Solve the simultaneous equation \(\frac{\text{x}}{2}\) - \(\frac{\text{y}}{5}\) = 1 and y - \(\frac{\text{x}}{3}\) = 8

We are given two simultaneous equations:

\[ \frac{x}{2} - \frac{y}{5} = 1 \] \[ y - \frac{x}{3} = 8 \]

Step 1: Make the second equation easier to work with.
We can write the second equation as:

\[ y - \frac{x}{3} = 8 \] Add \(\frac{x}{3}\) to both sides: \[ y = 8 + \frac{x}{3} \]

This expresses \(y\) in terms of \(x\).

Step 2: Substitute for \(y\) in the first equation.
Plug \(y = 8 + \frac{x}{3}\) into the first equation:

\[ \frac{x}{2} - \frac{y}{5} = 1 \] \[ \frac{x}{2} - \frac{1}{5}\left(8 + \frac{x}{3}\right) = 1 \] Expand: \[ \frac{x}{2} - \frac{8}{5} - \frac{x}{15} = 1 \]

Step 3: Get all terms involving \(x\) together and combine like terms.

• The terms with \(x\) are \(\frac{x}{2}\) and \(-\frac{x}{15}\)
• Combine them by expressing both over a common denominator, which is 30.

\[ \frac{x}{2} = \frac{15x}{30} \] \[ -\frac{x}{15} = -\frac{2x}{30} \] So, \[ \frac{15x}{30} - \frac{2x}{30} - \frac{8}{5} = 1 \] \[ \frac{13x}{30} - \frac{8}{5} = 1 \]

Step 4: Solve for \(x\).

• Add \(\frac{8}{5}\) to both sides to isolate the term with \(x\):

\[ \frac{13x}{30} = 1 + \frac{8}{5} \] \[ 1 = \frac{5}{5} \] So, \[ 1 + \frac{8}{5} = \frac{5}{5} + \frac{8}{5} = \frac{13}{5} \] Thus, \[ \frac{13x}{30} = \frac{13}{5} \]

• To solve for \(x\), multiply both sides by 30:

\[ 13x = 30 \times \frac{13}{5} \] \[ 13x = 13 \times 6 \] \[ 13x = 78 \] \[ x = \frac{78}{13} = 6 \]

Step 5: Substitute \(x = 6\) back into the expression for \(y\):

\[ y = 8 + \frac{x}{3} \] \[ y = 8 + \frac{6}{3} \] \[ y = 8 + 2 = 10 \]

Step 6: Check the solution in the original equations.

• First equation: \(\frac{x}{2} - \frac{y}{5} = 1\)

\[ \frac{6}{2} - \frac{10}{5} = 3 - 2 = 1 \]

• Second equation: \(y - \frac{x}{3} = 8\)

\[ 10 - \frac{6}{3} = 10 - 2 = 8 \]

Both equations are satisfied.

Conclusion:
The correct solution is \(x = 6\), \(y = 10\). This answer is correct because when these values are substituted into the original equations, they satisfy both equations. The main method used here is substitution — expressing one variable in terms of the other and then solving. This is a standard way to solve systems of linear equations.