Systems of equations are groups of equations, almost always having more than one variable. The number of equations in a system must be equal to the number of variables in the system, and more importantly the number of variables must equal the number of equations. In other words if there are three variables it will take three equations to solve for each variable. There are three basic ways to solve systems. They are graphing, substitution, and elimination. It does not matter which method is used, the results will be the same. The advantage of using a certain method at the right time is simply that of ease and the saving of time.

 

Graphing will be the first method covered. There are some problems with using this method, which will be explained later. In order to solve a system of equations by graphing, follow two steps.

1)      Graph each equation.

2)      See where they intersect and that is the answer.

 

 

Example

Solve y = 2x – 4

 

y = + 4

 

Graph the first equation, y = 2x – 4.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Using the same graph, graph the next equation, y = + 4, also.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To find the answer, look at where they cross.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

It is evident they cross at (3,2).

 

(3,2)

 

Answer!

 

 

 

 

Example

Solve 2x + y = 2

 

-y = -x – 5

 

In order to graph these two lines, get them both in slope-intercept form. To get 2x + y = 2 in slope intercept form, first subtract 2x from both sides. In the next equation, -y = -x – 5, divide everything by –1. This will get y all by itself on both sides.

 

y = -2x +2

 

y = x + 5

 

Graph both equations on the same graph.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(-1, 4)

 

 

Answer!

 

 

The previous examples may seem pretty simple and easy to follow but there are a few problems with this method. When trying to work one of these, one may realize that the graphs have to be exact or one will not get the correct answer. Another problem that might be encountered is the fact that if the answer does not fall exactly on two whole numbers (ex. (2, -1.3)), one will not be able to determine exactly what the answer is. It is important to understand the general concept behind graphing, which is that wherever the lines intersect, that is the answer. This understanding will come into more understanding later.

 

Substitution is the next method. It is a very good method that can be used to solve a lot of systems. In substitution there are five main steps to follow.

 

1)      Get a variable by itself.

2)      Substitute into the other equation.

3)      Solve.

4)      Plug back in.

5)      Solve.

 

Each of these steps will be explained throughout the next problem.

 

 

 

 

Example

Solve x = 3y + 11

 

4x – 2y = 14

 

Step 1 states that the first thing is to get a variable by itself. Look at both equations, one can clearly see that there is an x by itself in the first equation (x = 3y + 11).

 

Note- In some equations, like in the next example, there may not be a variable all by itself and the first step will be to get a variable by itself. In this case the variable is by itself so step 1 is already done.

 

With x by itself, move on to step 2, which is the substitution step. x = 3y + 11, substitute the 3y + 11 into the other equation (4x – 2y = 14) anywhere there is an x. Whenever doing this step make sure and use parentheses!

 

4(3y + 11) – 2y = 14

 

Looking at this equation, notice it is one equation with one variable. From here, solve for that variable, y. First distribute out the 4 to get rid of the parentheses.

 

12y + 44 – 2y = 14

 

Simplify by combining like terms on the same side.

 

10y + 44 = 14

 

Subtract 44 from both sides.

10y = -30

 

Divide both sides by 10.

 

y = -3

 

Now step 3 is complete. y is solved for. Look back to either of the original equations, in this case use the first one but it does not matter which one is used. Apply step 4, which is to plug the –3 in for the y. Use parentheses!

 

x = 3(-3) + 11

 

Solve for x.

 

x = 2

 

x = 2 and y = -3.

 

Write these in the correct order (alphabetically) to get the answer in proper form.

 

(2, -3)

 

Answer!

 

 

 

 

 

Example

 

Solve 3x – 2y = 3

 

2x + y = -5

 

Get a variable by itself. It does not matter which variable, just pick the easiest one. Looking at this system, one can see that if they solved for y in the second equation it will probably be the easiest way (all there is to do to get y by itself in the second equation is subtract 2x from both sides). Here is what the system looks like after that is done.

 

Note- It does not matter if one gets x or y by itself in step 1, or which equation to do it with. In the previous problem x was used, this one it is easiest to use y in the second equation. Just take the easiest way.

3x – 2y = 3

 

y = -2x –5

 

y is by itself so y = -2x –5. Moving on to step 2, substitute this value (-2x – 5) into the other equation, wherever there is y.

 

3x – 2(-2x – 5) = 3

 

Step 3 says to solve this equation. Distribute out the –2 to get rid of the parentheses.

 

3x + 4x + 10 = 3

 

Simplify by combining like terms on the same side of the equation.

 

7x + 10 = 3

 

Subtract 10 from both sides.

 

7x = -7

 

Divide both sides by 7.

 

x = -1

 

This completes step 3, x = -1. Plug this value back into one of the original equations. Since one of the originals is already changed into y = -2x –5, use it because it will be easiest to solve for y.

 

y = -2(-1) – 5

 

Solve for y.

 

y = -3

 

x = -1 and y = -3.

 

Writing in the correct order.

 

(-1, -3)

 

Answer!

 

 

 

 

 

Example

Solve 2a + b = -4

 

3a = 2b –6

 

Getting a variable by itself, it would be easiest to solve for b in the first equation. Although it initially may appear to solve for a in the second equation, all that needs done is to divide by 3. Dividing everything in the second equation by 3, will get a fraction and it is much easier not to work with fractions if they can be avoided.

 

b = -2a – 4

 

3a = 2b –6

 

Step 2 says to substitute or plug in the value for b (-2a - 4) into the other equation (3a = 2b – 6).

 

3a = 2(-2a – 4) –4

 

Solve for a by distributing the 2.

 

3a = -4a –8 – 6

 

Simplify by combining like terms on the same side.

 

3a = -4a –14

 

Add 4a to both sides to get all the a’s together.

 

7a = -14

 

Divide both sides by 7.

 

a = -2

 

Now plug this value (-2) in for a in the equation b = -2a – 4.

 

b = -2(-2) – 4 Solve for b.

 

b = 0

 

a = -2 and b = 0, write the results in proper (alphabetical) form.

 

(-2, 0)

 

Answer!

 

 

 

So far the graphing and substitution method have been covered. The last method of solving these types of systems is elimination. 6 steps can summarize elimination.

1)      Line up equations.

2)      Get opposites.

3)      Add equations.

4)      Solve.

5)      Plug back in.

6)      Solve.

 

Just like in the substitution method, the steps will be explained in detail working through a problem.

 

 

 

Example

 

Solve

 

5x - 3y = -5

 

4x + 3y = 23

 

Step 1 says to line up the equations. What does this mean? Here is a further explanation.

Lining up the equations means to write them on top of one another just like the example shows.

5x - 3y = -5

4x + 3y = 23

 

These two equations are already lined up because;

1)      the x’s are on top of the x’s.

2)      the y’s are on top of the y’s.

3)      the = signs are on top of each other.

4)      the constants, or numbers(-5 and –23), are on top of each other.

 

 

By saying they are “on top” of each other is the same as saying that they are in the same location in the equation. The following diagram may help to explain this.

5 x – 3 y = -5

 

 

4 x + 3 y = 23

 

 

It may be more easily seen through this diagram that all the rules previously mentioned are met. If they equations are not already lined up, it is usually not difficult to line them up.

 

 

 

5x- 3y = -5

4x + 3y = 23

 

Since these equations are already lined up move on to step 2 which is to get opposites. An explanation of opposites follows.

 

Opposites are two numbers that add up to 0. One has to be positive and one has to be negative. For example 5 and –5 add up to 0 and are therefore opposites. Other opposite pairs include, 2 and –2, 7 and –7, etc… Just because one is positive and one is negative does not make them opposites. 3 and –1 are not opposites because they do not add up to 0.

 

The numbers to look at are the numbers in front of the variables. Keep the numbers in line with the variable they are in front of. In other words, look at the 5 and 4 together because they are the numbers in front of the x’s and also look at the -3 and 3 because they are the numbers in front of the y’s. Do not interchange them, ever! Looking at the –3 and 3, one can tell that they are opposites, so step 2 is completed.

5x-3y = -5

4x + 3y = 23

 

Step 3 says to add the equations together. This is best shown with an illustration.

 

5x-3y = -5

+4x + 3y = 23

9x + 0y = 18

 

Add the 5x and 4x, as well as the –3y and +3y, and lastly the –5 and 23. This is the result of step 3. The entire goal was to get a zero in front of one of the variables. This was done with the 0 in front of the y. 0 times any number = 0, so simplify this equation.

 

9x = 18

 

Solve for x by dividing by 9. (Step 4)

 

x = 2

 

Continue on by plugging this value (2) in for x in one of the original equations. It does not matter which one to plug it into, just plug it into one that will be easy to solve, in this case use the first one (5x-3y = -5).

 

5(2) – 3y = -5

 

Solve for y by first multiplying.

 

10 – 3y = -5

 

Subtract 10 from each side.

 

-3y = -15

 

Divide both sides by –3

 

y = 5

 

x = 2 and y = 5, write the answer.

 

(2, 5)

 

Answer!

 

 

 

 

 

Example

Solve

2x + 4y = 4

x – 5y = -19

 

Looking at the system, notice that it is already lined up. So go to Step 2.

2x + 4y = 4

x – 5y = -19

 

Look at the numbers in front of the x’s (2 and 1) and the numbers in front of the y’s (4 and –5). It is easy to tell that neither group of numbers contains opposites. Get opposites. To do this, either multiply one of the equations or both of them to get opposites. The easiest way to do it in this problem is to multiply the bottom equation (x – 5y = -19), by -2. When this is done it will get a -2x for the first term.

 

-2(x – 5y = -19)

 

Multiply everything on both sides by –2.

 

-2x + 10y = 38

 

Now look at the entire system with the second equation multiplied by -2.

 

2x + 4y = 4

-2x + 10y = 38

 

Notice that there are now opposites with the 2 and –2 in front of the x’s. Moving on to step 3, add the equations.

2x + 4y = 4

+-2x + 10y = 38

0x +14y = 42

 

The x’s cancel out.

 

14y = 42

 

Solve for y by dividing by 14.

 

y = 3

 

Plug this value into one of the original equations to solve for x. x – 5y = -19 looks like the easiest one so plug the 3 in for the y.

 

x – 5(3) = -19

 

Multiply the 5 and 3.

 

x – 15 = -19

 

Add 15 to both sides.

 

x = -4

 

x = -4 and y = 3.

 

(-4, 3)

 

Answer!

 

 

 

 

 

 

 

Example

Solve

3x + 4y = 11

5x + 3y = 11

 

The equations are already lined up, so get opposites. Looking at the groups of numbers in front of the x’s (3 and 5) and the ones in front of the y’s (4 and 3), there are not opposites, so multiply. This problem is not as simple as the previous one. Without using fractions there is no easy way to get opposites, so take a different approach. Instead of just multiplying one of the equations, multiply both of them. Multiply the first equation by 5 and the second equation by -3. What this will do is get opposites in front of the x’s.

5(3x + 4y = 11)

-3(5x +3y = 11)

 

Multiply.

15x + 20y = 55

-15x – 9y = -33

 

There are opposites in front of the x’s (15 and –15).

 

Note- There are many different approaches to use for the previous step. One of those might include multiply the first equation by 3 and the second by –4. This would have resulted in opposites in front of the y’s. Either way is perfectly acceptable and will get the correct answer.

 

Add the equations.

 

15x + 20y = 55

+-15x – 9y = -33

0x +11y = 22

 

The x’s cancel out.

 

11y = 22

 

Solve for y by dividing by 11.

 

y = 2

 

Plug this value in for y in one of the original equations. It does not appear that one will be that much easier than another so plug into the first equation.

 

3x + 4(2) = 11

 

Multiply.

 

3x + 8 = 11

 

Subtract 8 from both sides.

 

3x = 3

 

Divide both sides by 3.

 

x = 1

 

x = 1 and y = 2

 

(1, 2)

Answer!

 

 

 

 

 

 

 

 

 

Example

Solve

 

2x = -3y + 9

3x + 4y = 12

 

Looking closely at this system one can see that they are not lined up. Even though the x’s are on top of the x’s the y’s on top of the y’s and the constants (numbers) are on top of each other, the equal signs are not on top of each other. ask the question “How do I move the equals sign?” The problem is not in the equals sign as it may appear but rather in the –3y term. It is on the wrong side of the equals sign. What needs to be done is to move it to the other side in the first equation. To do that, add 3y to both sides.

 

2x + 3y = 9

3x + 4y = 12

 

Now they are lined up. As discussed earlier, there are many different ways to get opposites from here. To show some variation, solve this one by multiplying the first equation by –4 and the second by 3. This will produce opposites in front of the y’s.

 

-4(2x +3y = 9)

3(3x +4y = 12)

 

Multiply.

 

-8x –12y = -36

9x +12y = 36

 

There are opposites in front of the y’s, so add the equations.

-8x –12y = -36

+9x +12y = 36

1x + 0y = 0

 

The y’s cancel out. One may think that they have made a mistake here because there is a zero for a constant, but there is no mistake at all, just continue on working the problem.

 

1x = 0

 

Solve for x by dividing both sides by 1.

x = 0

 

Plug this value back in for x in one of the original equations. Use the second equation this time (It does not matter which one is used).

 

3(0) + 4 y = 12

 

Multiply the 3(0) and get zero, so that term cancels out.

 

4y = 12

 

Divide both sides by 4.

 

y = 3

 

x = 0 and y = 3.

 

(0,3)

 

Answer!

 

To solve a system that has 2 variables, there are three different methods; graphing, substitution and elimination. It is very important to learn each method. The most important thing about graphing, even though it will rarely be used, is the concept that the point of intersection is the solution. With substitution and elimination it is important to be able to do both methods well. In some problems it might be easier to get a variable by itself and use substitution and in others it might be easier to get opposites for which elimination is the best way. This will save a lot of time in working these types of problems by choosing the easiest method to use. All three methods will get the same answer but one method will almost always be easier than another.

 

 

 

Example

Solve

5x + 2y = 6

-5x –2y = 12

 

Looking at this system, not only is it lined up, but there are already opposites. Thus, elimination would be the easiest method. With Step 1 and Step 2 already done, go to Step 3 which is to add the equations.

 

5x + 2y = 6

+-5x –2y = 12

0x +0y =18

 

Notice that not only do the x’s cancel out in this problem, but so do the y’s. This is not a mistake but rather a special kind of problem. Look at the result.

 

0 = 18

 

There is a dilemma here. How can one solve for a variable if there is not one here? Look at the result and ask the question “When is 0 = 18?” The answer is never. This means that there is no solution to this problem.

 

No solution

 

Answer!

 

How can that be? If one graphs these two lines, won’t they cross and that would be the answer? Putting both of these lines into slope intercept form and graphing them will produce the following graph.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

These lines will never cross. They will continue on like railroad tracks and never cross. These are called these parallel lines. The type of system where there are no solutions is called Inconsistent. Write down that there is no solution. If one is asked the question, “What will they graph look like?”, respond with “parallel lines.”

 

Example

Solve

4a – 2b = 6

b = 2a –3

 

Since one of the variables is already by itself, it would be best to use substitution to solve this system.

4a – 2b = 6

b = 2a –3

 

Plug 2a –3 into the first equation (4a – 2b = 6), for b.

 

4a –2(2a-3) = 6

 

Distribute the –2.

 

4a - 4a + 6 = 6

 

Combine the like terms on the same side (the a’s).

 

0a + 6 = 6

 

Subtract 6 from both sides.

 

0a = 0

 

because zero times anything is equal to 0.

 

0 = 0

 

One may think that this problem is exactly like the previous example, but it is different. The reason is “When is 0=0?” The answer is “always.” That means that there are many solutions. As a matter of fact, there are infinitely many solutions. Any point that is on either line is a solution. Here is how the answer is written.

 

{(x,y) : b = 2a-3}

 

Answer!

 

Note-This answer would be read as “Any x and y such that, it is on the line b = 2a-3.” It does not matter which equation is used to write the answer because they are same line. Here is what the graph of both of these lines look like. Substitute x in for a and y for b to graph it.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This type of system has a special name it is called Consistent and Dependent. The Consistent means that it has at least one solution (this name would apply to the previous examples where there was one point for an answer). Dependent means that it has multiple solutions.

 

To better understand how the names of these different types of systems are named and what their graph looks like here is a chart to summarize it.

 

Type of system	# of solutions	Graph looks like
Consistent	1	Lines intersect at 1 point
Consistent and Dependent	Infinitely many	2 lines lying on top of each other.
Inconsistent	0	Parallel lines

 

 

Sometime one may be asked to solve a system higher than degree 1. That means that there is an exponent that is higher than 1. These can have more than 1 answer. It is usually a good idea to try and use a graphing calculator if handy and graph it first to see how many times the graphs intersect and how many answers there will be. Use either substitution or elimination to solve these types of problems.

 

 

 

Example

Solve

x² + y² = 2

y = 2x + 1

 

Graph this first to see how may times, if any, they intersect. Here is what the graph will look like.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From the graph, notice that they cross twice.

 

To solve this problem use substitution since y is already solved for in the second equation. That means plug in 2x + 1 for y in the first equation.

x² + (2x + 1)² = 2

 

Raise 2x + 1 to the second power.

 

Note- This is one step where many students will make the mistake of saying that (2x + 1)²= 4x² +1. This is not true. Follow the next steps to see how to do it correctly.

 

x² + (2x + 1) (2x + 1)= 2

 

Multiply 2x + 1 by 2x + 1.

 

Remember this is done by multiplying every term in one by every term in the other.

 

x² + 4x² +2x +2x +1 = 2

 

Simplify by combing all like term on the same side of the equals sign.

 

5x² + 4x +1 = 2

 

To solve for x it is best to use the quadratic formula. The first step is to move the 2 to the other side so it is set = 0.

 

5x² + 4x - 1 = 0

 

Apply the quadratic formula (refer to that section if needed). Make sure and write the answer with the smallest number first.

 

 

Answer!