Question

solve for x. \\(\frac{4}{x - 2} + \frac{1}{x - 6} = \frac{4x - 5}{x^2 - 8x + 12}\\)\
\\(\bigcirc\\) 21\
\\(\bigcirc\\) \\(\frac{31}{9}\\)\
\\(\bigcirc\\) 11\
\\(\bigcirc\\) 3

Explanation:

Step1: Factor the denominator on the right

The denominator \(x^2 - 8x + 12\) factors to \((x - 2)(x - 6)\) since \(-2 \times -6 = 12\) and \(-2 + (-6) = -8\). So the equation becomes \(\frac{4}{x - 2} + \frac{1}{x - 6} = \frac{4x - 5}{(x - 2)(x - 6)}\).

Step2: Multiply through by the common denominator

Multiply each term by \((x - 2)(x - 6)\) (note that \(x
eq 2\) and \(x
eq 6\) to avoid division by zero) to get rid of the fractions:
\(4(x - 6) + 1(x - 2) = 4x - 5\)

Step3: Expand the left side

Expand \(4(x - 6)\) to \(4x - 24\) and \(1(x - 2)\) to \(x - 2\). So the left side becomes \(4x - 24 + x - 2\). Combine like terms: \(4x + x = 5x\) and \(-24 - 2 = -26\). So the left side is \(5x - 26\). The equation is now \(5x - 26 = 4x - 5\).

Step4: Solve for x

Subtract \(4x\) from both sides: \(5x - 4x - 26 = 4x - 4x - 5\), which simplifies to \(x - 26 = -5\). Then add 26 to both sides: \(x - 26 + 26 = -5 + 26\), so \(x = 21\). Wait, but let's check if this is valid. Wait, no, wait, let's re - check the expansion. Wait, \(4(x - 6)+(x - 2)=4x-24 + x - 2=5x - 26\), and the right side is \(4x - 5\). Then \(5x-26 = 4x - 5\). Subtract \(4x\): \(x-26=-5\), add 26: \(x = 21\). But wait, let's check if \(x = 21\) makes the original denominators non - zero. \(21-2 = 19
eq0\) and \(21 - 6=15
eq0\), so it's valid. Wait, but let's check our steps again. Wait, maybe I made a mistake. Wait, let's substitute \(x = \frac{31}{9}\)? Wait, no, let's do the algebra again. Wait, maybe I messed up the expansion. Wait, let's start over.

Wait, original equation: \(\frac{4}{x - 2}+\frac{1}{x - 6}=\frac{4x - 5}{(x - 2)(x - 6)}\)

Multiply both sides by \((x - 2)(x - 6)\):

\(4(x - 6)+(x - 2)=4x - 5\)

\(4x-24+x - 2=4x - 5\)

\(5x-26 = 4x - 5\)

\(5x-4x=-5 + 26\)

\(x = 21\). Wait, but 21 is one of the options. Wait, but let's check with \(x = 3\). Wait, maybe I made a mistake in the sign. Wait, no, let's check the original problem again. Wait, the denominator is \(x^2-8x + 12\), which is correct as \((x - 2)(x - 6)\). Let's check \(x = 3\):

Left side: \(\frac{4}{3 - 2}+\frac{1}{3 - 6}=\frac{4}{1}+\frac{1}{-3}=4-\frac{1}{3}=\frac{11}{3}\)

Right side: \(\frac{4\times3 - 5}{9 - 24 + 12}=\frac{12 - 5}{-3}=\frac{7}{-3}=-\frac{7}{3}\). Not equal.

\(x = 11\): Left side \(\frac{4}{11 - 2}+\frac{1}{11 - 6}=\frac{4}{9}+\frac{1}{5}=\frac{20 + 9}{45}=\frac{29}{45}\). Right side \(\frac{4\times11-5}{121-88 + 12}=\frac{44 - 5}{45}=\frac{39}{45}=\frac{13}{15}\). Not equal.

\(x=\frac{31}{9}\): Left side \(\frac{4}{\frac{31}{9}-2}+\frac{1}{\frac{31}{9}-6}=\frac{4}{\frac{31 - 18}{9}}+\frac{1}{\frac{31 - 54}{9}}=\frac{4\times9}{13}+\frac{9}{-23}=\frac{36}{13}-\frac{9}{23}=\frac{36\times23-9\times13}{13\times23}=\frac{828 - 117}{299}=\frac{711}{299}\approx2.378\)

Right side: \(\frac{4\times\frac{31}{9}-5}{(\frac{31}{9})^2-8\times\frac{31}{9}+12}=\frac{\frac{124}{9}-\frac{45}{9}}{\frac{961}{81}-\frac{248}{9}+12}=\frac{\frac{79}{9}}{\frac{961-2232 + 972}{81}}=\frac{\frac{79}{9}}{\frac{961 + 972-2232}{81}}=\frac{\frac{79}{9}}{\frac{1933 - 2232}{81}}=\frac{\frac{79}{9}}{\frac{-299}{81}}=\frac{79}{9}\times\frac{-81}{299}=\frac{79\times(-9)}{299}=\frac{-711}{299}\approx - 2.378\). Not equal.

Wait, but when \(x = 21\):

Left side: \(\frac{4}{21 - 2}+\frac{1}{21 - 6}=\frac{4}{19}+\frac{1}{15}=\frac{60 + 19}{285}=\frac{79}{285}\)

Right side: \(\frac{4\times21-5}{21^2-8\times21 + 12}=\frac{84 - 5}{441-168 + 12}=\frac{79}{285}\). Oh! I miscalculated the left side earlier. So \(x = 21\) is correct.

Answer:

21