Solve {l}{5x-3y=10}{6y=kx-42}

$$\left. \begin{array} { l } { 5 x - 3 y = 10 } \\ { 6 y = k x - 42 } \end{array} \right.$$

Solve for x, y

$$x=\frac{22}{k-10}$$

$$y=-\frac{10\left(k-21\right)}{3\left(k-10\right)}$$

$$k\neq 10$$

Steps Using Substitution

Consider the second equation. Subtract kx from both sides.

$$6y-kx=-42$$

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.

$$5x-3y=10,\left(-k\right)x+6y=-42$$

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

$$5x-3y=10$$

Add 3y to both sides of the equation.

$$5x=3y+10$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(3y+10\right)$$

Multiply \frac{1}{5} times 3y+10.

$$x=\frac{3}{5}y+2$$

Substitute \frac{3y}{5}+2 for x in the other equation, \left(-k\right)x+6y=-42.

$$\left(-k\right)\left(\frac{3}{5}y+2\right)+6y=-42$$

Multiply -k times \frac{3y}{5}+2.

$$\left(-\frac{3k}{5}\right)y-2k+6y=-42$$

Add -\frac{3ky}{5} to 6y.

$$\left(-\frac{3k}{5}+6\right)y-2k=-42$$

Add 2k to both sides of the equation.

$$\left(-\frac{3k}{5}+6\right)y=2k-42$$

Divide both sides by -\frac{3k}{5}+6.

$$y=\frac{10\left(k-21\right)}{3\left(10-k\right)}$$

Substitute \frac{10\left(-21+k\right)}{3\left(-k+10\right)} for y in x=\frac{3}{5}y+2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{5}\times \frac{10\left(k-21\right)}{3\left(10-k\right)}+2$$

Multiply \frac{3}{5} times \frac{10\left(-21+k\right)}{3\left(-k+10\right)}.

$$x=\frac{2\left(k-21\right)}{10-k}+2$$

Add 2 to \frac{2\left(-21+k\right)}{-k+10}.

$$x=-\frac{22}{10-k}$$

The system is now solved.

$$x=-\frac{22}{10-k},y=\frac{10\left(k-21\right)}{3\left(10-k\right)}$$

Steps Using Matrices

Consider the second equation. Subtract kx from both sides.

$$6y-kx=-42$$

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

$$5x-3y=10,\left(-k\right)x+6y=-42$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\-42\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-3\\-k&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right))\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right))\left(\begin{matrix}10\\-42\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right))\left(\begin{matrix}10\\-42\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-3\\-k&6\end{matrix}\right))\left(\begin{matrix}10\\-42\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5\times 6-\left(-3\left(-k\right)\right)}&-\frac{-3}{5\times 6-\left(-3\left(-k\right)\right)}\\-\frac{-k}{5\times 6-\left(-3\left(-k\right)\right)}&\frac{5}{5\times 6-\left(-3\left(-k\right)\right)}\end{matrix}\right)\left(\begin{matrix}10\\-42\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{10-k}&\frac{1}{10-k}\\\frac{k}{3\left(10-k\right)}&\frac{5}{3\left(10-k\right)}\end{matrix}\right)\left(\begin{matrix}10\\-42\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{10-k}\times 10+\frac{1}{10-k}\left(-42\right)\\\frac{k}{3\left(10-k\right)}\times 10+\frac{5}{3\left(10-k\right)}\left(-42\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{22}{10-k}\\-\frac{10\left(21-k\right)}{3\left(10-k\right)}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{22}{10-k},y=-\frac{10\left(21-k\right)}{3\left(10-k\right)}$$

Steps Using Elimination

Consider the second equation. Subtract kx from both sides.

$$6y-kx=-42$$

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.

$$5x-3y=10,\left(-k\right)x+6y=-42$$

To make 5x and -kx equal, multiply all terms on each side of the first equation by -k and all terms on each side of the second by 5.

$$\left(-k\right)\times 5x+\left(-k\right)\left(-3\right)y=\left(-k\right)\times 10,5\left(-k\right)x+5\times 6y=5\left(-42\right)$$

Simplify.

$$\left(-5k\right)x+3ky=-10k,\left(-5k\right)x+30y=-210$$

Subtract \left(-5k\right)x+30y=-210 from \left(-5k\right)x+3ky=-10k by subtracting like terms on each side of the equal sign.

$$\left(-5k\right)x+5kx+3ky-30y=-10k+210$$

Add -5kx to 5kx. Terms -5kx and 5kx cancel out, leaving an equation with only one variable that can be solved.

$$3ky-30y=-10k+210$$

Add 3ky to -30y.

$$\left(3k-30\right)y=-10k+210$$

Add -10k to 210.

$$\left(3k-30\right)y=210-10k$$

Divide both sides by -30+3k.

$$y=\frac{10\left(21-k\right)}{3\left(k-10\right)}$$

Substitute \frac{10\left(21-k\right)}{3\left(-10+k\right)} for y in \left(-k\right)x+6y=-42. Because the resulting equation contains only one variable, you can solve for x directly.

$$\left(-k\right)x+6\times \frac{10\left(21-k\right)}{3\left(k-10\right)}=-42$$

Multiply 6 times \frac{10\left(21-k\right)}{3\left(-10+k\right)}.

$$\left(-k\right)x+\frac{20\left(21-k\right)}{k-10}=-42$$

Subtract \frac{20\left(21-k\right)}{-10+k} from both sides of the equation.

$$\left(-k\right)x=-\frac{22k}{k-10}$$

Divide both sides by -k.

$$x=\frac{22}{k-10}$$

The system is now solved.

$$x=\frac{22}{k-10},y=\frac{10\left(21-k\right)}{3\left(k-10\right)}$$