# Applicable Mathematics/Systems of Equations

Home - 1 - 2 - 3 - 4 - 5 - 6- 7 - 8 - 9 - 10
A good deal of real world problems can be represented by various equations. Often, we will have more than one equation for a given problem.

# Substitution

Substitution uses letters, such as x, y, or z, as representations of unknown values. These letters are used in both equations and expressions as tools to solve many different types of problems. In some cases, the value of the letter is known. If so, by using the substitution method, a numerical value replaces the letter given. Then, after the letter is replaced by a number, the expression or equation is simplified.

## Introductory Examples
### People & Feet

In a room of people we know there are twice as many feet as people, and we can represent this with the equation $f=2p$, where $f$ represents the number of feet, and $p$ represents the number of people. Knowing there are 20 people, then we can make another equation $p=20$. Listing out the two equations we have:
| $f$ | $=2p$ |
| $p$ | $=20$ |
| Substituting for $p$, we get the equation ||
| $f$ | $=2\times 20$ |
| | $=40$ |
So we know there are 40 feet.

### Houses & Floorspace

Other situations can get more complex though. Suppose that your neighbor's house is 1.5 times as large as yours, but if you don't count your neighbor's 50 square unit basement, they are the same size. How many square units are the respective houses? Let $y$ be the area of your entire floor, and $n$ the area of your neighbor's floor. The problem can represented by the two equations.
| $n$ | $=1.5y$ |
| $y$ | $=n-50$ |
Here, you have two equations as before, but this time they both have two variables, while in the last example, one equation had one variable and the other had two. However, you can still use substitution though you just have to substitute with the equation's expression instead of a constant. Now we substitute $n$ in the second equation for the right-hand side of the first equation:
| $1.5y-50$ | $=y$ | Subtract y from both sides |
| $0.5y-50$ | $=0$ | Add 50 to both sides |
| $0.5y$ | $=50$ | Multiply both sides by 2 |
| $y$ | $=100$ |
Now we can substitute the area of your floor into the first equation to get the area of your neighbours floor:
| $n$ | $=1.5\times 100$ |
| | $=150$ |

## Systems of Linear Equations

Above we covered two real world examples (albeit simplified) that systems of equations are useful for solving. Equations composed of two or more linear functions are called Linear equations. Sets of these linear equations are called System of linear equations. Simply put, linear equations can only be solved if the number of Variables is equal to or less than the number of equations provided. The most common way of solving systems of equations is to use substitution, as shown above. However, we can represent the equations using matrices, allowing us to see patterns easier and perform operations more easily. Lets start with a set of 3 equations:
| $x+2y+3z$ | $=12$ |
| $2x+3y+z$ | $=17$ |
| $3x+y+2z$ | $=19$ |
The matrix on the left represents the coefficients of the variables, and is called a coefficient matrix. On the right the right-hand side of the equation is included in the matrix, giving us what is called an augmented matrix.
Solving the above equation would look like the table on the left below normally, whereas the matrix solution would look like the table on the right.
| $e_{1}$ | $x+2y+3z$ | $=12$ |
| $e_{2}$ | $2x+3y+1z$ | $=17$ |
| $e_{3}$ | $3x+y+2z$ | $=19$ |
| $e_{2}-2e_{1}$ | $-y-5z$ | $=-7$ |
| $e_{3}-3e_{1}$ | $-5y-7z$ | $=-17$ |
| $e_{5}-5e_{4}$ | $18z$ | $=18$ |
| $e_{6} \over 18$ | $z$ | $=1$ |
| $e_{4}$ | $-y-5(1)$ | $=-7$ |
| $-5-e_{4}$ | $y$ | $=2$ |
| $e_{1}$ | $x+2(2)+3(1)$ | $=12$ |
| $e_{2}-7$ | $x$ | $=5$ |
| ${\begin{bmatrix}1&2&3&\vdots &12\\2&3&1&\vdots &17\\3&1&2&\vdots &19\end{bmatrix}}$ | $r_{2}-2r_{1}\Rightarrow r_{1}$ |
| ${\begin{bmatrix}1&2&3&\vdots &12\\0&-1&-5&\vdots &-7\\3&1&2&\vdots &19\end{bmatrix}}$ | $r_{3}-3r_{1}\Rightarrow r_{3}$ |
| ${\begin{bmatrix}1&2&3&\vdots &12\\0&-1&-5&\vdots &-7\\0&-5&-7&\vdots &-17\end{bmatrix}}$ | $r_{3}-5r_{2}\Rightarrow r_{3}$ |
| ${\begin{bmatrix}1&2&3&\vdots &12\\0&-1&-5&\vdots &-7\\0&0&18&\vdots &18\end{bmatrix}}$ | $tobecontinued$ |