Problem 15 - Algebra and Trigonometry

Question

Solve and check each linear equation. $$\begin{array}{l}{25-[2+5 y-3(y+2)]=} \\\\{-3(2 y-5)-[5(y-1)-3 y+3]}\end{array}$$

Step-by-Step Solution

Answer

The solutions to the given equations are \(y = 29/2\) for the first equation and \(y = 17/8\) for the second equation.

Step 1: Simplify the first equation

Simplify and solve the equation within the bracket: \[25 - [2 + 5y - 3(y + 2)]\]Expand the equation to : \(25 - 2 - 5y + 3y + 6\) which simplifies to \(29 - 2y = 0\). Now, rearrange to isolate 'y': \(2y = 29, y = 29/2\).

Step 2: Check the first solution

Check the solution by replacing \(y\) with the solved value in the original equation. If both sides turn out equal, the obtained solution for \(y\) is correct.

Step 3: Simplify the second equation

Simplify and solve the equation within the bracket: \[-3(2y - 5) - [5(y - 1) - 3y +3]\]Expand the equation to: \(-6y + 15 - 5y + 5 + 3y - 3\). This simplifies to \(-8y + 17 = 0\). Now, rearrange to isolate 'y': \(8y = 17, y = 17/8\).

Step 4: Check the second solution

Check the solution by replacing \(y\) with the solved value in the original equation. If both sides turn out equal, the obtained solution for \(y\) is correct.

Key Concepts

Solving Linear Equations

The process of solving linear equations is about finding the value of the variable that makes the equation true. Linear equations often appear in the form of ax + b = c. Here, we see the variable \(y\) with various coefficients.

To solve such equations, we typically follow these steps:

• Simplify both sides of the equation if needed.
• Isolate the variable to one side of the equation.
• Perform arithmetic operations to solve for the variable.
• Check your solution by substituting the value back into the original equation.

When solving linear equations like the ones in our exercise, simplifying and isolating the variable (\(y\)) is key for obtaining a clear solution. We must ensure each step logically follows from the previous one.

Simplification

Simplification involves reducing the equation to its simplest form. When faced with brackets or parentheses, start by expanding the terms.

For example, consider the expression \(25 - [2 + 5y - 3(y+2)]\). You'll first want to expand everything inside the brackets and combine like terms. This looks like:

• \(2 + 5y\) becomes \(5y + 2\)
• \(-3(y+2)\) becomes \(-3y - 6\)

Combining these gives us \(25 - (5y + 2 - 3y - 6)\). By simplifying further, you eliminate all like terms and simplify to \(29 - 2y = 0\). Repeating this reduces complex expressions into straightforward equations that are easier to solve.

Isolating Variables

Isolating the variable means getting the variable (\(y\) in our case) on one side of the equation. After simplification, you'll typically end up with an expression like \(29 - 2y = 0\).

To isolate \(y\):

• Move constants to the other side. For instance, \(29 - 2y = 0\) becomes \(2y = 29\) once you move 29 to the other side.
• If the variable has a coefficient, divide by that number to isolate the variable. Here, divide both sides by 2, resulting in \(y = 29/2\).

By following this process, you aim to solve for \(y\) accurately, ensuring the equation balances on both sides.

Substitution Method

The substitution method is used to verify your solution. This step ensures your derived value for \(y\) satisfies the original equation.

Take the solved value of \(y\) (e.g., \(y = 29/2\)) and substitute it back into the standard equation you're working with.

• Plug in \(y\) into both parts of the original equation and simplify it.
• Check if both sides of the equation are equal afterward.

If they are equal, you have verified your solution is correct. By substituting back, it confirms that the calculations were accurate and ended with a valid solution.