Solve {l}{x+4y=-16}{3x-2(y-2)=-2}

$$\left\{ \begin{array} { l } { x + 4 y = - 16 } \\ { 3 x - 2 ( y - 2 ) = - 2 } \end{array} \right.$$

Solve for x, y

$$x=-4$$

$$y=-3$$

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.

$$x+4y=-16,3x-2\left(y-2\right)=-2$$

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

$$x+4y=-16$$

Subtract 4y from both sides of the equation.

$$x=-4y-16$$

Substitute -4y-16 for x in the other equation, 3x-2\left(y-2\right)=-2.

$$3\left(-4y-16\right)-2\left(y-2\right)=-2$$

Multiply 3 times -4y-16.

$$-12y-48-2\left(y-2\right)=-2$$

Multiply -2 times y-2.

$$-12y-48-2y+4=-2$$

Add -12y to -2y.

$$-14y-48+4=-2$$

Add -48 to 4.

$$-14y-44=-2$$

Add 44 to both sides of the equation.

$$-14y=42$$

Divide both sides by -14.

$$y=-3$$

Substitute -3 for y in x=-4y-16. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-4\left(-3\right)-16$$

Multiply -4 times -3.

$$x=12-16$$

Add -16 to 12.

$$x=-4$$

The system is now solved.

$$x=-4,y=-3$$

Steps Using Matrices

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

$$x+4y=-16,3x-2\left(y-2\right)=-2$$

Simplify the second equation to put it in standard form.

$$3x-2\left(y-2\right)=-2$$

Multiply -2 times y-2.

$$3x-2y+4=-2$$

Subtract 4 from both sides of the equation.

$$3x-2y=-6$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&4\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&4\\3&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&4\\3&-2\end{matrix}\right))\left(\begin{matrix}1&4\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&4\\3&-2\end{matrix}\right))\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&4\\3&-2\end{matrix}\right))\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&4\\3&-2\end{matrix}\right))\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-4\times 3}&-\frac{4}{-2-4\times 3}\\-\frac{3}{-2-4\times 3}&\frac{1}{-2-4\times 3}\end{matrix}\right)\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7}&\frac{2}{7}\\\frac{3}{14}&-\frac{1}{14}\end{matrix}\right)\left(\begin{matrix}-16\\-6\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7}\left(-16\right)+\frac{2}{7}\left(-6\right)\\\frac{3}{14}\left(-16\right)-\frac{1}{14}\left(-6\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-4,y=-3$$