To decide whether to use substitution or elimination to solve the system of equations:
-
Equations:
- \(8x - 15y = -32\) (Equation 1)
- \(6x + 3y = -5\) (Equation 2)
-
Checking for Convenience:
Substitution:
- In substitution, we typically solve one of the equations for one variable and then substitute it into the other equation.
- We could solve Equation 2 for \(y\): \[ 6x + 3y = -5 \implies 3y = -5 - 6x \implies y = -\frac{5 + 6x}{3} \]
- Substituting this expression for \(y\) back into Equation 1 would yield a complicated expression.
Elimination:
- In elimination, we can manipulate the equations to eliminate one variable by making the coefficients of one variable the same.
- For example, we can multiply Equation 2 by 5 to align the coefficients of \(y\): \[ 5(6x + 3y) = 5(-5) \implies 30x + 15y = -25 \]
- Then, we add this to Equation 1: \[ 8x - 15y + 30x + 15y = -32 - 25 \]
- This eliminates \(y\) and allows us to solve for \(x\) straightforwardly.
Conclusion: Considering the algebraic manipulation needed and the potential complexity of substitutions versus alignable coefficients, it appears that elimination is the more convenient method for this specific system of equations.