Solve {l}{(a+2012)-2(b-2013)=3}{3(a+2012)+4(b-2013)=5}

$$\left\{ \begin{array} { l } { ( a + 2012 ) - 2 ( b - 2013 ) = 3 } \\ { 3 ( a + 2012 ) + 4 ( b - 2013 ) = 5 } \end{array} \right.$$

Solve for a, b

$$a = -\frac{10049}{5} = -2009\frac{4}{5} = -2009.8$$

$$b = \frac{10063}{5} = 2012\frac{3}{5} = 2012.6$$

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.

$$a-2\left(b-2013\right)+2012=3,3\left(a+2012\right)+4\left(b-2013\right)=5$$

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

$$a-2\left(b-2013\right)+2012=3$$

Multiply -2 times b-2013.

$$a-2b+4026+2012=3$$

Add 4026 to 2012.

$$a-2b+6038=3$$

Subtract 6038 from both sides of the equation.

$$a-2b=-6035$$

Add 2b to both sides of the equation.

$$a=2b-6035$$

Substitute 2b-6035 for a in the other equation, 3\left(a+2012\right)+4\left(b-2013\right)=5.

$$3\left(2b-6035+2012\right)+4\left(b-2013\right)=5$$

Add -6035 to 2012.

$$3\left(2b-4023\right)+4\left(b-2013\right)=5$$

Multiply 3 times 2b-4023.

$$6b-12069+4\left(b-2013\right)=5$$

Multiply 4 times b-2013.

$$6b-12069+4b-8052=5$$

Add 6b to 4b.

$$10b-12069-8052=5$$

Add -12069 to -8052.

$$10b-20121=5$$

Add 20121 to both sides of the equation.

$$10b=20126$$

Divide both sides by 10.

$$b=\frac{10063}{5}$$

Substitute \frac{10063}{5} for b in a=2b-6035. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=2\times \frac{10063}{5}-6035$$

Multiply 2 times \frac{10063}{5}.

$$a=\frac{20126}{5}-6035$$

Add -6035 to \frac{20126}{5}.

$$a=-\frac{10049}{5}$$

The system is now solved.

$$a=-\frac{10049}{5},b=\frac{10063}{5}$$

Steps Using Matrices

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

$$a-2\left(b-2013\right)+2012=3,3\left(a+2012\right)+4\left(b-2013\right)=5$$

Simplify the first equation to put it in standard form.

$$a-2\left(b-2013\right)+2012=3$$

Multiply -2 times b-2013.

$$a-2b+4026+2012=3$$

Add 4026 to 2012.

$$a-2b+6038=3$$

Subtract 6038 from both sides of the equation.

$$a-2b=-6035$$

Simplify the second equation to put it in standard form.

$$3\left(a+2012\right)+4\left(b-2013\right)=5$$

Multiply 3 times a+2012.

$$3a+6036+4\left(b-2013\right)=5$$

Multiply 4 times b-2013.

$$3a+6036+4b-8052=5$$

Add 6036 to -8052.

$$3a+4b-2016=5$$

Add 2016 to both sides of the equation.

$$3a+4b=2021$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\3&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&-2\\3&4\end{matrix}\right))\left(\begin{matrix}1&-2\\3&4\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&4\end{matrix}\right))\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&4\end{matrix}\right))\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\3&4\end{matrix}\right))\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-\left(-2\times 3\right)}&-\frac{-2}{4-\left(-2\times 3\right)}\\-\frac{3}{4-\left(-2\times 3\right)}&\frac{1}{4-\left(-2\times 3\right)}\end{matrix}\right)\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&\frac{1}{5}\\-\frac{3}{10}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}-6035\\2021\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\left(-6035\right)+\frac{1}{5}\times 2021\\-\frac{3}{10}\left(-6035\right)+\frac{1}{10}\times 2021\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{10049}{5}\\\frac{10063}{5}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{10049}{5},b=\frac{10063}{5}$$