4

Here is what I did, but I'm not sure if it's right.

(3x)(5/x)=(3x)[(10/3x)+4]

15=(3x)(10/3x)+(3x)4

15=10+12x

5=12x

x=5/12

But I don't know what to do about the restrictions. Can someone help? Thank you in advance.

I think you have solved the problem correctly..If x is in denominator itself you should take care of x,it should not be zero
 

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