Question

for the following equation, a. write the value or values of the variable that make a denominator zero. these are the restrictions on the variable b. keeping the restrictions in mind, solve the equation
\\(\frac{2}{11x} + \frac{1}{4} = \frac{81}{22x} - \frac{1}{3}\\)

a. write the value or values of the variable that make a denominator zero.
0 (use a comma to separate answers as needed.)
b. what is the solution of the equation? select the correct choice below and, if necessary, fill in the answer box to complete your choice
a. the solution set is

b. the solution set is {x | x is a real number}
c. the solution set is \\(\varnothing\\).

Explanation:

Part a

Step1: Identify denominators

The denominators in the equation \(\frac{2}{11x}+\frac{1}{4}=\frac{81}{22x}-\frac{1}{3}\) are \(11x\), \(4\), \(22x\), and \(3\). The denominators that can be zero (since \(4\) and \(3\) are non - zero constants) are \(11x\) and \(22x\).

Step2: Solve for x when denominator is zero

Set \(11x = 0\) (or \(22x=0\), since if \(11x = 0\), then \(22x = 2\times11x=0\)). Solving \(11x=0\) gives \(x = 0\) (dividing both sides by \(11\)).

Step1: Find the least common denominator (LCD)

The denominators are \(11x\), \(4\), \(22x\), and \(3\). The prime factors of \(11\) is \(11\), of \(4\) is \(2^2\), of \(22\) is \(2\times11\), and of \(3\) is \(3\). The LCD of \(11x\), \(4\), \(22x\), and \(3\) is \(132x\) (since \(132x=12\times11x = 3\times4\times11x=6\times22x=44\times3x\)).

Step2: Multiply each term by the LCD

Multiply each term in the equation \(\frac{2}{11x}+\frac{1}{4}=\frac{81}{22x}-\frac{1}{3}\) by \(132x\):
\(\frac{2}{11x}\times132x+\frac{1}{4}\times132x=\frac{81}{22x}\times132x-\frac{1}{3}\times132x\)
Simplify each term:

• For \(\frac{2}{11x}\times132x\): \(\frac{2\times132x}{11x}=2\times12 = 24\) (the \(x\) terms cancel out, and \(132\div11 = 12\))
• For \(\frac{1}{4}\times132x\): \(\frac{132x}{4}=33x\)
• For \(\frac{81}{22x}\times132x\): \(\frac{81\times132x}{22x}=81\times6 = 486\) (the \(x\) terms cancel out, and \(132\div22 = 6\))
• For \(\frac{1}{3}\times132x\): \(\frac{132x}{3}=44x\)

So the equation becomes \(24 + 33x=486-44x\)

Step3: Solve the linear equation

Add \(44x\) to both sides: \(24+33x + 44x=486-44x+44x\), which simplifies to \(24 + 77x=486\)
Subtract \(24\) from both sides: \(77x=486 - 24=462\)
Divide both sides by \(77\): \(x=\frac{462}{77} = 6\)

Step4: Check the solution

We know from part (a) that \(x
eq0\). Substitute \(x = 6\) into the original equation:
Left - hand side (LHS): \(\frac{2}{11\times6}+\frac{1}{4}=\frac{2}{66}+\frac{1}{4}=\frac{1}{33}+\frac{1}{4}=\frac{4 + 33}{132}=\frac{37}{132}\)
Right - hand side (RHS): \(\frac{81}{22\times6}-\frac{1}{3}=\frac{81}{132}-\frac{1}{3}=\frac{81 - 44}{132}=\frac{37}{132}\)
Since \(LHS = RHS\) when \(x = 6\) and \(x = 6\) does not make any denominator zero, \(x = 6\) is a valid solution.

Answer:

\(0\)

Part b