Let's solve the given systems of linear equations step by step:
System (i):
[tex]\[
\begin{cases}
3x + 2y = 6, \\
2x - 3y = 17
\end{cases}
\][/tex]
To solve this system, we use the method of elimination or substitution. By solving, we find:
- From the first equation: [tex]\(3x + 2y = 6\)[/tex]
- From the second equation: [tex]\(2x - 3y = 17\)[/tex]
After solving these simultaneously, we get:
[tex]\[
x = 4, \quad y = -3
\][/tex]
System (iv):
[tex]\[
\begin{cases}
\frac{x}{2} + \frac{y}{3} = 8, \\
\frac{5x}{4} - 3y = -3
\end{cases}
\][/tex]
To solve this system, we first clear the fractions:
- Multiply the first equation by 6 to clear denominators:
[tex]\[ 3x + 2y = 48 \][/tex]
- Multiply the second equation by 4 to clear denominators:
[tex]\[ 5x - 12y = -12 \][/tex]
After solving these simultaneously, we find:
[tex]\[
x = 12, \quad y = 6
\][/tex]
Equation (vi):
[tex]\[
\frac{x + y}{2} + \frac{3x - 5y}{4} = 2
\][/tex]
Solve this by eliminating the fractions. Multiply through by the least common multiple of the denominators (4):
[tex]\[
2(x + y) + (3x - 5y) = 8
\][/tex]
Simplify and combine like terms:
[tex]\[
2x + 2y + 3x - 5y = 8 \\
5x - 3y = 8
\][/tex]
Let’s solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[
x = \frac{3y + 8}{5}
\][/tex]
So the solution for equation (vi) is a parametric form:
[tex]\[
x = \frac{3y}{5} + \frac{8}{5}
\][/tex]
Thus, the solutions are:
- System (i): [tex]\((x, y) = (4, -3)\)[/tex]
- System (iv): [tex]\((x, y) = (12, 6)\)[/tex]
- Equation (vi): [tex]\((x, y) = \left(\frac{3y}{5} + \frac{8}{5}, y\right)\)[/tex]
These are the detailed solutions for the given equations and systems.
