Simply put, two-step equations – word problems are two step equations expressed using words instead of just numbers and mathematical symbols. They are just a bit more complicated than one-step equations with word problems and they demand just a bit more effort to solve. If you are not confident in your abilities to solve two-step equations with word problems, you can go to one-step equations – word problems and practice some more before continuing with this lesson. But if you feel ready, we will show you how to solve it using this example:
Hermione’s Bikes rents bikes for $10 plus $4per hour. Janice paid $30 to rent a bike. For how many hours did she rent the bike?
First thing we have to do in this assignment is to find the variable and see what its connection is with the other values. The thing we do not know is the number of hours Janice rented the bike for and we have been asked to find that out. That means the number of hours is our variable.
The cost of renting a bike is 10$ to take the bike and 4$ for every hour it spends in our possession. The final sum Janice paid is $30. Let us write that down as an equation.
4 * x + 10 = 30
Now, in order to make things neater and more clear, let us move all the numbers (except for the number 4 – we have to get rid of it in a different way) to the right side of the equation. Like this:
4 * x = 30 – 10
To simplify things further, let us perform the subtraction.
4 * x = 20
The next thing to do is to get rid of the number 4 in front of the variable. We will do that by dividing the whole equation by 4.
4 * x = 20 |:4
x = 5
Now that we have calculated the value of the variable, we can tell that Janice rented that bike for 5 hours. If you want to check the result – you can. If you multiply $4 that Janice paid per hour by the 5 hours she spend with that bike and then add the $10 she had to pay regardless of the time she spent with the bike, you will get a total sum of $30 that is indeed the full sum she paid.
These word problems are called two-step because you have to perform two mathematical operations in order to solve them. In this case – addition (subtraction) and multiplication (division). To practice solving two-step equations – word problems, feel free to use the worksheets below.
Two-step equations – word problems exams for teachers
|Exam Name||File Size||Downloads||Upload date|
|Two-step equations – Word problems – Integers – very easy||0 B||37458||January 1, 1970|
|Two-step equations – Word problems – Integers – easy||0 B||35353||January 1, 1970|
|Two-step equations – Word problems – Integers – medium||0 B||48886||January 1, 1970|
|Two-step equations – Word problems – Integers – hard||0 B||26477||January 1, 1970|
|Two-step equations – Word problems – Integers – very hard||0 B||16401||January 1, 1970|
|Two-step equations – Word problems – Decimals – very easy||0 B||4635||January 1, 1970|
|Two-step equations – Word problems – Decimals – easy||0 B||5537||January 1, 1970|
|Two-step equations – Word problems – Decimals – medium||0 B||8037||January 1, 1970|
|Two-step equations – Word problems – Decimals – hard||0 B||7861||January 1, 1970|
|Two-step equations – Word problems – Decimals – very hard||0 B||5917||January 1, 1970|
Two-step equations – word problems worksheets for students
|Worksheet Name||File Size||Downloads||Upload date|
|Two-step equations – Word problems – Integers||0 B||38222||January 1, 1970|
|Two-step equations – Word problems – Decimals||0 B||10936||January 1, 1970|