Subtraction with Addition in Parentheses Practice Problems

Master subtraction of whole numbers with addition in parentheses through step-by-step practice problems. Learn the distributive rule a-(b+c)=a-b-c with examples.

📚Practice Subtraction with Addition in Parentheses
  • Apply the rule a-(b+c)=a-b-c to solve subtraction problems with parentheses
  • Subtract each term in parentheses separately from the first number
  • Solve expressions like 25-(10+5) using order of operations
  • Work with negative numbers in subtraction with addition in parentheses
  • Master two different methods: distribute subtraction or solve parentheses first
  • Build confidence with step-by-step guided practice problems

Understanding Subtracting Whole Numbers with Addition in Parentheses

Complete explanation with examples

Subtraction of whole numbers with addition in parentheses refers to a situation where we must perform the mathematical operation of subtraction on the sum of some terms that are in parentheses.
In this case, we must remember that the subtraction will be performed on each and every term separately.

The rule is as follows:

a−(b+c)=a−b−ca - (b +c) = a - b - c

a - (b +c) = a - b - c

Each of the terms has its own numerical value.

Detailed explanation

Practice Subtracting Whole Numbers with Addition in Parentheses

Test your knowledge with 40 quizzes

\( 99:(33:10)= \)

Examples with solutions for Subtracting Whole Numbers with Addition in Parentheses

Step-by-step solutions included
Exercise #1

15:(2×5)= 15:(2\times5)= ?

Step-by-Step Solution

First we need to apply the following formula:

a:(b×c)=a:b:c a:(b\times c)=a:b:c

Therefore, we get:

15:2:5= 15:2:5=

Now, let's rewrite the exercise as a fraction:

1525= \frac{\frac{15}{2}}{5}=

Then we'll convert it to a multiplication of two fractions:

152×15= \frac{15}{2}\times\frac{1}{5}=

Finally, we multiply numerator by numerator and denominator by denominator, leaving us with:

1510=1510=112 \frac{15}{10}=1\frac{5}{10}=1\frac{1}{2}

Answer:

112 1\frac{1}{2}

Video Solution
Exercise #2

10:(10:5)= 10:(10:5)=

Step-by-Step Solution

To solve the expression 10:(10:5) 10 : (10 : 5) , we will apply the order of operations systematically.

Step 1: Evaluate the inner division 10:5 10 : 5 .
When we compute 10:5 10 : 5 , we are finding how many times 5 fits into 10. This calculation can be expressed as:
105=2 \frac{10}{5} = 2 .

Step 2: Substitute the result from step 1 into the outer division.
Now, we substitute 10:(10:5) 10 : (10 : 5) with 10:2 10 : 2 . Once again, we apply division:
102=5 \frac{10}{2} = 5 .

Therefore, the solution to the expression 10:(10:5) 10 : (10 : 5) is 5 5 .

Answer:

5 5

Video Solution
Exercise #3

18:(6×3)= 18:(6\times3)=

Step-by-Step Solution

To solve the expression 18÷(6×3) 18 \div (6 \times 3) , we need to follow the order of operations, which specifies that multiplication should be performed before division. Therefore, we proceed as follows:

  • Step 1: Calculate the operation inside the parentheses: (6×3)(6 \times 3).
    We multiply 66 by 33 to get 1818.
  • Step 2: Replace the multiplication expression in the original division: 18÷1818 \div 18.
  • Step 3: Perform the division: 18÷18=118 \div 18 = 1.

Thus, the result of the expression 18÷(6×3) 18 \div (6 \times 3) is 1\mathbf{1}.

Answer:

1

Video Solution
Exercise #4

2−(1+1)= 2-(1+1)=

Step-by-Step Solution

To solve the expression 2−(1+1) 2 - (1 + 1) , follow these steps:

  • First, evaluate the expression inside the parentheses: 1+1 1 + 1 .
  • This gives 2 2 .
  • Now replace the parentheses with this result, transforming the expression to 2−2 2 - 2 .
  • The result of 2−2 2 - 2 is 0 0 .

Therefore, the solution to the expression is 0 0 .

Answer:

0

Video Solution
Exercise #5

19−(5+11)= 19-(5+11)=

Step-by-Step Solution

To solve the problem 19−(5+11)19 - (5 + 11), we will follow these steps:

  • Step 1: Evaluate the expression inside the parentheses. This means we need to calculate 5+115 + 11.
  • Step 2: Once the sum inside the parentheses is found, subtract this sum from 19.

Let's work through each step:

Step 1: Calculate 5+115 + 11 which equals 16.

Step 2: Substitute 16 in place of 5+115 + 11 in the original expression. You have 19−1619 - 16.

Now, solve 19−1619 - 16, which equals 3.

Therefore, the solution to the problem is 33.

Answer:

3

Video Solution

Frequently Asked Questions

What is the rule for subtraction with addition in parentheses?

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The rule is a-(b+c)=a-b-c. When subtracting a sum in parentheses, you subtract each term separately from the first number. For example, 12-(3+2) becomes 12-3-2=7.

How do you solve 25-(10+5) step by step?

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Method 1: Apply the rule 25-(10+5)=25-10-5=15-5=10. Method 2: Solve parentheses first 25-(15)=25-15=10. Both methods give the same answer.

What are common mistakes when subtracting addition in parentheses?

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Common mistakes include: 1) Only subtracting the first term in parentheses, 2) Adding instead of subtracting when distributing the negative sign, 3) Forgetting to change signs when removing parentheses, 4) Not following order of operations correctly.

How do negative numbers work in subtraction with parentheses?

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When you have expressions like -4-(-14+(-23)), first solve inside parentheses: -4-(-37). Then apply sign rules: subtracting a negative becomes addition, so -4-(-37)=-4+37=33.

Why does a-(b+c) equal a-b-c?

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This follows the distributive property of subtraction over addition. The negative sign in front of parentheses distributes to each term inside, changing their signs. So -(b+c) becomes -b-c, making a-(b+c)=a-b-c.

What grade level learns subtraction with addition in parentheses?

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This concept is typically introduced in middle school (grades 6-8) as part of order of operations and pre-algebra. Students learn it after mastering basic arithmetic and before moving to more complex algebraic expressions.

How is this different from regular subtraction?

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Regular subtraction involves two numbers: a-b. Subtraction with addition in parentheses involves multiple terms: a-(b+c+d). You must either solve the parentheses first or distribute the subtraction to each term inside the parentheses.

What real-world problems use subtraction with parentheses?

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Examples include calculating change when buying multiple items, finding temperature differences, budget calculations where you subtract total expenses, and measuring distances when accounting for multiple segments or detours.

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