Solve the system by substitution. \[ \begin{aligned} y & =3 x+28 \\ y & =-4 x\end{aligned} \]

Solve the system of equations \( y=3x+28;y=-4x \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=3x+28\\y=-4x\end{array}\right.\) - step1: Substitute the value of \(y:\) \(3x+28=-4x\) - step2: Move the variable to the left side: \(3x+28+4x=0\) - step3: Add the terms: \(7x+28=0\) - step4: Move the constant to the right side: \(7x=0-28\) - step5: Remove 0: \(7x=-28\) - step6: Divide both sides: \(\frac{7x}{7}=\frac{-28}{7}\) - step7: Divide the numbers: \(x=-4\) - step8: Substitute the value of \(x:\) \(y=3\left(-4\right)+28\) - step9: Simplify: \(y=16\) - step10: Calculate: \(\left\{ \begin{array}{l}x=-4\\y=16\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=-4\\y=16\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(-4,16\right)\) The solution to the system of equations by substitution is \( (x,y) = (-4,16) \).