Question
In Problems 25-54, solve each system. Use any method you wish. $\left\{\begin{array}{r}2 y^2-3 x y+6 y+2 x+4=0 \\ 2 x-3 y+4=0\end{array}\right.$
Step 1
We choose the second equation for simplicity: \[ 2x - 3y + 4 = 0. \] Solve for \( x \): \[ 2x = 3y - 4 \] \[ x = \frac{3y - 4}{2}. \] Show more…
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