1. $$ \left\{\begin{array}{l}y=3 x^{2}+6 x-24 \ y+33=18 x\end{array} ight. $$

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1. $$ \left\{\begin{array}{l}y=3 x^{2}+6 x-24 \ y+33=18 x\end{array}ight. $$

UpStudy AI · Accepted Answer

Solve the system of equations $$ y=3x^{2}+6x-24;y+33=18x $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}y=3x^{2}+6x-24\y+33=18x\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$3x^{2}+6x-24+33=18x$$
- step2: Simplify:
  $$3x^{2}+6x+9=18x$$
- step3: Move the expression to the left side:
  $$3x^{2}+6x+9-18x=0$$
- step4: Subtract the terms:
  $$3x^{2}-12x+9=0$$
- step5: Factor the expression:
  $$3\left(x-3ight)\left(x-1ight)=0$$
- step6: Divide the terms:
  $$\left(x-3ight)\left(x-1ight)=0$$
- step7: Separate into possible cases:
  $$\begin{align}&x-3=0\&x-1=0\end{align}$$
- step8: Solve the equation:
  $$\begin{align}&x=3\&x=1\end{align}$$
- step9: Calculate:
  $$x=3\cup x=1$$
- step10: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=3\y=3x^{2}+6x-24\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=3x^{2}+6x-24\end{array}ight.$$
- step11: Calculate:
  $$\left\{ \begin{array}{l}x=3\y=21\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=-15\end{array}ight.$$
- step12: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=-15\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=21\end{array}ight.$$
- step13: Check the solution:
  $$\left\{ \begin{array}{l}x=1\y=-15\end{array}ight.\cup \left\{ \begin{array}{l}x=3\y=21\end{array}ight.$$
- step14: Rewrite:
  $$\left(x,yight) = \left(1,-15ight)\cup \left(x,yight) = \left(3,21ight)$$
The solution to the system of equations is $$ (x,y) = (1,-15) $$ or $$ (x,y) = (3,21) $$.
The solutions are $$ (1, -15) $$ and $$ (3, 21) $$.

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