Download How to solve systems of equations (two variables, two equations

How to solve systems of equations (two variables, two equations)-Substitution
Method
2𝑥 + 3𝑦 = 2
�. This guide
−5𝑥 + 7𝑦 = 3
will show you how to solve these equations using the substitution method. The substitution
method works by taking one of our equations and solving for one of the variables in the
equation.
There are many ways to solve systems of equations of the type �
𝑥 + 2𝑦 = −1
Example 1 Solve the system of equations � 3𝑥 + 4𝑦 = 1 �
STEP 1: Choose one of the equations and solve for one of the variables.
We can work with either of the equation 𝑥 + 2𝑦 = −1 or 3𝑥 + 4𝑦 = 1 and solve for either 𝑥
or 𝑦. We usually work with something that will require less work. I will choose the equation
𝑥 + 2𝑦 = 1. I will go ahead and solve for 𝑥 because 𝑥 is easiest to solve for. All I need to do is
subtract 2𝑦 to the other side.
𝑥 + 2𝑦 − 2𝑦 = −1 − 2𝑦
𝑥 = −2𝑦 − 1
STEP 2: Replace one of the variables in the equation you did not choose in STEP 1
with the variable you solved for in STEP 1 and simplify.
What we mean by this is take the equation 3𝑥 + 4𝑦 = 1 and replace the 𝑥 in this equation with
what you got in step 1. Specifically, replace 𝑥 in 3𝑥 + 4𝑦 = 1 with 𝑥 = −2𝑦 + 1
3(−2𝑦 − 1) + 4𝑦 = 1
−6𝑦 − 3 + 4𝑦 = 1
−2𝑦 − 3 = 1
STEP 3: Solve the equation you got in STEP 2:
−2𝑦 − 3 + 3 = 1 + 3
−2𝑦
−2
=
4
−2
( add 3 to both sides)
( divide both sides by -2)
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu
𝑦 = −2
STEP 4: Solve for the other variable. Take the result you got in STEP 3 and place
the result in either one of the equations and solve for the variable.
We know that 𝑦 = −2 so we need𝑥. We can plug in 𝑦 = −2 into either 𝑥 + 2𝑦 = −1 or
3𝑥 + 4𝑦 = 1. It does not matter.
I will plug into 3𝑥 + 4𝑦 = 1:
3𝑥 + 4(−2) = 1
3𝑥 − 8 = 1
3𝑥 − 8 + 8 = 1 + 8
3𝑥 = 9
𝑥=3
STEP 5: Write the solution as an ordered pair.
We will write our answer 𝑥 = 3, 𝑦 = −2 as (3, −2)
6𝑥 + 3𝑦 = 21
Example 2 Solve the system of equations � 5𝑥 + 7𝑦 = 40 �
STEP 1: (from now on I will only refer to steps above)
Just like our first example, we can work either equation. However, the first equation is easier
to work with. Why? Notice that all the coefficients and 21 are divisible by 3. If I solve for one
of the variables in this equation it will be easier for me because I will be able to divide by three
and possibly avoid fractions. The second equation has no common factors so you will probably
wind up with fractions, which makes us do a little more work.
Taking 6𝑥 + 3𝑦 = 21, I will solve for y by subtracting 6𝑥 from both sides and dividing by three.
−6𝑥 + 6𝑥 + 3𝑦 = −6𝑥 + 21
3𝑦 = −6𝑥 + 21
STEP 2:
𝑦 = −2𝑥 + 7
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu
I will take the equation 5𝑥 + 7𝑦 = 40 and replace 𝑦 with 𝑦 = −2𝑥 + 7.
5𝑥 + 7(−2𝑥 + 7) = 40
5𝑥 − 14𝑥 + 49 = 40
−9𝑥 + 49 = 40
STEP 3:
−9𝑥 + 49 − 49 = 40 − 49
−9𝑥 = −9
𝑥=1
STEP 4:
We know that 𝑥 = 1 so we need 𝑦. We can plug in 𝑥 = 1 into either 6𝑥 + 3𝑦 = 21 or
5𝑥 + 7𝑦 = 40. It does not matter
I will plug into 5𝑥 + 7𝑦 = 40:
5(1) + 7𝑦 = 40
5 + 7𝑦 = 40
−5 + 5 + 7𝑦 = −5 + 40
7𝑦 = 35
STEP 5
𝑦=5
We will write our answer 𝑥 = 1, 𝑦 = 5 as (1,5)
6𝑥 + 5𝑦 = 59
Example 3 Solve the system of equations � 4𝑥 + 3𝑦 = 37 �
STEP 1:
In this example, neither of these equations is “easiest” to work with in part because none of
these equations have a number that every term is divisible by like Example 2. I will use
4𝑥 + 3𝑦 = 37 but 6𝑥 + 5𝑦 = 59 would have worked just fine.
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu
For the equation 4𝑥 + 3𝑦 = 37 neither 𝑥 or 𝑦 are easy to solve for. I will solve for 𝑥 but solving
for 𝑦 is perfectly fine.
4𝑥 + 3𝑦 − 3𝑦 = 37 − 3𝑦
4𝑥 = −3𝑦 + 37
𝑥=
STEP 2:
−3
𝑦+
4
37
4
I will take the equation 6𝑥 + 5𝑦 = 59 and replace 𝑥 with 𝑥 =
−3
6�
4
−18
𝑦+
4
𝑦+
−3
37
4
� + 5𝑦 = 59
222
4
4
𝑦+
37
4
+ 5𝑦 = 59
I will next multiply the least common denominator (L.C.D.) which is 4 so I can get rid of the
denominator:
4●
STEP 3:
−18
4
𝑦 + 4●
222
4
+ 4●5𝑦 = 4●59
−18𝑦 + 222 + 20𝑦 = 236
2𝑦 + 222 = 236
2𝑦 + 222 = 236
2𝑦 = 14
STEP 4:
𝑦=7
We know that 𝑦 = 7 so we need 𝑥. We can plug in 𝑦 = 7 into either 4𝑥 + 3𝑦 = 37 or
6𝑥 + 5𝑦 = 59. It does not matter.
I will plug into 6𝑥 + 5𝑦 = 59:
6𝑥 + 5(7) = 59
6𝑥 + 35 = 59
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu
6𝑥 + 35 − 35 = 59 − 35
6𝑥 = 24
𝑥=4
STEP 5
We will write our answer 𝑥 = 4, 𝑦 = 7 as (4,7)
The next two examples will showcase two systems of equations with infinite number of
solutions or no solutions, respectively.
𝑥 + 2𝑦 = 10
Example 4 Solve the system of equations � 2𝑥 + 4𝑦 = 20 �
STEP 1:
We can work with either of the equations 𝑥 + 2𝑦 = 10 or 2𝑥 + 4𝑦 = 20 and solve for either 𝑥
or 𝑦. I will choose the equation 𝑥 + 2𝑦 = 10. I will go ahead and solve for 𝑥 because 𝑥 is
easiest to solve for. All I need to do is subtract 2𝑦 to the other side.
𝑥 + 2𝑦 − 2𝑦 = 10 − 2𝑦
STEP 2:
𝑥 = −2𝑦 + 10
I will take the equation 2𝑥 + 4𝑦 = 20 and replace 𝑥 with 𝑥 = −2𝑦 + 10:
2(−2𝑦 + 10) + 4𝑦 = 20
−4𝑦 + 20 + 4𝑦 = 20
20 = 20
STOP!!!! We don’t have any variables left!! What to do? When you come to a case where you
have no variables left and two numbers that equal each other we have an infinite number of
solutions and we are done. Our solution is: infinite number of solutions
𝑥 + 2𝑦 = 5
Example 5 Solve the system of equations � 2𝑥 + 4𝑦 = 20 �
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu
STEP 1:
We can work with either of the equations 𝑥 + 2𝑦 = 5 or 2𝑥 + 4𝑦 = 20 and solve for either 𝑥
or 𝑦. I will choose the equation 𝑥 + 2𝑦 = 5. I will go ahead and solve for 𝑥 because 𝑥 is easiest
to solve for. All I need to do is subtract 2𝑦 to the other side.
𝑥 + 2𝑦 − 2𝑦 = 5 − 2𝑦
STEP 2:
𝑥 = −2𝑦 + 5
I will take the equation 2𝑥 + 4𝑦 = 20 and replace 𝑥 with 𝑥 = −2𝑦 + 5:
2(−2𝑦 + 5) + 4𝑦 = 20
−4𝑦 + 10 + 4𝑦 = 20
10 = 20
STOP!!!! We don’t have any variables left!! What to do? When you come to a case where you
have no variables left and two numbers that don’t equal each other, we have no solutions and
we are done. Our solution is: no solutions or ∅
Created by Ivan Canales, Senior Tutor. San Diego City College. icanales@sdccd.edu