Problem 51 - 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. $$ 3[4-2(y+2)]=2 y-4[1+2(1+y)] $$

Step-by-Step Solution

Answer

Question: Determine whether the given equation is a conditional equation, an identity, or a contradiction, and then solve it: 3[4-2(y+2)]=2y-4[1+2(1+y)] Answer: The given equation is a conditional equation, and the solution is y = -6.

Step 1: Expand and simplify the equation

To begin, let's distribute the numbers in the brackets using the distributive property: $$ 3[4-2(y+2)]=2 y-4[1+2(1+y)] $$ $$ 3(4)-3(2)(y+2)=2y-4(1)-4(2)(1+y) $$ Now, work on the parentheses and simplify the equation: $$ 12 - 6(y+2) = 2y - 4 - 8(1+y) $$

Step 2: Continue simplifying both sides of the equation

Now, we'll continue to simplify the equation by distributing and combining like terms: $$ 12 - 6y - 12 = 2y - 4 - 8 - 8y $$ Combine the constants and the 'y' terms: $$ -6y = 2y - 8y - 4 - 8 $$

Step 3: Solve for y

Add 8y to both sides of the equation to isolate the 'y' terms on one side: $$ 2y = -12 $$ Now, divide both sides by 2 to solve for y: $$ y = -6 $$ Since we have found a specific value for y (-6) that makes the equation true, the equation is a conditional equation. The final solution is: $$ y = -6 $$ This is a conditional equation as it takes specific values for y to make it true. In this case, y = -6.

Key Concepts

Algebraic Identities

Algebraic identities are equations that hold true for all values of the variables involved. They play a fundamental role in simplifying algebraic expressions and solving equations. A common example is the distributive property, which states that for any real numbers a, b, and c, the identity \( a(b + c) = ab + ac \) always holds. In our exercise, when expanding the brackets, we implicitly used this property. However, the equation we are examining is a conditional equation rather than an identity, because it doesn't hold true for all values of \( y \), but for a specific value that satisfies the equation, in this case, \( y = -6 \). If we were unable to find a specific value for \( y \) that made the equation true, or if every possible value of \( y \) satisfied the equation, then we would classify it as an identity or a contradiction, respectively.

Contradiction Equations

Contradiction equations are those that have no solution, meaning no value for the variable will satisfy the equation. These types of equations lead to an absurd statement such as \( 4 = 5 \) which we know is never true. During the course of solving an equation, if we end up with a contradictory statement, then we can conclude that the original equation has no solutions. Contradictory equations can look deceptively normal at first glance, and it's only upon simplifying and attempting to resolve them that their true nature becomes apparent. However, in our exercise, since we found that \( y = -6 \), it indicates that the equation isn't a contradiction after all, because there is at least one value that satisfies the equation.

Distributive Property

The distributive property is a cornerstone in algebra that allows us to multiply a single term by each term within a parenthesis. Formally, it can be expressed as \( a(b + c) = ab + ac \). This property is essential because it lets us simplify expressions and solve equations by expanding brackets, as seen in our exercise.

In the initial step of solving the textbook problem, we used the distributive property to expand \( 3[4-2(y+2)] \) and \( -4[1+2(1+y)] \), which is crucial for further simplification. Throughout the solving process, the distributive property helps in breaking down expressions into more manageable pieces, thus leading to the eventual isolation and solving for \( y \). Understanding and applying the distributive property correctly is vital in working through algebraic equations efficiently.