# How do you solve Systems of Equations by Substitution Word Problems

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System of Equations are set of equations with same set of unknown variables.

For a system of equations to be solvable, the number of equations should be at least equal to the number of unknown values.

To be able to solve for unknown values using Substitution method in word problems, we may follow these steps:

- Identify the given values.

- Set a variables (letters) for corresponding unknown value.

- Set up equations based on the given condition or applicable formula.

- Using the one of the equations, isolate one variable on one side (equation A).

- Substituted equation A to the other equation. Usually, a variable will be replaced by a number or algebraic expression value.

- Then repeat the previous step if there are more equations involve until the final equation is in terms of one variable. This will allow you to solve the first unknown value that can be use to solve the other succeeding unknown value using the other equations.

Here is an example of word problem that applies Substitution Method in solving System of Equations.

Suppose Shiela bought 10 outfits. If a blouse has a retail price of $3.50 and a short has a retail price of $ 4.00. How many blouses and short did she buy when her total expenses is $36.50.

Given:

Total outfits = 10 =number of blouses + number of shorts.

Retail price: 1 blouse = $3.50 and 1 short =$4.00

Total expenses =$36.50

Unknown values: x= number of blouses and y = number of shorts

System of Equations: x+y = 10 (total outfits)

3.50x+4.00y = 36.50 (total expenses)

Using x+y=10, isolate x:

x = 10 -y

Substitute x =10-y in 3.50x+4.00y =36.50:

3.50(10-y) +4.00y = 36.50

Simplify: 35-3.50y +4.00y =36.50

35 + 0.50y =36.50

35 + 0.50y **-35** =36.50 **-35**

0.50y =1.50

0.50y/0.50 =1.50/0.50

y = 3

Substitute solved value y=3 in x= 10-y = 10-3 =7.

Conclusion: Shiela bought 7 blouses and 3 shorts.