Certainly! Let's solve the given system of equations step-by-step.
The system of equations is:
[tex]\[
\begin{aligned}
1) \quad & 2.5y + 3x = 27 \\
2) \quad & 5x - 2.5y = 5
\end{aligned}
\][/tex]
Step 1: Add the two equations to eliminate [tex]\( y \)[/tex].
[tex]\[
(2.5y + 3x) + (5x - 2.5y) = 27 + 5
\][/tex]
Simplify the left-hand side by combining like terms:
[tex]\[
2.5y - 2.5y + 3x + 5x = 32
\][/tex]
This simplifies to:
[tex]\[
8x = 32
\][/tex]
So, the result of adding the two equations is:
[tex]\[
8x = 32
\][/tex]
Step 2: Solve for [tex]\( x \)[/tex] from the combined equation [tex]\( 8x = 32 \)[/tex].
[tex]\[
x = \frac{32}{8} = 4
\][/tex]
Step 3: Substitute [tex]\( x = 4 \)[/tex] back into one of the original equations to solve for [tex]\( y \)[/tex]. Let's use the first equation:
[tex]\[
2.5y + 3x = 27
\][/tex]
Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[
2.5y + 3 \cdot 4 = 27
\][/tex]
This simplifies to:
[tex]\[
2.5y + 12 = 27
\][/tex]
Subtract 12 from both sides:
[tex]\[
2.5y = 27 - 12
\][/tex]
[tex]\[
2.5y = 15
\][/tex]
Step 4: Solve for [tex]\( y \)[/tex]:
[tex]\[
y = \frac{15}{2.5} = 6
\][/tex]
Conclusion: The solution to the system of equations is [tex]\( x = 4 \)[/tex] and [tex]\( y = 6 \)[/tex].
The system has a unique solution:
[tex]\( \boxed{x = 4, y = 6} \)[/tex]
