Problem 28 - Elementary Algebra

Question

Perform the following additions and subtractions. Subtract \(4 s-3\) from \(7 s+8\)

Step-by-Step Solution

Answer

Answer: The result is \(3s + 11\).

Step 1: Write down the given expressions

We are given the expressions \(7s + 8\) and \(4s - 3\) and asked to subtract \(4s - 3\) from \(7s + 8\).

Step 2: Change the subtraction to addition

To make it easier to combine like terms, we can change the subtraction to addition by multiplying the second expression by \(-1\). This gives us: \(7s + 8 - (4s - 3) = 7s + 8 + (-1)(4s - 3)\).

Step 3: Distribute the \(-1\) to each term in the second expression

Now let's distribute the \(-1\) to both terms in the second expression: \(7s + 8 + (-1)(4s) + (-1)(-3) = 7s + 8 - 4s + 3\).

Step 4: Combine like terms

Now we can combine the like terms. In this case, we have terms with variable \(s\) and constant terms: \((7s - 4s) + (8 + 3) = 3s + 11\).

Step 5: Write the final result

The subtraction of the expressions results in \(3s + 11\). So, \(7s + 8 - (4s - 3) = 3s + 11\).

Key Concepts

Subtraction

Subtraction is one of the basic operations in algebra that you will often use when working with algebraic expressions. When subtracting one expression from another, we essentially reverse the signs of the expression being subtracted.
In the given exercise, we are tasked to subtract \(4s - 3\) from \(7s + 8\). To effectively subtract, we convert this operation into an addition by reversing the signs of the terms in \(4s - 3\).

This means:

• Multiplying each term in \(4s - 3\) by \(-1\),
• Resulting in \(7s + 8 + (-4s + 3)\).

This transformation allows us to seamlessly transition into combining terms, simplifying the expression, and eventually arriving at the final solution.

Like Terms

Like terms are terms that contain the same variable raised to the same power. Identifying and combining like terms is a fundamental skill in algebra, which makes simplification straightforward.
In our exercise, after converting subtraction to addition, the expression is transformed into \(7s + 8 + (-4s + 3)\). Here,

• \(7s\) and \(-4s\) are like terms because they have the same variable \(s\).
• \(8\) and \(3\) are like terms as they are both constant terms.

To simplify the expression, combine the coefficients of the \(s\) terms, and add the constants:

\((7s - 4s) + (8 + 3) = 3s + 11\). This process is essential for reducing expressions to their simplest forms.

Distributive Property

The distributive property is a powerful algebraic tool that allows you to multiply a single term by two or more terms inside a parenthesis. It makes algebraic manipulation more manageable, especially when dealing with expressions that involve subtraction and addition.
In this context, let's revisit how the distributive property was used in solving the exercise:To change subtraction to addition, we multiplied \(-1\) across the terms in the expression \(4s - 3\).

• This results in transforming it into \(-1 \times 4s\) and \(-1 \times (-3)\).
• Using the distributive property gives us \(-4s + 3\).

This step sets the stage for efficiently combining like terms and simplifying the overall expression. Understanding the distributive property's role in such transformations is vital for tackling more complex algebraic problems.