Problem 47 - Elementary Algebra

Question

For the following problems, solve each conditional equation. If the equation is not conditional, identify it as an identity or a contradiction. $$ 4(4 y+2)=3 y+2[1-3(1-2 y)] $$

Step-by-Step Solution

Answer

Based on the step by step solution provided, determine the type of equation and the solution for the equation: 4(4y+2) = 3y+2[1-3(1-2y)] Answer: The equation is a conditional equation with the solution y = -12.

Step 1: Distribute and simplify the expression on the right side

To begin, let's simplify the right side of the equation. We need to distribute \(-2\) inside the brackets and then simplify the resulting expression: $$ 4(4y+2)=3y+2[1-3(1-2y)] $$ $$ 4(4y+2)=3y+2(1-3+6y) $$ $$ 4(4y+2)=3y+2(6y-2). $$

Step 2: Distribute and combine like terms

Now, we'll distribute \(4\) to both terms inside the left parenthesis, and \(2\) to both terms inside the right parenthesis, and then combine like terms: $$ 16y+8=3y+12y-4 $$ $$ 16y+8=15y-4. $$

Step 3: Isolate the variable y

Next, we need to get \(y\) by itself on one side of the equation. To do this, subtract \(15y\) from both sides and subtract \(8\) from both sides: $$ 16y+8-15y-8=15y-4-15y-8 $$ $$ y=-12. $$ Since we got a unique value for \(y\), the equation is a conditional equation with the solution \(y=-12\).

Key Concepts

Solving Equations

Understanding how to solve equations is a fundamental part of algebra. Equations are mathematical statements with expressions set equal to each other. Solving an equation means finding the value of the variable that makes this statement true. In the given exercise, we start by simplifying both sides of the equation. This simplification involves distributing constants over terms and combining like terms to make the equation simpler.

• Distribute: Apply the distributive property, which states that multiplication over addition can be expanded, like \( a(b + c) = ab + ac \).
• Combine Like Terms: This involves summing terms that have the same variable or are constant.
• Isolate the Variable: Use basic addition, subtraction, multiplication, and division to get the variable alone on one side of the equation.

The ultimate aim is obtaining a statement where a variable equals a specific number, such as \( y = -12 \), which indicates the solution to the equation.

Algebraic Identities

Algebraic identities are special equations that hold true for all values of the involved variables. Unlike solving a regular equation for a specific solution, recognizing an identity involves showing that both sides of the equation represent the same thing, regardless of the value of the variable.
To identify an identity:

• Simplify both sides: Make sure both sides have been combined and simplified fully.
• Compare both sides: If they simplify to identical expressions involving a variable, it's an identity.

For example, an identity would be something like \( x(x-1) = x^2 - x \), which is true for any real number \( x \). In the exercise, we did not encounter an identity, but understanding them helps in solving various algebra problems.

Contradictions in Algebra

Contradictions in algebra occur when simplifying or solving an equation leads to an untrue statement, like \( 0 = 1 \). This signifies that no value of the variable can satisfy the equation.
To identify a contradiction:

• Follow the same steps for solving an equation.
• Check if simplifying results in a false numeric equality.

Recognizing contradictions is important because it tells us something is wrong or impossible in the setup of the problem. In our exercise, the equation was conditional and not a contradiction, meaning it had a valid solution. However, knowing about contradictions can alert you when no solution exists.