Linear Systems

A Linear System is a set of \(2\) or more Linear Equations. It is recommended you first view this lesson to better understand how linear equations work. We can solve a linear system to determine the solution(s). The number of solutions of a linear system depends on the relation between the equations within that system.

In a linear system, there are \(3\) unique cases for intersection:

1. Lines don't intersect; this means that they are parallel to one another:


2. Lines only intersect at one point; this is the most common form of linear system:


3. Lines intersect infinetely at every point; this means that the lines/equations are the exact same:

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Methods for Solving Linear Systems

Solving a system of \(2\) linear equations involves finding the Point of Intersection between those lines. This is referred to as the solution of the system of linear equations.

Finding the point of intersection can be achieved either algebraically or graphically. We can verify our results algebraically by substituting the solution into all linear expressions. If both sides are equal, the solution is correct.

There are \(3\) main methods for solving linear systems:

Graphing Method

• The simplest but least accurate way of portraying the relationships between the set of lines.
• All you need to do is graph the set of lines based on their resepctive equations to see if or where they intersect
• Before graphing the lines, we can create a table of values to identify where each point of the respective lines lie

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Substitution Method

To solve a linear system by substitution, follow these steps:

1. Solve one of the equations for one variable in terms of the other variable
2. Substitute the expression from step 1 into the other equation and solve for the remaining variable
3. Substitute back into one of the original equations to find the value of the other variable
4. Check your solution by substituting into both original equations, or into statements of a word problem

Example

Using the substitution method, solve the linear system with linear equations \(8x - y = 10\) and \(3x - y = 9\).

First, we can solve one linear equation for \(x\) in terms of \(y\):

\(8x - y = 10\)

\(y = 8x - 10\)

Next, we can substitute that expression for \(y\) in the other linear equation:

\(3x - (8x - 10) = 9\)

\(3x - 8x + 10 = 9\)

Then, we can solve this system to determine the \(x\)-coordinate of the Point of Intersection:

\(3x - 8x = 9 - 10\)

\(-5x = -1\)

\(\cfrac{\cancel{-5}x}{\cancel{-5}} = \cfrac{-1}{-5}\)

\(x = 0.2\)

Finally, we can plug P\(x\), \(0.2\), into either equation to determine the corresponding \(y\)-coordinate and the Point of Intersection:

\(8(0.2) - y = 10\)

\(1.6 - y = 10\)

\(- y = 10 - 1.6\)

\(-y = 8.4\)

\(\cfrac{\cancel{-}y}{\cancel{-1}} = \cfrac{8.4}{-1}\)

\(y = -8.4\)

Therefore, we can determine that the Point of Intersection is \(\boldsymbol{(0.2, -8.4)}\).

We can verify our results by substituting the point into the other equation:

\(3(0.2) - (-8.4) = 9\)

\(0.6 + 8.4 = 9\)

\(9 = 9\)

We can confirm that our solution is correct.

NOTE: You can also start by solving for \(x\) in terms of \(y\) using the same steps outlined above.

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Elimination Method

To solve a linear system by substitution, follow these steps:

1. Arrange the two equations so that like terms are aligned
2. Choose the variables you wish to eliminate
3. If necessary, multiply one or both equations by a value so that they have the same or opposite coefficient in front of the variable you want to eliminate
4. Add or subtract (as needed)
5. Solve for the remaining variable
6. Substitute into one of the original equations to find the value of the other variable
7. Check your solution by substutiting into the original equations, or into the word problem
8. If you are solving a word problem, write the answer in words

Example

Using the elimination method, we can solve the linear system with linear equations \(4 + y -3x = 0\) and \(x + y = 8\)

First, we can align all the terms in both equations so they share the same form (ie slope-intercept form):

\(y_1 = 3x - 4\)

\(y_2 = -x + 8\)

Next, we can add or subtract a multiple of one equation to (or from) the other equation so that the terms cancel each other out:

\(3(y = -x + 8)\)

\(3y = -3x + 24\)

\((3y + y) = (-3x + 3x) + (24 - 4)\)

\(4y = 20\)

\(\cfrac{\cancel{4}y}{\cancel{4}} = \cfrac{20}{4}\)

\(y = 5\)

Then, we can solve for \(x\) by substituting the \(y\)-value, \(5\), into one of the original equations:

\(4 + 5 -3x = 0\)

\(-3x = -9\)

\(\cfrac{\cancel{-3}x}{\cancel{-3}} = \cfrac{-9}{-3}\)

\(x = 3\)

Therefore, we can determine that the Point of Intersection is \(\boldsymbol{(3, 5)}\).

We can confirm our results by substituting the point into the other equation:

\(x + y = 8\)

\(3 + 5 = 8\)

\(8 = 8\)

We can confirm that our solution is correct.

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