Systems of Equations—Solving with Elimination Study Guide
Introduction to Systems of Equations—Solving with Elimination
To be a scholar of mathematics you must be born with talent, insight, concentration, taste, luck, drive and the ability to visualize and guess.
—Paul R. Halmos (1916–2006) American Mathematician
In this lesson, you will learn how to solve systems of equations with two or three variables using elimination.
Now that we know how to solve a system of equations using substitution, let's look at another method: elimination. The goal of elimination is to combine two or more equations together to eliminate a variable, using addition, subtraction, multiplication, or division.
We will start with the same example we looked at in Lesson 18, 3x + y = 7 and x – y = 13. Instead of beginning by writing one variable in terms of the other, we will combine these two equations to eliminate either x or y, leaving us with one equation and one variable.
We can add the equations or subtract one equation from the other. We can also multiply or divide one or both equations by a constant before adding or subtracting. Look closely at these equations:
- 3x + y = 7
- x – y = 13
The first equation has a y term and the second equation has a –y term. If we add these two equations together, the y terms will drop out and we will be left with one equation and one variable, x. Add the left side of the first equation to the left side of the second equation and add the right side of the first equation to the right side of the second equation:
Now we can solve for x by dividing both sides of the equation by 4, and we find that x = 5. Once we have the value of one variable, just as with the substitution method, we can use it to find the value of the other variable. Substitute 5 for x in either equation and solve for y:
- 3x + y = 7
- 3(5) + y = 7
- 15 + y = 7
- y = –8
If adding the equations or subtracting one equation from the other will not cause one variable to drop out, we must multiply one equation by a constant and then add or subtract. Look at the following system of equations:
- 2x – 3y = 5
- x + 2y = 13
First, we must decide which variable we want to eliminate. Eliminating y will take a couple of steps; before we can add the two equations, the coefficients of y in each equation will have to be the same. We can do that by multiplying the first equation by 2 and by multiplying the second equation by 3. However, we can eliminate x a little more easily. If we multiply just the second equation by 2, we can subtract it from the first equation, and that will eliminate x. We must multiply every term on both sides of the equation by 2:
- 2(x + 2y) = 2(13)
- 2x + 4y = 26
We do not need to multiply the first equation by anything at all. It is important to understand that what we do to one equation we do not have to do to the other equation. If we choose to multiply an equation by a constant, we must multiply every term in the equation by the constant. Because every term grows in exactly the same way, the value of the equation does not change. The equation 2x + 4y = 26 is still true after being multiplied by 2, and the equation 2x – 3y = 5 is still true after not being changed at all.
Now, we can subtract the second equation from the first:
We can divide both sides of the equation by –7, and we find that y = 3. Substitute 3 for y in either equation and solve for x:
- x + 2y = 13
- x + 2(3) = 13
- x + 6 = 13
- x = 7
The solution to this system of equations is x = 7, y = 3.
Now that you know two methods for solving systems of equations, you can use either one to solve any system. If an equation in a system has a variable alone on one side of the equation, such as y = 2x + 4, substitution might be the easier route to take. If the two equations have an identical variable term, such as –3x, elimination might be the easier method. Either method will work every time!
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