Solve 2x+3y=18text{stit}6x+ky=50

$$2 x + 3 y = 18 \text { stit } 6 x + k y = 50$$

Solve for x, y

$$x=-\frac{3\left(25-3k\right)}{k-9}$$

$$y=-\frac{4}{k-9}$$

$$k\neq 9$$

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.

$$2x+3y=18,6x+ky=50$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2x+3y=18$$

Subtract 3y from both sides of the equation.

$$2x=-3y+18$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-3y+18\right)$$

Multiply \frac{1}{2} times -3y+18.

$$x=-\frac{3}{2}y+9$$

Substitute -\frac{3y}{2}+9 for x in the other equation, 6x+ky=50.

$$6\left(-\frac{3}{2}y+9\right)+ky=50$$

Multiply 6 times -\frac{3y}{2}+9.

$$-9y+54+ky=50$$

Add -9y to ky.

$$\left(k-9\right)y+54=50$$

Subtract 54 from both sides of the equation.

$$\left(k-9\right)y=-4$$

Divide both sides by -9+k.

$$y=-\frac{4}{k-9}$$

Substitute -\frac{4}{-9+k} for y in x=-\frac{3}{2}y+9. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\left(-\frac{4}{k-9}\right)+9$$

Multiply -\frac{3}{2} times -\frac{4}{-9+k}.

$$x=\frac{6}{k-9}+9$$

Add 9 to \frac{6}{-9+k}.

$$x=\frac{3\left(3k-25\right)}{k-9}$$

The system is now solved.

$$x=\frac{3\left(3k-25\right)}{k-9},y=-\frac{4}{k-9}$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$2x+3y=18,6x+ky=50$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&3\\6&k\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}18\\50\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\6&k\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&3\\6&k\end{matrix}\right))\left(\begin{matrix}2&3\\6&k\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\6&k\end{matrix}\right))\left(\begin{matrix}18\\50\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\6&k\end{matrix}\right))\left(\begin{matrix}18\\50\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\6&k\end{matrix}\right))\left(\begin{matrix}18\\50\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{k}{2k-3\times 6}&-\frac{3}{2k-3\times 6}\\-\frac{6}{2k-3\times 6}&\frac{2}{2k-3\times 6}\end{matrix}\right)\left(\begin{matrix}18\\50\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{k}{2\left(k-9\right)}&-\frac{3}{2\left(k-9\right)}\\-\frac{3}{k-9}&\frac{1}{k-9}\end{matrix}\right)\left(\begin{matrix}18\\50\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{k}{2\left(k-9\right)}\times 18+\left(-\frac{3}{2\left(k-9\right)}\right)\times 50\\\left(-\frac{3}{k-9}\right)\times 18+\frac{1}{k-9}\times 50\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3\left(3k-25\right)}{k-9}\\-\frac{4}{k-9}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{3\left(3k-25\right)}{k-9},y=-\frac{4}{k-9}$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$2x+3y=18,6x+ky=50$$

To make 2x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 2.

$$6\times 2x+6\times 3y=6\times 18,2\times 6x+2ky=2\times 50$$

Simplify.

$$12x+18y=108,12x+2ky=100$$

Subtract 12x+2ky=100 from 12x+18y=108 by subtracting like terms on each side of the equal sign.

$$12x-12x+18y+\left(-2k\right)y=108-100$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$18y+\left(-2k\right)y=108-100$$

Add 18y to -2ky.

$$\left(18-2k\right)y=108-100$$

Add 108 to -100.

$$\left(18-2k\right)y=8$$

Divide both sides by 18-2k.

$$y=\frac{4}{9-k}$$

Substitute \frac{4}{9-k} for y in 6x+ky=50. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x+k\times \frac{4}{9-k}=50$$

Multiply k times \frac{4}{9-k}.

$$6x+\frac{4k}{9-k}=50$$

Subtract \frac{4k}{9-k} from both sides of the equation.

$$6x=\frac{18\left(25-3k\right)}{9-k}$$

Divide both sides by 6.

$$x=\frac{3\left(25-3k\right)}{9-k}$$

The system is now solved.

$$x=\frac{3\left(25-3k\right)}{9-k},y=\frac{4}{9-k}$$