Sales Toll Free No: 1-855-666-7446

Substitution Method

Top

When there comes a situation to solve system of equations one can use either substitution method or elimination method to solve the given set of equations.

Solving systems of equations by graphing is often not accurate if the solutions are not integers. Sometimes you won't have a graphing calculator to use.

There are several methods for solving systems of equations without graphing. One of these is called the substitution method. For a given system of equations it is used to eliminate the variables. Substitution allows you to create a one variable equation. In this page we will discuss how to solve problems using substitution method.

Solve by Substitution Method

Back to Top
Substitution method comes in handy when we need to solve a linear system algebraically. It is said to be one of the important method useful for solving two equations in two unknowns. In this method we can substitute one(x or y) value with the other to find the other value. Solve for one variable in one equation and substitute the value of that variable into the other equation. This makes more sense with an example.Some examples on how to use substitution method to solve problems are clearly presented below.

Example 1: Solve using the substitution method.
x + y = 6
x = y + 2

Solution: The second equation states that x and y + 2 are equivalent expressions. Thus in the first equation, we can substitute y + 2 for x.

x + y = 6

y + 2 + y = 6       Substituting y + 2 for x in the first equation.

Since this equation now has only one variable, we can solve for y.

2y + 2 = 6     Collect all the like terms

2y = 4

y = 2

Next substitute 2 for y in either of the original equations.

x + y = 6

x + 2 = 6

x = 4

We check x = 4 and y  = 2 in both equations.

Check : x + y = 6

4 + 2 = 6

6 = 6

x = y + 2

4 = 2 + 2

4 = 4
Therefore the solution of the system is (4, 2).
Sometimes neither equations has a variable alone on one side. We can solve one equation for one of the variables and proceed as before. There is more than one way to solve a system using substitution. Solving for a variable with a coefficient of 1 or - 1 is a good place to start. No matter what variable you solve for first, you should always get the same answer.Example 2: Solve using the substitution method.

x - 2y = 6

3x + 2y = 4

Solution: Solve the first equation for x

x  = 6 + 2y

Substitute 6 + 2y for x in the second equation.

3(6 + 2y) + 2y = 4

18 + 6y + 2y = 4

18 + 8y = 4

8y = -14

y = - $\frac{7}{4}$

We go back to either of the original equations and substitute - $\frac{7}{4}$ for y. It will be easier to solve for x in the first equation.

x - 2 (- $\frac{7}{4}$) = 6

x + $\frac{7}{2}$ = 6

x = $\frac{5}{2}$

We check ($\frac{5}{2}$, - $\frac{7}{4}$) in both equations.

x - 2y = 6

$\frac{5}{2}$ - 2(- $\frac{7}{4}$) = 6

= $\frac{5}{2}$ + $\frac{7}{2}$ = 6

6 = 6

3x + 2y = 4

3. $\frac{5}{2}$ + 2(- $\frac{7}{4}$) = 4

= $\frac{15}{2}$ - $\frac{7}{2}$ = 4

 4 = 4

Therefore the solution of the system is ($\frac{5}{2}$, - $\frac{7}{4}$)

Examples

Back to Top
Given below are some problems based on substitution method for a clearer understanding of the topic.

Example: An art class is planning a trip to a museum. There are 22 people going on the trip. There are four drivers and two types of vehicles, vans and cars. The vans seat six people, and the cars seat four people, including drivers. How many vans and cars does the class need for the trip? Use the system below .

Solution: Let v = The number of vans and

c = The number of cars.

Drivers : v + c = 4.

People : 6v + 4c = 22

You can this solve this system by using the substitution method.

v + c = 4

Solve the first equation for v.

v = - c + 4

Substitute - c + 4 for v in the second equation.

6 (- c + 4) + 4c = 22

-6c + 24 + 4c = 22

- 2c + 24 = 22            solve for c.

- 2c = -2

c = 1

v + 1 = 4        Substitute 1 for c in the first equation.

V = 3,   on solving

Since c = 1 and v = 3 the art class should use 1 car and 3 vans.

Example 2: Solve the system using substitution:

x + y = 6

5x + 5y = 10

Solution:
  
x + y = 6           Solve the first equation for x.

x = 6 - y    Substitute 6 - y for x in the second equation.

5(6 - y) + 5y = 10

30 - 5y + 5y = 10

Solve for y

30 = 10, This is not true.

Since 30 = 10 is a false statement, the system has no solution.

Example 3: Solve the system y = x + 6.1 and y = - 2x - 1.4 using the method of substitution.

Solution: Start with one equation

y = - 2x - 1.4

x + 6.1 = - 2x - 1.4

Substitute x + 6.1 for y in that equation.

3x = - 7.5

x = - 2.5

Substitute - 2. 5 for x in either equation and solve for y.

y = ( -2.5) + 6.1

y = 3.6

Since x = - 2. 5 and y = 3.6, the solution is ( - 2.5, 3.6).

Check to see if ( -2.5, 3.6) satisfies the other equation.

3. 6 = -2 ( - 2.5) - 1. 4

3.6 = 5 - 1.4

3.6 = 3.6

If required you can even use your graphing calculator to check your solved algebraic solution.