To solve the system of linear equations and find the value of [tex]\(\frac{x}{y}\)[/tex], let's proceed with the following steps:
Step 1: Write down the given equations:
[tex]\[
\frac{3x}{2} - \frac{2y}{3} = 5 \quad \text{(Equation 1)}
\][/tex]
[tex]\[
\frac{5x}{6} + \frac{7y}{12} = \frac{17}{2} \quad \text{(Equation 2)}
\][/tex]
Step 2: Clear the fractions by multiplying both sides of each equation by the least common multiple (LCM) of the denominators.
For Equation 1, the denominators are 2 and 3. The LCM is 6.
[tex]\[
6 \left( \frac{3x}{2} \right) - 6\left(\frac{2y}{3}\right) = 6 \cdot 5
\][/tex]
[tex]\[
9x - 4y = 30 \quad \text{(Equation 3)}
\][/tex]
For Equation 2, the denominators are 6 and 12. The LCM is 12.
[tex]\[
12 \left( \frac{5x}{6} \right) + 12 \left( \frac{7y}{12} \right) = 12 \left( \frac{17}{2} \right)
\][/tex]
[tex]\[
10x + 7y = 102 \quad \text{(Equation 4)}
\][/tex]
Step 3: Solve the system of linear equations using the substitution or elimination method.
We'll use the elimination method. Multiply Equation 3 by 7 and Equation 4 by 4 to align the coefficients of [tex]\( y \)[/tex] for elimination:
[tex]\[
7(9x - 4y) = 7 \cdot 30
\][/tex]
[tex]\[
63x - 28y = 210 \quad \text{(Equation 5)}
\][/tex]
[tex]\[
4(10x + 7y) = 4 \cdot 102
\][/tex]
[tex]\[
40x + 28y = 408 \quad \text{(Equation 6)}
\][/tex]
Step 4: Add Equations 5 and 6 to eliminate [tex]\( y \)[/tex]:
[tex]\[
63x - 28y + 40x + 28y = 210 + 408
\][/tex]
[tex]\[
103x = 618
\][/tex]
[tex]\[
x = \frac{618}{103}
\][/tex]
[tex]\[
x = 6
\][/tex]
Step 5: Substitute [tex]\(x = 6\)[/tex] back into one of the original equations to find [tex]\(y\)[/tex]:
Using Equation 3:
[tex]\[
9x - 4y = 30
\][/tex]
[tex]\[
9(6) - 4y = 30
\][/tex]
[tex]\[
54 - 4y = 30
\][/tex]
[tex]\[
-4y = 30 - 54
\][/tex]
[tex]\[
-4y = -24
\][/tex]
[tex]\[
y = \frac{-24}{-4}
\][/tex]
[tex]\[
y = 6
\][/tex]
Step 6: Calculate [tex]\(\frac{x}{y}\)[/tex]:
[tex]\[
\frac{x}{y} = \frac{6}{6} = 1
\][/tex]
Therefore, the value of [tex]\(\frac{x}{y}\)[/tex] is [tex]\( \boxed{1} \)[/tex].
