Problem 44 - Algebra A Combined Function

Question

Solve. See Examples 1 through 7 $$ \frac{5(x-1)}{4}=\frac{3(x+1)}{2} $$

Step-by-Step Solution

Answer

The solution is \( x = -11 \).

Step 1: Clear the Fractions

To eliminate the fractions in the equation \( \frac{5(x-1)}{4} = \frac{3(x+1)}{2} \), we can multiply every term by the Least Common Denominator (LCD) of 4 and 2, which is 4. This gives us:\\[\begin{aligned}&4\cdot\frac{5(x-1)}{4} = 4\cdot\frac{3(x+1)}{2} \&5(x-1) = 2\cdot3(x+1)\end{aligned}\]

Step 2: Distribute and Simplify

Next, we will distribute the coefficients across the terms inside the parentheses:\\[\begin{aligned}5(x-1) &= 5x - 5 \2\cdot3(x+1) &= 6(x+1) = 6x + 6\end{aligned}\]So the equation becomes: \( 5x - 5 = 6x + 6 \).

Step 3: Isolate Variables on One Side

To isolate the variable \( x \), subtract \( 5x \) from both sides of the equation to bring all \( x \) terms to one side:\[\begin{aligned}5x - 5 - 5x &= 6x + 6 - 5x \-5 &= x + 6\end{aligned}\]

Step 4: Solve for x

Now, subtract 6 from both sides to solve for \( x \):\[\begin{aligned}-5 - 6 &= x + 6 - 6 \-11 &= x\end{aligned}\]Thus, \( x = -11 \).

Key Concepts

Solving Equations

Solving equations in algebra involves finding the unknown variable's value that makes the equation true. In our exercise, we deal with an equation featuring fractions. Solving these types of equations typically requires eliminating the fractions to simplify the process. The equation is:\[ \frac{5(x-1)}{4} = \frac{3(x+1)}{2} \]To solve it, we must manipulate both sides systematically to find the value of the variable. Each manipulation should keep the equation balanced and bring us closer to isolating the variable. Keep this in mind:

• Clear fractions by using the least common denominator.
• Use properties like the distributive property strategically.
• Gradually isolate the variable.

This approach will help direct you through various algebraic equations systematically. Always ensure your steps maintain the equation's integrity until you reach the potential solution.

Least Common Denominator

The least common denominator (LCD) is a crucial step in simplifying equations with fractions like:\[ \frac{5(x-1)}{4} = \frac{3(x+1)}{2} \]The LCD is the smallest multiple shared by the denominators of all fractions in the equation. In this case, the denominators are 4 and 2, and the smallest shared multiple is 4. Why is this important?

• The LCD allows us to clear fractions entirely by multiplying both sides of the equation by this number.
• This transforms our equation into a simpler form with entirely integer coefficients, making further operations more straightforward.

By multiplying every term by the LCD, we move to a simpler equation:\[ 5(x-1) = 2 \cdot 3(x+1) \]Now, there are no fractions to worry about, enabling us to use other algebraic techniques more effectively.

Distributive Property

The distributive property is a fundamental algebraic property used to simplify and solve equations involving products within parentheses. It states that:\[ a(b + c) = ab + ac \]In our example:\[ 5(x-1) = 5x - 5 \]Here, the distributive property is applied to "push" the multiplier across the terms in the parentheses. This allows:

• The equation to be reduced to a simpler form for easier manipulation.
• Each term to be rewritten in a way that facilitates further solution steps, like isolating variables later on.

The same method is applied to:\[ 6(x+1) \]Which becomes \( 6x + 6 \). Ensuring you use the distributive property correctly is key to maintaining the equation's balance and moving towards isolating the variable.

Isolating Variables

Isolating the variable is one of the final steps in solving equations. Here, the goal is to get the unknown variable, such as \( x \), alone on one side of the equation. Once we applied the distributive property, our equation was:\[ 5x - 5 = 6x + 6 \]To isolate \( x \), follow these steps:

• First, subtract \( 5x \) from both sides to collect the \( x \) terms on one side.
• After simplification, you'll have \(-5 = x + 6\).
• Finally, subtract 6 from both sides to completely isolate \( x \).

This gives us \(-11 = x\), meaning our solution is \( x = -11 \). The key to this process is performing the same operations on both sides of the equation to keep it balanced.