Solve {l}{x+3(y-1)=5}{5x+13-y=5}

$$\left. \begin{array} { l } { x + 3 ( y - 1 ) = 5 } \\ { 5 x + 13 - y = 5 } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=3$$

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.

$$x+3\left(y-1\right)=5,5x-y+13=5$$

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

$$x+3\left(y-1\right)=5$$

Multiply 3 times y-1.

$$x+3y-3=5$$

Add 3 to both sides of the equation.

$$x+3y=8$$

Subtract 3y from both sides of the equation.

$$x=-3y+8$$

Substitute -3y+8 for x in the other equation, 5x-y+13=5.

$$5\left(-3y+8\right)-y+13=5$$

Multiply 5 times -3y+8.

$$-15y+40-y+13=5$$

Add -15y to -y.

$$-16y+40+13=5$$

Add 40 to 13.

$$-16y+53=5$$

Subtract 53 from both sides of the equation.

$$-16y=-48$$

Divide both sides by -16.

$$y=3$$

Substitute 3 for y in x=-3y+8. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-3\times 3+8$$

Multiply -3 times 3.

$$x=-9+8$$

Add 8 to -9.

$$x=-1$$

The system is now solved.

$$x=-1,y=3$$

Steps Using Matrices

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

$$x+3\left(y-1\right)=5,5x-y+13=5$$

Simplify the first equation to put it in standard form.

$$x+3\left(y-1\right)=5$$

Multiply 3 times y-1.

$$x+3y-3=5$$

Add 3 to both sides of the equation.

$$x+3y=8$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&3\\5&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\-8\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&3\\5&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&3\\5&-1\end{matrix}\right))\left(\begin{matrix}1&3\\5&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\5&-1\end{matrix}\right))\left(\begin{matrix}8\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\5&-1\end{matrix}\right))\left(\begin{matrix}8\\-8\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\5&-1\end{matrix}\right))\left(\begin{matrix}8\\-8\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{-1-3\times 5}&-\frac{3}{-1-3\times 5}\\-\frac{5}{-1-3\times 5}&\frac{1}{-1-3\times 5}\end{matrix}\right)\left(\begin{matrix}8\\-8\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}&\frac{3}{16}\\\frac{5}{16}&-\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}8\\-8\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}\times 8+\frac{3}{16}\left(-8\right)\\\frac{5}{16}\times 8-\frac{1}{16}\left(-8\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=3$$