Solving inequalities by addition and subtraction worksheet

Addition & Subtraction

Solve simple inequalities involving addition and subtraction

The simplest equations can be solved with just one operation. To solve for the variable, students use either addition or subtraction with these problems.

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The Benefits of Solving Inequalities Worksheets:

Analyze boundaries:

For situations where exact equality is not achieved, inequalities can help students assess boundary values of the required result. Even though students do not get the precise result, they can identify upper or lower boundaries.

Flexibility in prediction:

This leaves room for flexibility for scenarios where changes might occur, therefore lowering the possibility of error that may have been generated from the prediction of a fixed outcome.

Real world application: 

Solving inequalities worksheets can show students how these boundaries and predictions can be implemented in our daily lives through mathematical solving of linear inequalities.

While solving simple inequalities or linear inequalities in mathematics, there are four possible results for the comparison of the left-hand side (LHS) to the right-hand side (RHS) of an inequality sign:

1. Less than (<): LHS is always lesser in value than RHS
2. Less than or equal to (?): LHS is lesser or equal in value to RHS
3. Greater than (>): LHS is always greater in value than RHS
4. Greater than or equal to (?): LHS is greater or equal in value to RHS

Students are provided with real-life examples and can learn to solve inequalities through these worksheets.

Example:

Ravi was asked to buy a maximum of 80 chocolates. These chocolates are sold in packets, each of which contains 12 chocolates. Find the number of packets he needs to buy to satisfy the requirement.

Solution:

Assume that Ravi has to buy x number of packets to meet the requirements. Each packet contains 12 chocolates.

Thus we can write the above problem in the form of linear inequality as:

12 * number of packets < 80

Or,

12x < 80

=> 12x/12 < 80/12 (Dividing both sides with 12)

=> x < 20/3

Therefore the number of packets Ravi needs to buy to meet the requirement is 20/3. 

Or, x < 6.66 (as 20/3 is equal to 6.66)

Because the number of packets cannot be in a fraction, we must round it up to 6 packets. 

This is how a scenario can be represented in the form of inequalities.

Inequalities can be solved by either of the following methods:

• Addition or subtraction of the same number from both sides of the inequality sign without altering the sign
• Multiplication or division of the same non-zero positive number to both sides of the sign. In the case of multiplication or division with a negative number, the inequality sign will be reversed.

Printable Solving inequalities Worksheets

Solving inequalities worksheets can be downloaded or printed so students can practice with them at their own pace and convenience. The printable worksheets contain varying levels of difficulty,  from simple inequalities to more complex ones, to assist students in understanding this concept.

Question 1:

Solve the following inequality : x + 23 – 20 \(\geq\) 678

Solution:

x + 23 – 20 \(\geq\) 678

x + 3678

x + 3 – 3 \(\geq\) 678 -3       (Subtraction property of inequality)

x \(\geq\) 675

Question 2:

Seven more than a number is more than 50.

Solution: 

Let the number be ‘x’

According to the question, 

7 + x > 50

7 – 7 + x > 50 – 7                               (Subtraction property of inequality)

x > 43

This means that the number under consideration is more than 43.

Question 3

The sum of a number and 39 is less than 50, what is the number?

Solution:

Let the number be x, the equation representing the question is as follows:

x + 39 <  50

x + 39 – 39 <  50 – 39                      (Subtraction property of inequality)

x <  11

So, the  number is less than 11.

Question 4

For a dance piece, a certain number of dancers were required. The requirement is that the final number must be less than 100. The choreographer has already chosen 12 dancers, how many dancers can be added to the troupe? Also, represent this situation graphically.

Let the number of dancers to be added be x. 

Since 12 dancers were already chosen, the number of dancers is as follows

x + 12 < 100

Simplifying the equation, we get

x + 12 – 12 < 100 – 12                       (Subtraction property of inequality)

x < 88

Therefore the number of dancers to be selected should be lesser than 88.

Question 5

Ice-cream was distributed for a family reunion. To understand the approximate quantity that was distributed, Alan remembers the following “From the total amount of ice cream, four scoops were distributed  but the final quantity must be greater than or equal to two scoops.” Based on the statement, what was the minimum number of scoops that was initially present? Also plot a graph showing the same.

Solution:

Let the initial number of scoops of ice cream be x.

x – 4 \(\geq\) 2

Add 4 on both sides

x – 4 + 4 \(\geq\) 2 + 4      (Addition property of inequality)

x \(\geq\) 6

This means that a minimum of 6 scoops were initially available.

Now, graph the inequality