A system of linear equations is a group of two or more linear equations with two or more variables. The solution to a system of linear equations is an ordered pair that creates true statements for all the linear equations within the system. For example, look at the following system:
\( \begin{align*} & y= 2x+3 \\ & 2x + y = -1 \\ \end{align*} \)
The SOLUTION to the system is the point or ordered pair \( (-1, 1) \). If you substitute the point into both equations, you arrive at two true statements. For the first equation, \( y = 2x+3 \), if you substitute in the point \( 1 = 2(-1) + 3 \), you obtain the statement \( 1 = 1 \), which is true. For the second equation, \( 2x + y = -1 \), if you substitute in the ordered pair, \( 2(-1) + 1 = -1, \) this becomes \( -1 = -1 \), which is also true a true statement.
To find the solution to a system of linear equations, there are three methods that can be used: graphing, substitution, and using linear combinations (sometimes called elimination).
Solving a System of Linear Equations by Graphing
A system of linear equations can be solved by graphing by locating the point of intersection of the two lines. Looking at the graph below (using the same system of equations as above), you can see that the graph of the lines of the two equations intersect at the point \( (-1, 1) \).
To find the solution to a system of linear equations, there are three methods that can be used: graphing, substitution, and using linear combinations (sometimes called elimination).
Solving a System of Linear Equations by Graphing
A system of linear equations can be solved by graphing by locating the point of intersection of the two lines. Looking at the graph below (using the same system of equations as above), you can see that the graph of the lines of the two equations intersect at the point \( (-1, 1) \).
Learn how to solve a system by graphing by clicking here: Solving a Linear System by Graphing .
Solving a System of Linear Equations using Substitution
When using substitution to solve a system of equations, you do just that, you substitute! But this time, the process going may look a little different than the substituting you've done before. The substitution may in involve a quantity that is an expression instead of just a constant value. In order to use this method, one of the two equations has to have what is called an "isolated variable". Looking again at the system above there are two equations: \( y = 2x + 3 \) and \( 2x + y = -1 \). The first equation \( y = 2x + 3 \) has an isolated variable since the "\( y \)" is all alone on one side of the equal sign. To solve this system we can substitute the expression "\( 2x +3 \)" in place of "\( y \)" in the second linear equation. Sometimes you might have an equation where "\( x \)" is the isolated variable as well, it just depends on the system. If the system does NOT have an equation with an isolated variable, then you will need to rewrite one of the equations so that there is an isolated variable.
Learn how to solve a system using substitution by clicking here: Solving a Linear System Using Substitution.
When using substitution to solve a system of equations, you do just that, you substitute! But this time, the process going may look a little different than the substituting you've done before. The substitution may in involve a quantity that is an expression instead of just a constant value. In order to use this method, one of the two equations has to have what is called an "isolated variable". Looking again at the system above there are two equations: \( y = 2x + 3 \) and \( 2x + y = -1 \). The first equation \( y = 2x + 3 \) has an isolated variable since the "\( y \)" is all alone on one side of the equal sign. To solve this system we can substitute the expression "\( 2x +3 \)" in place of "\( y \)" in the second linear equation. Sometimes you might have an equation where "\( x \)" is the isolated variable as well, it just depends on the system. If the system does NOT have an equation with an isolated variable, then you will need to rewrite one of the equations so that there is an isolated variable.
Learn how to solve a system using substitution by clicking here: Solving a Linear System Using Substitution.
Solving a System of Linear Equations using Linear Combinations (Elimination)
In order to solve a linear system using linear combinations, you essentially are "combining" the two equations together using addition. Sometimes this method is also called elimination because one of the variables will be "eliminated" by adding or subtracting to equal zero. For this to happen, either the the coefficients for the x-terms from both equations or the coefficients for the \(y\)-terms in both equations need to be opposites of each other. For example, if one equation has an \(x\)-term of "\( 2x \)", then the other equation would need to have an \(x\)-term of "\( -2x \)". When the linear equations are combined the \(x\) variable is eliminated, \( 2x + -2x =0 \). When using linear combinations, make sure that each equation is in standard form.
Learn how to solve a system using linear combinations by clicking here: Solving a Linear System Using Linear Combinations.
Here's a question you might be asking: When should I use each method? Any of the three methods can be used to solve a linear system, however, sometimes one method might be more efficient than another depending on what the system consists of. Use these guidelines to help choose the method that might be most appropriate for any given system:
These guidelines will help with the majority of the systems you will need to solve. But keep in mind that there may be a system where the most efficient method is not so obvious, so you may need to rewrite one, or both, of the equations in order to find the solution to the system.
In order to solve a linear system using linear combinations, you essentially are "combining" the two equations together using addition. Sometimes this method is also called elimination because one of the variables will be "eliminated" by adding or subtracting to equal zero. For this to happen, either the the coefficients for the x-terms from both equations or the coefficients for the \(y\)-terms in both equations need to be opposites of each other. For example, if one equation has an \(x\)-term of "\( 2x \)", then the other equation would need to have an \(x\)-term of "\( -2x \)". When the linear equations are combined the \(x\) variable is eliminated, \( 2x + -2x =0 \). When using linear combinations, make sure that each equation is in standard form.
Learn how to solve a system using linear combinations by clicking here: Solving a Linear System Using Linear Combinations.
Here's a question you might be asking: When should I use each method? Any of the three methods can be used to solve a linear system, however, sometimes one method might be more efficient than another depending on what the system consists of. Use these guidelines to help choose the method that might be most appropriate for any given system:
- Graphing: use this method if both equations are in slope-intercept form ( \( y = mx + b \))
- Substitution: use when one of the two equations already has an isolated variable
- Linear Combinations (Elimination): use when both of the equations are in standard form
These guidelines will help with the majority of the systems you will need to solve. But keep in mind that there may be a system where the most efficient method is not so obvious, so you may need to rewrite one, or both, of the equations in order to find the solution to the system.
Quick Check
Determine which method would be most efficient to use to solve each system. Then, solve the system using that method.
1) \( \begin{align*} & x + 3y = 1 \\ & 5x + 6y = 14\\ \end{align*} \)
2) \( \begin{align*} & y = 3x + 2 \\ & y = -4x + 2 \\ \end{align*} \)
3) \( \begin{align*} & y = 5x - 7 \\ & -4x + y = -1 \\ \end{align*} \)
Determine which method would be most efficient to use to solve each system. Then, solve the system using that method.
1) \( \begin{align*} & x + 3y = 1 \\ & 5x + 6y = 14\\ \end{align*} \)
2) \( \begin{align*} & y = 3x + 2 \\ & y = -4x + 2 \\ \end{align*} \)
3) \( \begin{align*} & y = 5x - 7 \\ & -4x + y = -1 \\ \end{align*} \)