Algebra Challenge: Solve the System of Equations 5y + 3x = 15 and -2y - 4x = -34

To solve the given system of equations using the elimination method, we proceed as follows:

• Step 1: Align the equations to eliminate one variable.
  We have: $\begin{aligned} 5y + 3x &= 15 \quad \text{(Equation 1)} \\ -2y - 4x &= -34 \quad \text{(Equation 2)} \end{aligned}$

• Step 2: Make the coefficients of $y$ equal in magnitude by manipulating the equations.
  Multiply Equation 1 by 2 and Equation 2 by 5:
  $\begin{aligned} (2) \cdot (5y + 3x) &= 2 \cdot 15 \quad \Rightarrow \quad 10y + 6x = 30 \\ (5) \cdot (-2y - 4x) &= 5 \cdot (-34) \quad \Rightarrow \quad -10y - 20x = -170 \end{aligned}$

• Step 3: Add the modified equations to eliminate $y$.
  $\begin{aligned} (10y + 6x) + (-10y - 20x) &= 30 + (-170) \\ 0y - 14x &= -140 \\ -14x &= -140 \end{aligned}$

• Step 4: Solve for $x$.
  Dividing by $-14$:
  $\begin{aligned} x &= \frac{-140}{-14} \\ x &= 10 \end{aligned}$

• Step 5: Substitute $x = 10$ back into one of the original equations to solve for $y$.
  Using Equation 1:
  $\begin{aligned} 5y + 3(10) &= 15 \\ 5y + 30 &= 15 \\ 5y &= 15 - 30 \\ 5y &= -15 \\ y &= \frac{-15}{5} \\ y &= -3 \end{aligned}$

Therefore, the solution to the system of equations is $x = 10$ and $y = -3$.