Solve {l}{cx+3y+(3-c)=0,12x+}{cy-c=0}

$$\left. \begin{array}{l}{ c x + 3 y + ( 3 - c ) = 0,12 x + }\\{ c y - c = 0 }\end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x\in \mathrm{C}\text{, }y=-1\text{, }&\text{unconditionally}\\x=\frac{c-6}{c}\text{, }y=1\text{, }&c\neq 0\end{matrix}\right.$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$cy-c=0,3y+cx+3-c=0$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$cy-c=0$$

Add c to both sides of the equation.

$$cy=c$$

Divide both sides by c.

$$y=1$$

Substitute 1 for y in the other equation, 3y+cx+3-c=0.

$$3+cx+3-c=0$$

Add 3 to 3-c.

$$cx+6-c=0$$

Subtract 6-c from both sides of the equation.

$$cx=c-6$$

Divide both sides by c.

$$x=\frac{c-6}{c}$$

The system is now solved.

$$y=1,x=\frac{c-6}{c}$$

Solve for x, y

$$\left\{\begin{matrix}\\x\in \mathrm{R}\text{, }y=-1\text{, }&\text{unconditionally}\\x=\frac{c-6}{c}\text{, }y=1\text{, }&c\neq 0\end{matrix}\right.$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$cy-c=0,3y+cx+3-c=0$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$cy-c=0$$

Add c to both sides of the equation.

$$cy=c$$

Divide both sides by c.

$$y=1$$

Substitute 1 for y in the other equation, 3y+cx+3-c=0.

$$3+cx+3-c=0$$

Add 3 to 3-c.

$$cx+6-c=0$$

Subtract 6-c from both sides of the equation.

$$cx=c-6$$

Divide both sides by c.

$$x=\frac{c-6}{c}$$

The system is now solved.

$$y=1,x=\frac{c-6}{c}$$