There are two ways **to solve a system of equations algebraically**: the **elimination method** and the **substitution method**. Though the elimination method can be used at anytime, there are certain problems that lend them self to the use of the substitution method. The topic of this post is solving a system of two equations with the substitution method. The problems covered in this post are basic, that is if you think algebra 2 is a basic course! By basic, I mean there is no problem solving. We are given the equations and told to solve them. Once solving the system is learned, then we can look at problem solving related to solving a system of equations.

Example 1 – **Solve the system of equations using the substitution method**.

As you can see in the figure below, I have the problem worked out. I have the given equations label A and B and I will refer to them as such as I solve the problem in paragraph format. This is one way to write mathematics. If your teachers are like me, then I am sure many of them spend time getting you to explain your answers. Here is what I think would be a viable way for you to write out how to solve a problem. You may notice the green numbers to the right of the problem. This is so you can follow along with needing to count the lines I am referring to in the problem. Let’s get started solving a system of two equations.

In the problem to the right, lines 1 and 2 of the represent the given system of equations. I labeled the first equation A and the second equation B.

I could have solve equation A for x or y because each term had a coefficient of 1 and I could have solved equation B for y because it had a coefficient of 1. From my diagram, line 3 contains equation A, which I decided to solve for y and the equivalent equation A is on line 4.

Using the substitution property of equality I replace the y in equation B with the 10 – x (line 5) and the new equation B is contained in line 6. Simplifying the equation by combining the 2x and –x gives line 7. The subtraction property of equality allows us to isolate the x with a value of 5 (line 8).

Since I was able to find an x value, I will be able to find a y value. In line 9, I used equation A to solve for y. I substituted 5 in for y , line 10. Subtracting 5 from both sides of the equation in line 10 will give the result for y. It happens to be 5.

Thus, this system is** independent and consistent** and has one solution of (5,5).

The video below covers the problem above and additional problem. Coming soon, parts 2 and 3 of this lesson on solving systems of equations, I will tackle to problem solving situations.

Filed under: Algebra 2, Equations and Graphs, Solving Systems of Equations | Tagged: 2 variable system of equations, Algebra 2, solve a system of equations with 2 variables, solve a system of equations with substitution method, Solving a system of equations, solving a system with 2 equations |

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