Linear Systems of Equations – Methods and Examples

Linear systems of equations are systems that contain equations with variables of the first degree. All the equations of these systems share the same solution. Depending on their characteristics, these systems can have different types of solutions.

There are three main methods for solving linear systems of equations: by the graphical method, by the substitution method, and by the elimination method.

ALGEBRA
linear systems of equations

Relevant for

Learning how to solve linear systems of equations.

See methods

ALGEBRA
linear systems of equations

Relevant for

Learning how to solve linear systems of equations.

See methods

Linear systems of equations and solutions

A linear system of equations is simply a set of two or more linear equations that have to be solved simultaneously. Here, we will specifically look at systems that have two equations and two unknowns.

Solution to a system

A solution to a linear system of equations is an ordered pair that makes both equations true. A solution is what the equations have in common, it is the point where their lines intersect. If an ordered pair is a solution to one equation but not the other, then it is not a solution to the system of equations.

consistent system is a system that has at least one solution.

An inconsistent system is a system that has no solutions.

There are three possible results that we can find when working with these systems:

1. One solution

2. No solution

3. Infinite solutions.


Systems of equations with one solution

If the system of two equations with two unknowns has one solution, it is an ordered pair that is a solution to both equations. This means that when we plug the solution into the equations, both equations turn out to be true.

If we obtain a solution as a final answer to the system of equations, this means that this system is consistent.

The following graph shows a system of two equations with two unknowns that has one solution:

system of equation with one solution

Systems of equations with no solutions

If the lines that represent the equations are parallel, these lines will never intersect. That means that these equations have no points in common. In this situation, the system of equations has no solution.

If, when solving our system of equations, we discover that it has no solution, this means that the system is inconsistent.

In the following graph, we can see a system of two equations with two unknowns that have no solution:

system of equations with no solution

Systems of equations with infinite solutions

If the two lines that represent the equations end up overlapping each other, then there are an infinite number of solutions. In this situation, they would end up being the same line, so any solution that works with the first equation will also work with the second.

If we obtain an infinite number of solutions to our system of equations, then it means that the system is consistent.

In the following graph, we can observe a system of two equations with two unknowns that has an infinite number of solutions:

system of equations with infinite solutions

Solve systems of equations graphically

We can follow the following steps to solve linear systems of equations graphically:

Step 1: Graph the first equation.

You can use any method to graph a line. If you need it, you can check out our guide on how to graph linear functions.

Step 2: Graph the second equation in the same coordinate system as the first.

You can graph the second equation using the same or a different method than the first equation. The difference is that we graph it in the same coordinate system.

Step 3: Find the solution.

If the lines intersect in only one place, then the point of intersection is the solution to the system of equations.

If the lines are parallel, then they never intersect and therefore there is no solution.

If the lines are on top of each other, then there are an infinite number of solutions.

Step 4: Check the pair of solutions in both equations.

Plug the solution into both equations. If both equations are true, then the solution is true. If any of the equations is false, then that ordered pair is not the correct solution.

EXAMPLE 1

Solve the system of equations graphically: \left\{ {\begin{array}{*{20}{c}} {x+y=4} \\ {2x-y=-1} \end{array}} \right.

Step 1:Graph the first equation.

We can rewrite the equation in the form y=mx+b, where is the slope and b is the y-intercept.

 x+y=4

y=4-x

Therefore, the y-intercept is 4 and the slope is -1: 

system of equations by graphing 1

Step 2: Graph the second equation.

We use the same method as the previous equation:

 2x-y=-1

y=2x+1

Therefore, the y-intercept is 1 and the slope is 2: 

system of equations by graphing 2

Step 3: Find the solution.

We now determine if the equations intersect and where they intersect.

We see that the equations intersect at the point (1, 3).

Step 4: Check the pair of solutions in both equations.

We can easily verify that by substituting the values of x=1 and y=3, both equations are true:

 x+y=4

1+3=4

4=4

2x-y=-1

2(1)-3=-1

-1=-1

EXAMPLE 2

Solve the system of equations graphically: \left\{ {\begin{array}{*{20}{c}} {x+2y=7} \\ {3x-y=7} \end{array}} \right.

Step 1: Graph the first equation.

We can rewrite the equation in the form y=mx+b, where is the slope and b is the y-intercept.

 x+2y=7

y=-\frac{1}{2}x+\frac{7}{2}

Therefore, the y-intercept is 7/2 and the slope is -1/2: 

system of equations by graphing 3

Step 2: Graph the second equation.

We use the same method as the previous equation:

 3x-y=7

y=3x-7

Therefore, the y-intercept is -7 and the slope is 3: 

system of equations by graphing 4

Step 3: Find the solution.

We have to determine if the equations intersect and where they intersect.

We see that the equations intersect at the point (3, 2).

Step 4: Check the pair of solutions in both equations.

We can easily verify that by substituting the values of x=3 and y=2, both equations are true:

 x+2y=7

3+2(2)=7

7=7

3x-y=7

3(3)-2=7

7=7


Solve systems of equations by substitution

Follow these steps to solve systems of equations using the substitution method:

Step 1: Simplify if possible.

This includes removing parentheses or other grouping signs and combining like terms.

If we have fractions, we can multiply by the least common multiple.

Step 2: Solve an equation for one variable.

It doesn’t matter which equation we choose or for which variable we solve.

We want this to be as easy as possible. If one of the equations already has a variable isolated, we can use that equation.

Step 3: Substitute what you got in step 2 into the other equation.

Make sure to plug in the equation you didn’t use in step 2. This will give us an equation with only one variable.

Step 4: Solve for the remaining variable.

Solve the equation resulting from step 3 for the remaining variable. If you need help with this, you can look at our guide on how to solve equations with one unknown.

Step 5: Solve for the second variable.

Substitute the value you found in step 4 into either of the two equations and solve for the other variable.

Step 6: Check the solution in both equations.

Plug the values ​​of the unknowns into both equations. If both equations are true, the values ​​are the correct solution.

EXAMPLE 1

Solve the system of equations using the substitution method: \left\{ {\begin{array}{*{20}{c}} {3x-5y=15} \\ {y=2x+4} \end{array}} \right.

Step 1: Simplify if possible.

Both equations are already simplified.

Step 2: Solve an equation for one variable.

We can see that the second equation is already solved for the variable y:

y=2x+4

Step 3: Substitute what you got in step 2 into the other equation.

We substitute the expression y=2x+4 in the first equation:

3x-5y=15

3x-5(2x+4)=15

3x-10x-20=15

Step 4: Solve for the remaining variable:

3x-10x-20=15

-7x=35

x=-5

Step 5: Solve for the second variable.

We substitute x=-5 in the second equation: 

y=x+4

y=-5+4

y=-1

Step 6: Check the solution in both equations.

EXAMPLE 2

Solve the system of equations using the substitution method: \left\{ {\begin{array}{*{20}{c}} {x-2y=3} \\ {2x-3y=7} \end{array}} \right.

Step 1: Simplify if possible.

Both equations are already simplified.

Step 2: Solve an equation for one variable.

We can solve the first equation for x:

x-2y=3

x=3+2y

Step 3: Substitute what you got in step 2 into the other equation.

We substitute the expression x=3+2y in the second equation: 

2x-3y=7

2(3+2y)-3y=7

6+4y-3y=7

Step 4: Solve for the remaining variable:

6+4y-3y=7

y=1

Step 5: Solve for the second variable.

We substitute y=1 in the first equation: 

x-2y=3

x-2(1)=3

x=5

Step 6: Check the solution in both equations.

Try solving the following practice problems

Solve the system 2x-4y=-6, y=3x-1.

Choose an answer






Solve the system 3x+2y=2, y=-x+2.

Choose an answer







Solve systems of equations by elimination

Follow these steps to solve a linear system of equations using the substitution method:

Step 1: Simplify if possible and write the equations in the form Ax+By = C.

This includes removing parentheses or other grouping signs and combining like terms.

If there are fractions, we can multiply by the least common multiple.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for either x or y.

We are going to add the equations and we need one of the variables to be eliminated after doing the addition. One way to guarantee this is by having opposite coefficients in one of the variables since the sum of opposites is 0.

It doesn’t matter which variable we decide to eliminate. We want to make this as easy as possible. If one of the variables already has opposite coefficients, we can simply add the equations. If we have different coefficients, we need to multiply by numbers so that the coefficients become opposite.

For example, if we have 2x in one equation and 3x in the second, we can multiply the first by -3 and the second by 2, thus we get -6 in the first and 6 in the second.

Step 3: Add the equations.

When adding the equations, one of the variables will be eliminated and we will obtain an equation with only one variable.

Step 4: Solve for the remaining variable.

Solve the equation resulting from step 3 for the remaining variable. If you need help with this, you can look at our guide on how to solve equations with one unknown.

Step 5: Solve for the second variable.

Substitute the value you found in step 4 into either of the two equations and solve for the other variable.

Step 6: Check the solution in both equations.

Plug the values ​​of the unknowns into both equations. If both equations are true, the values ​​are the correct solution.

EXAMPLE 1

Solve the system of equations using the elimination method: \left\{ {\begin{array}{*{20}{c}} {2x+2y=10} \\ {-2x+3y=5} \end{array}} \right.

Step 1: Simplify if possible and write in the form Ax+By=C.

Both equations are already simplified and in the form Ax+By=C.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We already have opposite coefficients in the variable x.

Step 3: Add the equations. 

2x+2y=10

+   \hspace{1cm}    -2x+3y=5               

___________________

5y=15

Step 4: Solve for the remaining variable:

5y=15

y=3

Step 5: Solve for the second variable.

We substitutey=3 in the first equation: 

2x+2y=10

2x+2(3)=10

2x+6=10

2x=4

x=2

Step 6: Check the solution in both equations.

EXAMPLE 2

Solve the system of equations using the elimination method:\left\{ {\begin{array}{*{20}{c}} {2x=y+3} \\ {-x+3y=11} \end{array}} \right.

Step 1: Simplify if possible and write in the form Ax+By=C.

Both equations are already simplified. Now, we write them in the form Ax+By=C:

\left\{ {\begin{array}{*{20}{c}} {2x-y=3} \\ {-x+3y=11} \end{array}} \right.

Step 2: Multiply one or both equations by a number that will create opposite coefficients for x or for y.

We multiply the second equation by 2:

\left\{ {\begin{array}{*{20}{c}} {2x-y=3} \\ {-2x+6y=22} \end{array}} \right.

Step 3: Add the equations. 

2x-y=3

+   \hspace{1cm}    -2x+6y=22               

___________________

5y=25

Step 4: Solve for the remaining variable:

5y=25

y=5

Step 5: Solve for the second variable.

We substitute y=5 in the first equation: 

2x-y=3

2x-5=3

2x=8

x=4

Step 6: Check the solution in both equations.

Try solving the following practice problems

Solve the system 2x-3y=3, -2x+y=-5.

Choose an answer






Solve the system 3x+2y=-9, 2x+y=-5.

Choose an answer







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