Problem 33 - Elementary Algebra

Question

Solve each of the equations. $$\frac{2(x-1)}{3}=\frac{3(x+2)}{5}$$

Step-by-Step Solution

Answer

The solution is \(x = 28\).

Step 1: Clear the Fractions

To eliminate fractions, multiply both sides of the equation by the least common multiple of the denominators. The denominators are 3 and 5, so the least common multiple is 15. Multiply each side of the equation by 15 to clear the fractions: \[15 \cdot \frac{2(x-1)}{3} = 15 \cdot \frac{3(x+2)}{5}\]

Step 2: Simplify Each Side

Simplify both sides of the equation after multiplying:For the left side:\[15 \frac{2(x-1)}{3} = 5 \cdot 2(x-1) = 10(x-1)\]For the right side:\[15 \cdot \frac{3(x+2)}{5} = 3 \cdot 3(x+2) = 9(x+2)\]

Step 3: Distribute

Apply the distributive property to remove the parentheses:Left side:\[10(x-1) = 10x - 10\]Right side:\[9(x+2) = 9x + 18\]

Step 4: Move Variables to One Side

To simplify, move all terms containing \(x\) to one side and constant terms to the opposite side. Subtract \(9x\) from both sides:\[10x - 9x - 10 = 18\]

Step 5: Combine Like Terms

Combine like terms:\[x - 10 = 18\]

Step 6: Solve for x

Add 10 to both sides to isolate \(x\):\[x = 18 + 10\]\[x = 28\]

Key Concepts

Solving Equations

To solve an algebraic equation, the main goal is to find the value of the unknown variable that makes the equation true. In our example, the equation to solve is \( \frac{2(x-1)}{3} = \frac{3(x+2)}{5} \). The process typically involves the following steps:

• Eliminate any fractions by finding a common denominator and multiplying both sides of the equation by it.
• Use distribution to eliminate parentheses.
• Gather all terms with the unknown variable on one side, and constants on the other.
• Simplify the equation by combining like terms.
• Finally, isolate the variable by undoing any operations affecting it.

By following these steps systematically, you can find the solution to the equation without making mistakes.

Distribution

The distributive property is a key concept in algebra that allows you to simplify expressions by multiplying a single term by all the terms inside a parenthesis. In the equation \( \frac{2(x-1)}{3} = \frac{3(x+2)}{5} \), we see distribution at work.
For the expression \(10(x-1)\), distribution means multiplying 10 by both \(x\) and -1, which gives us \(10x - 10\). Similarly, for \(9(x+2)\), it involves multiplying 9 by both \(x\) and 2, resulting in \(9x + 18\).

• Distribution is helpful for removing parentheses and simplifying expressions.
• It ensures that every term inside the parentheses is accounted for in the math operation.
• Helps in setting up equations that are easier to manipulate and solve.

Understanding distribution is crucial, as it lays the groundwork for more complex algebraic operations.

Fractions in Equations

Fractions can make equations look complicated, but they can be simplified by clearing them through multiplication. In our original equation \( \frac{2(x-1)}{3} = \frac{3(x+2)}{5} \), the fraction aspect is tackled initially.

• The least common multiple of the denominators (3 and 5 in this case) is found, which is 15.
• All terms are multiplied by 15, which cancels out the denominators and results in whole numbers, simplifying the equation.
• This step transforms the equation into one that is more manageable without fractions, reducing chances of errors.

Clearing fractions makes it smoother to handle equations, especially as the math progresses, allowing a clear path to solve for the unknown variable.

Like Terms in Algebra

Combining like terms is an essential skill in algebra that makes solving equations simpler and clearer. Like terms are terms that have the same variable raised to the same power, making them combinable. In our problem,
After distributing, you end up with expressions like \(10x - 10\) and \(9x + 18\). When solving \(10x - 9x - 10 = 18\), the terms \(10x\) and \(-9x\) can be combined since they both contain the variable \(x\). Therefore, you can simplify them as \(x\).

• Identify terms with the same variable and exponent.
• Add or subtract these terms to simplify the equation further.
• Helps in isolating the variable, paving the way to finding the solution.

Combining like terms efficiently reduces complexity and makes your solution path clearer, helping you stay organized as you solve the equation step-by-step.