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Solving Systems of Linear Equations

Solving Systems of Linear Equations

There are several ways to solve linear equations. For example, some systems are not as complex as others. We will discuss the easiest to the harder one that can be solved by basics of Algebra.

Solving Systems of Equations with Two variables

Given: 2X -3Y = -2
4X + Y = 24

In this case the easiest way is to solve by the method of substitution, on the equation 4X + Y = 24, Solve for Y and you should get Y = 24 – 4X. Now take equation 2X -3Y = -2 and substitute the Y for its value
2X -3(24 – 4X) = -2
2X – 72 + 12X = -2
14X – 72 = -2
14X = 70
X = 5

Now that you know the value of X, you can substitute on any of the original equations.

2X -3Y = -2
2(5) -3Y = -2
10 – 3Y = -2
-3Y = -12
Y = 4

Let’s try another method, where you cancel one variable of the equation. We will use the same equation from the previous problem.

Given: 2X -3Y = -2
4X + Y = 24

You can choose whichever variable you want to cancel, at the end of the day you should be able to get the answer, I would like to choose to cancel variable Y. The way is that you cancel is by multiplying a both sides of the equations and add both equations.

2X -3Y = -2
4X + Y = 24

2X -3Y = -2
3(4X + Y) = 3(24)

2X -3Y = -2
12X + 3Y = 72
Add both equations and you should get a new equation, this equation will let you get the values for X and Y

14X + 0Y = 70
14X = 70
X = 5

Now that you know the value of X, just substitute on any of the given equations

4X + Y = 24
4(5) + Y = 24
20 + Y = 24
Y = 4

We will cover system of linear equations with three variables next time!