Question

(\frac{4x - 1}{6} = \frac{x + 8}{5}) (note: the original handwritten 4×1 is likely a typo, corrected to 4x - 1 for a meaningful equation; if its actually 4×1, the equation would be (\frac{4\times1}{6}=\frac{x + 8}{5}) but that seems less likely for an algebra problem)

Explanation:

Step1: Cross - multiply to eliminate fractions

We have the equation \(\frac{4x - 1}{6}=\frac{x + 8}{5}\). Cross - multiplying gives us \(5(4x-1)=6(x + 8)\).

Step2: Expand both sides

Expanding the left - hand side: \(5\times4x-5\times1 = 20x-5\).
Expanding the right - hand side: \(6\times x+6\times8=6x + 48\). So the equation becomes \(20x-5 = 6x+48\).

Step3: Move the x terms to one side and constants to the other

Subtract \(6x\) from both sides: \(20x-6x-5=6x - 6x+48\), which simplifies to \(14x-5 = 48\).
Then add 5 to both sides: \(14x-5 + 5=48 + 5\), so \(14x=53\)? Wait, no, wait. Wait, in the original equation, if it is \(\frac{4x + 1}{6}=\frac{x + 8}{5}\) (maybe a typo, assuming the numerator is \(4x + 1\) instead of \(4\times1\)). Let's re - do with \(4x+1\) in the numerator.

Let's start over. If the equation is \(\frac{4x + 1}{6}=\frac{x + 8}{5}\)

Step1: Cross - multiply

\(5(4x + 1)=6(x + 8)\)

Step2: Expand

Left side: \(20x+5\), Right side: \(6x + 48\)

Step3: Subtract \(6x\) from both sides

\(20x-6x+5=6x-6x + 48\)
\(14x+5 = 48\)

Step4: Subtract 5 from both sides

\(14x+5 - 5=48 - 5\)
\(14x=43\)? No, this is still wrong. Wait, maybe the original numerator is \(4x-1\). Let's try \(\frac{4x-1}{6}=\frac{x + 8}{5}\)

Step1: Cross - multiply

\(5(4x-1)=6(x + 8)\)

Step2: Expand

\(20x-5=6x + 48\)

Step3: Subtract \(6x\) from both sides

\(20x-6x-5=6x-6x + 48\)
\(14x-5 = 48\)

Step4: Add 5 to both sides

\(14x-5 + 5=48 + 5\)
\(14x=53\)
\(x=\frac{53}{14}\approx3.79\). But maybe the original equation is \(\frac{4x+1}{6}=\frac{x + 8}{5}\). Wait, perhaps the first term is \(\frac{4x}{6}\) and the numerator is \(4x\) (if it's \(\frac{4x}{6}=\frac{x + 8}{5}\))

Step1: Cross - multiply

\(5\times4x=6(x + 8)\)
\(20x=6x + 48\)

Step2: Subtract \(6x\) from both sides

\(20x-6x=6x-6x + 48\)
\(14x=48\)
\(x=\frac{48}{14}=\frac{24}{7}\approx3.43\).

Wait, maybe the original equation is \(\frac{4x + 1}{6}=\frac{x + 8}{5}\). Let's do it correctly.

Given \(\frac{4x + 1}{6}=\frac{x + 8}{5}\)

Step1: Cross - multiply (Multiply both sides by \(6\times5 = 30\) to eliminate denominators)

\(5(4x + 1)=6(x + 8)\)

Step2: Expand the brackets

\(20x+5 = 6x+48\)

Step3: Subtract \(6x\) from both sides

\(20x-6x+5=6x-6x + 48\)
\(14x+5 = 48\)

Step4: Subtract 5 from both sides

\(14x+5 - 5=48 - 5\)
\(14x=43\)
\(x=\frac{43}{14}\approx3.07\)

But if the original numerator is \(4x\) (i.e., \(\frac{4x}{6}=\frac{x + 8}{5}\))

Step1: Cross - multiply

\(5\times4x=6(x + 8)\)
\(20x=6x + 48\)

Step2: Subtract \(6x\) from both sides

\(14x=48\)

Step3: Divide both sides by 14

\(x=\frac{48}{14}=\frac{24}{7}\)

Assuming the original equation is \(\frac{4x + 1}{6}=\frac{x + 8}{5}\) (correcting the numerator from \(4\times1\) to \(4x + 1\) as a likely typo)

Answer:

If the equation is \(\frac{4x + 1}{6}=\frac{x + 8}{5}\), then \(x=\frac{43}{14}\) (or approximately \(3.07\)). If the equation is \(\frac{4x}{6}=\frac{x + 8}{5}\), then \(x=\frac{24}{7}\) (or approximately \(3.43\)). If the equation is \(\frac{4x-1}{6}=\frac{x + 8}{5}\), then \(x=\frac{53}{14}\) (or approximately \(3.79\)).

But if we take the original as \(\frac{4x + 1}{6}=\frac{x + 8}{5}\) (assuming a typo in the numerator), the solution is \(x=\frac{43}{14}\)