Solve {l}{2(y+7)-x=2}{x/2+y=1}

$$\left\{ \begin{array} { l } { 2 ( y + 7 ) - x = 2 } \\ { \frac { x } { 2 } + y = 1 } \end{array} \right.$$

Solve for y, x

$$x=7$$

$$y = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

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.

$$2\left(y+7\right)-x=2,y+\frac{1}{2}x=1$$

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

$$2\left(y+7\right)-x=2$$

Multiply 2 times y+7.

$$2y+14-x=2$$

Subtract 14 from both sides of the equation.

$$2y-x=-12$$

Add x to both sides of the equation.

$$2y=x-12$$

Divide both sides by 2.

$$y=\frac{1}{2}\left(x-12\right)$$

Multiply \frac{1}{2} times x-12.

$$y=\frac{1}{2}x-6$$

Substitute \frac{x}{2}-6 for y in the other equation, y+\frac{1}{2}x=1.

$$\frac{1}{2}x-6+\frac{1}{2}x=1$$

Add \frac{x}{2} to \frac{x}{2}.

$$x-6=1$$

Add 6 to both sides of the equation.

$$x=7$$

Substitute 7 for x in y=\frac{1}{2}x-6. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{1}{2}\times 7-6$$

Multiply \frac{1}{2} times 7.

$$y=\frac{7}{2}-6$$

Add -6 to \frac{7}{2}.

$$y=-\frac{5}{2}$$

The system is now solved.

$$y=-\frac{5}{2},x=7$$

Steps Using Matrices

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

$$2\left(y+7\right)-x=2,y+\frac{1}{2}x=1$$

Simplify the first equation to put it in standard form.

$$2\left(y+7\right)-x=2$$

Multiply 2 times y+7.

$$2y+14-x=2$$

Subtract 14 from both sides of the equation.

$$2y-x=-12$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-12\\1\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-12\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-12\\1\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}-12\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{2}}{2\times \frac{1}{2}-\left(-1\right)}&-\frac{-1}{2\times \frac{1}{2}-\left(-1\right)}\\-\frac{1}{2\times \frac{1}{2}-\left(-1\right)}&\frac{2}{2\times \frac{1}{2}-\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}-12\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}&\frac{1}{2}\\-\frac{1}{2}&1\end{matrix}\right)\left(\begin{matrix}-12\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\left(-12\right)+\frac{1}{2}\\-\frac{1}{2}\left(-12\right)+1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{2}\\7\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-\frac{5}{2},x=7$$