Problem 21 - Algebra and Trigonometry

Question

\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ 4\left(y-\frac{1}{2}\right)-y=6(5-y) $$

Step-by-Step Solution

Answer

The solution is \(y = \frac{32}{9}\).

Step 1: Expand Both Sides

Start by expanding both sides of the equation. The left-hand side is expanded as:\[4(y - \frac{1}{2}) = 4y - 2\]The equation now becomes:\[4y - 2 - y = 6(5 - y)\]

Step 2: Simplify the Equation

Combine like terms on the left side to simplify:\[4y - y - 2 = 3y - 2\]Now expand the right-hand side:\[6(5 - y) = 30 - 6y\]Substitute back to get:\[3y - 2 = 30 - 6y\]

Step 3: Isolate Variable y

Add \(6y\) to both sides to move all terms with \(y\) to one side:\[3y + 6y - 2 = 30\]This simplifies to:\[9y - 2 = 30\]

Step 4: Solve for y

Add 2 to both sides to solve for \(y\):\[9y = 32\]Next, divide both sides by 9:\[y = \frac{32}{9}\]

Step 5: Verify the Solution

Substitute \(y = \frac{32}{9}\) back into the original equation to verify if it satisfies the equation:\[4\left(\frac{32}{9} - \frac{1}{2}\right) - \frac{32}{9} = 6\left(5 - \frac{32}{9}\right)\]This simplifies to a true statement, confirming that \(y = \frac{32}{9}\) is indeed the correct solution.

Key Concepts

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra that involves finding the value of the variable that makes the equation true. Linear equations can often be recognized by their highest variable power of 1, meaning they form a straight line when graphed.

The process typically starts with expanding any terms that are multiplied out, breaking them down into simpler parts. This is important because it reduces the equation to its basic components, making it easier to handle. For example, in the given exercise, we had to expand both sides:

1. Left: \(4(y - \frac{1}{2})\) into \(4y - 2\)
2. Right: \(6(5-y)\) into \(30 - 6y\)

With both sides expanded, our goal is to simplify by combining like terms, especially the variable terms, to arrange them onto one side of the equation. Finally, solve for the variable by isolating it, which often involves operations like addition, subtraction, multiplication, or division to "undo" the equation around the variable. The solution aims to bring the equation to a format like \(y =\) some number, as seen when we isolated \(y\) to get \(y = \frac{32}{9}\). This solution is our answer, provided it makes the original equation true upon substitution.

Algebra

Algebra is the branch of mathematics that uses letters and symbols to represent numbers and quantities in formulas and equations. The beauty of algebra is in its ability to solve problems and find unknown variables.

In our current problem, we were tasked with handling an equation that involves some fundamental algebra skills like expanding expressions and combining like terms. The exercise started with adjusting the equation by expansion, which is a common algebraic technique to get rid of parentheses.

Once we expanded and simplified the equation, it became easier to move forward by rearranging terms and solving for the unknown variable. Mastery of these skills allows algebra students to tackle more complex equations and develop a more dynamic understanding of mathematical relationships. This problem is an excellent demonstration of how these algebraic operations intertwine to simplify and solve equations, ensuring all parts of the equation are balanced and correct.

Equation Verification

Equation verification is the crucial final step in solving equations. It confirms that the solution you found is indeed correct. This involves substituting your solution back into the original equation to check if the left side equals the right side.

In our exercise, after determining \(y = \frac{32}{9}\), we must verify this solution by substituting it back into the original equation. This is where both sides of the equation should simplify to the same value, ensuring the solution works. For example:

• Substituting \(y = \frac{32}{9}\) into the left side of the equation, simplifies it into an expression.
• Similarly, substituting \(y = \frac{32}{9}\) into the right side also simplifies it into a similar expression.

If both simplify to the same result, as they did in this exercise, this verification confirms that \(y = \frac{32}{9}\) is indeed the correct and true solution. Thus, verifying the solution is a vital step that reinforces confidence in the result, ensures accuracy, and aids in the understanding of the problem-solving process.