Questions: Solve the system of equations using elimination: -5x-8y=-8 and 6x+9y=12.

Solution Steps

To solve the system of equations using elimination, we need to eliminate one of the variables by making the coefficients of that variable equal in both equations. We can then subtract or add the equations to eliminate that variable and solve for the other variable. Once we have one variable, we can substitute it back into one of the original equations to find the other variable.

We start with the given system of equations: \[ \begin{cases} -5x - 8y = -8 \\ 6x + 9y = 12 \end{cases} \]

To eliminate one of the variables, we can multiply the equations by suitable constants so that the coefficients of one of the variables are equal in magnitude. Let's eliminate \(x\).

Multiply the first equation by \(6\) and the second equation by \(5\): \[ \begin{cases} 6(-5x - 8y) = 6(-8) \\ 5(6x + 9y) = 5(12) \end{cases} \] This simplifies to: \[ \begin{cases} -30x - 48y = -48 \\ 30x + 45y = 60 \end{cases} \]

Add the two equations to eliminate \(x\): \[ (-30x - 48y) + (30x + 45y) = -48 + 60 \] This simplifies to: \[ -3y = 12 \]

Solve the equation for \(y\): \[ y = \frac{12}{-3} = -4 \]

Substitute \(y = -4\) into the first original equation to solve for \(x\): \[ -5x - 8(-4) = -8 \] This simplifies to: \[ -5x + 32 = -8 \] \[ -5x = -8 - 32 \] \[ -5x = -40 \] \[ x = \frac{-40}{-5} = 8 \]

Final Answer