Solve {l}{x+y=16}{3(x+2)+(y+2)=34+2}

$$\left\{ \begin{array} { l } { x + y = 16 } \\ { 3 ( x + 2 ) + ( y + 2 ) = 34 + 2 } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=10$$

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.

$$x+y=16,3\left(x+2\right)+y+2=36$$

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

$$x+y=16$$

Subtract y from both sides of the equation.

$$x=-y+16$$

Substitute -y+16 for x in the other equation, 3\left(x+2\right)+y+2=36.

$$3\left(-y+16+2\right)+y+2=36$$

Add 16 to 2.

$$3\left(-y+18\right)+y+2=36$$

Multiply 3 times -y+18.

$$-3y+54+y+2=36$$

Add -3y to y.

$$-2y+54+2=36$$

Add 54 to 2.

$$-2y+56=36$$

Subtract 56 from both sides of the equation.

$$-2y=-20$$

Divide both sides by -2.

$$y=10$$

Substitute 10 for y in x=-y+16. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-10+16$$

Add 16 to -10.

$$x=6$$

The system is now solved.

$$x=6,y=10$$

Steps Using Matrices

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

$$x+y=16,3\left(x+2\right)+y+2=36$$

Simplify the second equation to put it in standard form.

$$3\left(x+2\right)+y+2=36$$

Multiply 3 times x+2.

$$3x+6+y+2=36$$

Add 6 to 2.

$$3x+y+8=36$$

Subtract 8 from both sides of the equation.

$$3x+y=28$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\28\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&1\\3&1\end{matrix}\right))\left(\begin{matrix}1&1\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\3&1\end{matrix}\right))\left(\begin{matrix}16\\28\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\3&1\end{matrix}\right))\left(\begin{matrix}16\\28\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\3&1\end{matrix}\right))\left(\begin{matrix}16\\28\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-3}&-\frac{1}{1-3}\\-\frac{3}{1-3}&\frac{1}{1-3}\end{matrix}\right)\left(\begin{matrix}16\\28\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{1}{2}\\\frac{3}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}16\\28\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 16+\frac{1}{2}\times 28\\\frac{3}{2}\times 16-\frac{1}{2}\times 28\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=10$$