Problem 38
Question
Solve. See Examples 1 through 7 $$ \frac{4(5-w)}{3}=-w $$
Step-by-Step Solution
Verified Answer
The solution is \(w = 20\).
1Step 1: Distribute the 4
Start by distributing the 4 inside the parentheses. This means you multiply 4 by each term inside the parentheses.\[ 4(5 - w) = 4 \times 5 - 4 \times w = 20 - 4w \]
2Step 2: Rewrite the Equation
Substitute the distributed expression back into the equation. The equation now looks like this:\[ \frac{20 - 4w}{3} = -w \]
3Step 3: Eliminate the Fraction
Multiply every term in the equation by 3 to eliminate the fraction. This gives:\[ 20 - 4w = -3w \]
4Step 4: Collect Like Terms
Add \(4w\) to both sides of the equation to move all the terms involving \(w\) to one side.\[ 20 = 4w - 3w \]
5Step 5: Simplify
Simplify the equation:\[ 20 = w \]
6Step 6: Verify the Solution
Plug \(w = 20\) back into the original equation to ensure it is correct:\[ \frac{4(5 - 20)}{3} = -20 \]\[ \frac{4(-15)}{3} = -20 \]\[ \frac{-60}{3} = -20 \]\[ -20 = -20 \]The solution is correct.
Key Concepts
Distributive PropertyFractions in EquationsLike TermsAlgebraic Verification
Distributive Property
The distributive property is a fundamental concept in algebra that helps to simplify expressions and solve equations. In simple terms, it allows you to multiply a single term by each term within a set of parentheses. For example, if you have a term like \(4(5-w)\), applying the distributive property means you multiply 4 by both 5 and \(-w\).
This results in:
Using the distributive property correctly is crucial for breaking down more complex expressions into manageable pieces.
This results in:
- \(4 \times 5 = 20\)
- \(4 \times (-w) = -4w\)
Using the distributive property correctly is crucial for breaking down more complex expressions into manageable pieces.
Fractions in Equations
When you are working with equations that include fractions, it's typically easier to solve them by eliminating the fractions early on. This simplifies the process and reduces potential errors.
In the expression \(\frac{20 - 4w}{3} = -w\), the fraction can be eliminated by multiplying every term by the denominator, which is 3 in this case. This operation serves to clear the fraction for a clearer path to solving for \(w\).
Let's see the steps:
In the expression \(\frac{20 - 4w}{3} = -w\), the fraction can be eliminated by multiplying every term by the denominator, which is 3 in this case. This operation serves to clear the fraction for a clearer path to solving for \(w\).
Let's see the steps:
- Multiply both sides of the equation by 3, aiming to eliminate the fraction: \(3 \times \frac{20-4w}{3} = 3 \times (-w)\)
- This simplifies to: \(20 - 4w = -3w\).
Like Terms
In algebra, identifying and combining like terms is a key skill for simplifying expressions and solving equations. Like terms are terms that have the same variable raised to the same power. For instance, in the equation \(20 - 4w = -3w\), \(-4w\) and \(-3w\) are like terms because they both include the variable \(w\).
The process of combining like terms in this context involves moving all similar terms to one side of the equation to simplify it.
Here's how it works:
The process of combining like terms in this context involves moving all similar terms to one side of the equation to simplify it.
Here's how it works:
- Add \(4w\) to both sides: \(20 = -3w + 4w\)
- This simplifies to: \(20 = w\)
Algebraic Verification
Verifying your solution is an essential step to ensure accuracy in solving algebraic equations. It involves substituting the found solution back into the original equation to see if it holds true.
For instance, with the solution \(w = 20\) from the equation \(\frac{4(5-w)}{3} = -w\), verify by substituting \(w = 20\) back:
For instance, with the solution \(w = 20\) from the equation \(\frac{4(5-w)}{3} = -w\), verify by substituting \(w = 20\) back:
- Original equation: \(\frac{4(5 - 20)}{3} = -20\)
- Simplifies to: \(\frac{4(-15)}{3} = -20\)
- Which further simplifies to: \(\frac{-60}{3} = -20\)
- And confirms: \(-20 = -20\)
Other exercises in this chapter
Problem 38
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