Solving Systems by Substitution, Exercises of Pre-Calculus

There are 5 steps to this method: 1) Solve for one variable in one equation. 2) Substitute the variable into the other equation. 3) Solve for the first ...

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Solving(Systems(by(Substitution(
The second method is the substitution method. The word substitution means to
replace something with something else. This can easily be done if you have solved for
one of the variables. If you already have a variable by itself, then this is a good method
to choose because the first step is done. There are 5 steps to this method:
1) Solve for one variable in one equation.
2) Substitute the variable into the other equation.
3) Solve for the first variable.
4) Substitute the answer for the first variable in either equation to find the other
variable.
5) Write the answer as an ordered pair.
Let’s look at an example where substitution works nicely:
Example(1(
34 1
21
xy
xy
-=
=-
Notice that x is already solved for in the second equation, so the first step is already
done. We can replace the x in the first equation with the 2y-1 that the x equals in the
second equation to get:
3(2y - 1)4y = 1 Now simplify by distributing and combining like terms.
6y – 34y = 1
2y 3 = 1 Now add 3 to move the 3.
2y 3 + 3 = 1 + 3
2y = 4 Now divide both sides by 2 to get:
y = 2
We have now finished step 3 since we have solved for the
first variable. Step 4 has us substitute this answer of y = 2
into either equation to solve for the other variable. Let’s
substitute into the second equation of x = 2y 1.
x = 2(2) 1 = 41 = 3.
Now that we have the second variable, x = 3, we just need to write our answer as an
ordered pair (x, y) = (3, 2). Figure 1 shows the solution of Example 1.
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Solving Systems by Substitution

The second method is the substitution method. The word substitution means to

replace something with something else. This can easily be done if you have solved for

one of the variables. If you already have a variable by itself, then this is a good method

to choose because the first step is done. There are 5 steps to this method:

  1. Solve for one variable in one equation.

  2. Substitute the variable into the other equation.

  3. Solve for the first variable.

  4. Substitute the answer for the first variable in either equation to find the other

variable.

  1. Write the answer as an ordered pair.

Let’s look at an example where substitution works nicely:

Example 1

x y

x y

Notice that x is already solved for in the second equation, so the first step is already

done. We can replace the x in the first equation with the 2y- 1 that the x equals in the

second equation to get:

3 (2y - 1) – 4y = 1 Now simplify by distributing and combining like terms.

6y – 3 – 4y = 1

2y – 3 = 1 Now add 3 to move the – 3.

2y – 3 + 3 = 1 + 3

2y = 4 Now divide both sides by 2 to get:

y = 2

We have now finished step 3 since we have solved for the

first variable. Step 4 has us substitute this answer of y = 2

into either equation to solve for the other variable. Let’s

substitute into the second equation of x = 2y – 1.

x = 2(2) – 1 = 4 – 1 = 3.

Now that we have the second variable, x = 3, we just need to write our answer as an

ordered pair (x, y) = (3, 2). Figure 1 shows the solution of Example 1.

Figure 1 : Solution of Example 1

Example 2

y x

x y

Notice that the first equation already has a variable solved for, so step one is done and

we can substitute

x + into the second equation for the y-variable to get:

x x

æ ö

ç ÷

è ø

Now, let’s change our subtraction to adding the opposite,

distribute, then combine like terms.

x x

x

x

x

x

æ ö

ç

æ ö

    • = ç ÷

è ø

è ø

÷

Notice that again the variables disappeared and the statement left is true. We did not do

anything wrong, this means that the equations are the same equation. The equations

just look different. There are an infinite number of solutions to this system. Figure 3

shows the solution of Example 3.

Figure 3 : Solution of Example 3

When solving algebraically using the substitution method or the next method, the

Elimination/Addition Method, and you have the situations that we encountered in

examples 2 or 3 with the substitution method, you will have either infinite or no solutions

depending on whether the remaining statement is true or false.