Solve {r}{5x+3y=c}{kx+4(5x-c/3)=1}

$$\left\{ \begin{array} { r } { 5 x + 3 y = c } \\ { k x + 4 ( \frac { 5 x - c } { 3 } ) = 1 } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{4c+3}{3k+20}\text{, }y=-\frac{5-ck}{3k+20}\text{, }&k\neq -\frac{20}{3}\\x=\frac{3\left(-4y-1\right)}{20}\text{, }y\in \mathrm{C}\text{, }&c=-\frac{3}{4}\text{ and }k=-\frac{20}{3}\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 3.

$$3kx+4\left(5x-c\right)=3$$

Use the distributive property to multiply 4 by 5x-c.

$$3kx+20x-4c=3$$

Add 4c to both sides.

$$3kx+20x=3+4c$$

Combine all terms containing x,y.

$$\left(3k+20\right)x=3+4c$$

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.

$$\left(3k+20\right)x=4c+3,5x+3y=c$$

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

$$\left(3k+20\right)x=4c+3$$

Divide both sides by 20+3k.

$$x=\frac{4c+3}{3k+20}$$

Substitute \frac{3+4c}{20+3k} for x in the other equation, 5x+3y=c.

$$5\times \frac{4c+3}{3k+20}+3y=c$$

Multiply 5 times \frac{3+4c}{20+3k}.

$$\frac{5\left(4c+3\right)}{3k+20}+3y=c$$

Subtract \frac{5\left(3+4c\right)}{20+3k} from both sides of the equation.

$$3y=\frac{3\left(ck-5\right)}{3k+20}$$

Divide both sides by 3.

$$y=\frac{ck-5}{3k+20}$$

The system is now solved.

$$x=\frac{4c+3}{3k+20},y=\frac{ck-5}{3k+20}$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{4c+3}{3k+20}\text{, }y=-\frac{5-ck}{3k+20}\text{, }&k\neq -\frac{20}{3}\\x=\frac{3\left(-4y-1\right)}{20}\text{, }y\in \mathrm{R}\text{, }&c=-\frac{3}{4}\text{ and }k=-\frac{20}{3}\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 3.

$$3kx+4\left(5x-c\right)=3$$

Use the distributive property to multiply 4 by 5x-c.

$$3kx+20x-4c=3$$

Add 4c to both sides.

$$3kx+20x=3+4c$$

Combine all terms containing x,y.

$$\left(3k+20\right)x=3+4c$$

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.

$$\left(3k+20\right)x=4c+3,5x+3y=c$$

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

$$\left(3k+20\right)x=4c+3$$

Divide both sides by 20+3k.

$$x=\frac{4c+3}{3k+20}$$

Substitute \frac{3+4c}{20+3k} for x in the other equation, 5x+3y=c.

$$5\times \frac{4c+3}{3k+20}+3y=c$$

Multiply 5 times \frac{3+4c}{20+3k}.

$$\frac{5\left(4c+3\right)}{3k+20}+3y=c$$

Subtract \frac{5\left(3+4c\right)}{20+3k} from both sides of the equation.

$$3y=\frac{3\left(ck-5\right)}{3k+20}$$

Divide both sides by 3.

$$y=\frac{ck-5}{3k+20}$$

The system is now solved.

$$x=\frac{4c+3}{3k+20},y=\frac{ck-5}{3k+20}$$