Solve {l}{2m-2n=4}{2n+(m-1)*(-1)=3}

$$\left\{ \begin{array} { l } { 2 m - 2 n = 4 } \\ { 2 n + ( m - 1 ) \times ( - 1 ) = 3 } \end{array} \right.$$

Solve for m, n

$$m=6$$

$$n=4$$

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.

$$2m-2n=4,-\left(m-1\right)+2n=3$$

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

$$2m-2n=4$$

Add 2n to both sides of the equation.

$$2m=2n+4$$

Divide both sides by 2.

$$m=\frac{1}{2}\left(2n+4\right)$$

Multiply \frac{1}{2} times 4+2n.

$$m=n+2$$

Substitute n+2 for m in the other equation, -\left(m-1\right)+2n=3.

$$-\left(n+2-1\right)+2n=3$$

Add 2 to -1.

$$-\left(n+1\right)+2n=3$$

Multiply -1 times n+1.

$$-n-1+2n=3$$

Add -n to 2n.

$$n-1=3$$

Add 1 to both sides of the equation.

$$n=4$$

Substitute 4 for n in m=n+2. Because the resulting equation contains only one variable, you can solve for m directly.

$$m=4+2$$

Add 2 to 4.

$$m=6$$

The system is now solved.

$$m=6,n=4$$

Steps Using Matrices

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

$$2m-2n=4,-\left(m-1\right)+2n=3$$

Simplify the second equation to put it in standard form.

$$-\left(m-1\right)+2n=3$$

Multiply -1 times m-1.

$$-m+1+2n=3$$

Subtract 1 from both sides of the equation.

$$-m+2n=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}4\\2\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right))\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right))\left(\begin{matrix}4\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right))\left(\begin{matrix}4\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-2\\-1&2\end{matrix}\right))\left(\begin{matrix}4\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2\times 2-\left(-2\left(-1\right)\right)}&-\frac{-2}{2\times 2-\left(-2\left(-1\right)\right)}\\-\frac{-1}{2\times 2-\left(-2\left(-1\right)\right)}&\frac{2}{2\times 2-\left(-2\left(-1\right)\right)}\end{matrix}\right)\left(\begin{matrix}4\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}1&1\\\frac{1}{2}&1\end{matrix}\right)\left(\begin{matrix}4\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}4+2\\\frac{1}{2}\times 4+2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}6\\4\end{matrix}\right)$$

Extract the matrix elements m and n.

$$m=6,n=4$$