Home / United States / Math Classes / 8th Grade Math / Solving the Special Systems of Linear Equations

A system of linear equations is a collection of two or more linear equations that have multiple variables. A system of linear equations can have no solutions, one solution, or infinitely many solutions depending on the nature of the equations. Learn how to solve a system of linear equations with the help of some examples....Read MoreRead Less

When two or more linear equations work together, it is called a system of linear equations.

The solution of a system of linear equations is the point of intersection of the graph of each equation on the coordinate plane.

In most cases, a system of linear equations has only one solution, but it can also have no solution (parallel lines) or infinite solutions (same line). All three cases are discussed in this article.

When the graphs intersect at a point, a system of linear equations has one solution.

When the graphs are parallel, a system of linear equations has no solution.

When the graphs are on the same line, a system of linear equations has infinite solutions.

When there are no points where lines intersect or equation graphs are parallel, a system of equations has no solution.

Let’s take an example. Find the solution for the system of linear equations:

\(y=-5x+7 \) Equation 1.

\(y=-5x-4 \) Equation 2.

**Method 1:** Graph the two equations.

The lines have the same slope (-5) but different y-intercepts (7 and -4), respectively. The lines are thus parallel. There is no point that is a solution to both equations because parallel lines do not intersect.

As a result, the system has no solution.

**Method 2:** Solve the given systems of linear equations by inspection.

You should be aware that you can rewrite the system as follows:

Take all the variable terms on one side and constant terms on another side.

\(-5x+y=7 \) Revised Equation 1

\(-5x+y=-4 \) Revised Equation 2

\(-5x+y \) cannot be both 7 and -4. As a result, there is no solution for the system.

A system of equations can have an infinite number of solutions when a solution set of infinite points exists for which the L.H.S. and R.H.S. of an equation become equal, or when straight lines overlap in the graph.

Let’s take an example. Find the solution for the system of linear equations:

\(-2x+y=1 \) Equation 1.

\(-4x+2y=2 \) Equation 2.

**Method 1:** Graph the two equations.

The lines have the same y-intercept, 1, and the same slope, 2. Thus, the two equations represent the same line.

The solutions are all the points on the line. As a result, the system has an infinite number of solutions.

**Method 2:** Solve the given system of linear equations by the elimination method.

Divide equation 2 by 2, so that the x terms have a coefficient of -2.

\(-2x+y=1 \) \(-2x+y=1 \) Equation 1.

\(-4x+2y=2 \) **Divide by 2**. \(-2x+y=1 \) Revised equation 2.

On subtracting revised equation 2 from equation 1, we get

0=0.

Hence the two equations are equivalent.

The solutions of the system are all the points on the line. As a result, the system has an infinite number of solutions.

**Example 1:**

Solve the system of linear equations.

\(y=-x+1\) Equation 1.

\(y=-x+3\) Equation 2.

**Solution:**

**Method 1:** Graph the equations.

The lines have the same slope (-1) but different y-intercepts (1 and 3), respectively. The lines are thus parallel. There is no point that is a solution to both equations because parallel lines do not intersect.

As a result, the system has no solution.

**Method 2:** Solve the given systems of linear equations by inspection.

Rewrite the equations as follows:

\(x+y=1\) Revised Equation 1

\(x+y=3\) Revised Equation 2

\(x+y\) cannot be equal to both 1 and 3. Hence, there is no solution for the system.

**Example 2: Solve the system of linear equations:**

\(x+y=4\) Equation 1.

\(2x+2y=8\) Equation 2.

**Solution:**

**Method 1:** Graph the two equations.

The lines have the same y-intercept, 4, and the same slope, -1. Thus, the graph of the two equations is the same line.

The system’s solutions are all the points on the line. As a result, the system has an infinite number of solutions.

**Method 2:** Solve the given systems of linear equations by elimination.

Divide equation 2 by 2, so that the x terms have a coefficient of -2.

\(x+y=4\) \(x+y=4\) Equation 1.

\(2x+2y=8\) **Divide by 2**. \(x+y=4\) Revised equation 2.

Subtract revised equation 2 from equation 1 we get 0 = 0.

Hence the two equations are equivalent.

The system’s solutions are all the points on the line. As a result, the system has an infinite number of solutions.

**Example 3:**

You are throwing a birthday bash. You spend $120.00 on a total of 60 turkey and veggie burgers. Both types of burgers cost $2.00 each. How many of each burger do you purchase?

**Solution:**

To write a system of linear equations, we use a verbal model. Let \(x\) stand for the number of turkey burgers, and \(y\) for the number of veggie burgers.

Number of turkey burgers, \(x\) + Number of veggie burgers, \(y\) = Total number of burgers

Cost per turkey burger X \(x\) + Cost per veggie burger X \(y\) = Total cost

The system is:

\(x+y=60\) Equation 1

\(2x+2y=120\) Equation 2

Using elimination method.

\(x+y=4\) \(x+y=4\) Equation 1.

\(2x+2y=8\) **Divide by 2**. \(x+y=4\) Revised equation 2.

Subtract revised equation 2 from equation 1, we get 0 = 0.

Hence the two equations are equivalent.

The system’s solutions are all the points on the line. As a result, the system has an infinite number of solutions.

Hence, there is insufficient data to determine the number of each burger.

**Example 4:**

You are planning a get together with your friends. You spend $160.00 on a total of 60 Chicago pizza and Greek pizza. Both pizzas cost $2.00 each. How many of each pizza do you plan to purchase?

**Solution:**

To write a system of linear equations, we use a verbal model. Let x stand for the number of Chicago pizzas, and y for the number of Greek pizzas.

Number of Chicago pizzas, \(x\) + Number of Greek pizzas, \(y\) = Total number of pizza

Cost per Chicago pizza X \(x\) + Cost per Greek pizza X \(y\) = Total cost

The system is:

\(x+y=60\) Equation 1

\(2x+2y=160\) Equation 2

Use the graphical method to solve this problem.

The lines have the same slope (-1) but different y-intercepts (60 and 80), respectively. The lines are thus parallel. There is no point that is a solution to both equations because parallel lines do not intersect.

As a result, the system has no solution.

Hence, there is insufficient data to determine the number of each burger.

Frequently Asked Questions

All one-variable linear equations have only one solution.

The following are the three main methods for solving a system of equations:

- The graphical method
- The substitution method
- The elimination method