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Let's start by solving the first system of equations:

**System a:**
\[
\begin{align*}
2x + 3y &= -1 \\
3x + 4y &= 0
\end{align*}
\]

We'll use the elimination method by eliminating \(y\). First, we'll manipulate the equations to align the coefficients of \(y\).

**Multiply the first equation by 4:**
\[
8x + 12y = -4
\]

**Multiply the second equation by 3:**
\[
9x + 12y = 0
\]

Now, subtract the first modified equation from the second:
\[
(9x + 12y) - (8x + 12y) = 0 - (-4)
\]
\[
x = 4
\]

Plug \(x = 4\) back into the first original equation:
\[
2(4) + 3y = -1
\]
\[
8 + 3y = -1
\]
\[
3y = -9
\]
\[
y = -3
\]

The solution for system (a) is \((x, y) = (4, -3)\).

**System b:**
\[
\begin{align*}
3x + 2y &= 13 \\
-x + y &= -1
\end{align*}
\]

For system (b), let's isolate \(y\) in the second equation:
\[
y = -1 + x
\]

Substitute this expression for

Taller 1 – Algebra lineal. Resolver el sistema de ecuaciones usando métodos como sustitución, igualación, reducción y la regla de Cramer.

Learn how to solve systems of linear equations using methods such as substitution, elimination, equalization, and Cramer's Rule. Includes step-by-step solutions for algebraic problems involving determinants and systems with two variables.

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