Problem 80 - Elementary Algebra Within Reach

Question

Solve the equation and check your solution. (Some of the equations have no solution.) $$\frac{x+1}{6}=\frac{3 x}{10}$$

Step-by-Step Solution

Answer

The solution for the equation \( \frac{x+1}{6} = \frac{3x}{10}\) is \(x = \frac{5}{4}\)

Step 1: Cross Multiplication

In this step, we perform cross multiplication to get rid of fractions. With cross multiplication, the equation \(\frac{x+1}{6} = \frac{3x}{10}\) becomes \(10*(x+1) = 6*3x\), which simplifies to \(10x+10 = 18x\).

Step 2: Solving for x

Now, by transferring 10x to the other side, the equation becomes: \(10 = 18x - 10x\), which further simplifies to \(10 = 8x\). Divide both sides by 8, we get \(x = \frac{10}{8}\), simplifying that further we get \(x = \frac{5}{4}\).

Step 3: Checking the solution

Substitute the obtained value of the variable \(x = \frac{5}{4}\) back into the original equation. Let’s check if both sides are equal: \(\frac{5/4+1}{6} = \frac{3*5/4}{10}\). Both sides of the equation simplify to \(\frac{1}{2}\), verifying that our solution is correct.

Key Concepts

Equation Solving

Equation solving is a fundamental skill in algebra that involves finding the value of the variable that makes the equation true. In our example, we have the equation \(\frac{x+1}{6} = \frac{3x}{10}\). This involves manipulating the equation to isolate the variable \(x\) on one side. When solving equations, several operations may be used to simplify and to combine like terms. Here, we used cross multiplication, a technique that allows us to eliminate the fractions and simplify the process of solving the equation. By performing these operations step by step, we maintain the balance of the equation while inching closer to our solution.

Cross Multiplication

Cross multiplication is a technique used to clear fractions when working with rational equations like \(\frac{x+1}{6} = \frac{3x}{10}\). The process involves multiplying each side of the equation by the denominator of the other side.

• Write the expression as two fractions equal to each other.
• Multiply the numerator of the first fraction by the denominator of the second.
• Multiply the numerator of the second fraction by the denominator of the first.

This results in the equation \(10(x+1) = 6 \cdot 3x\), which simplifies the equation to a linear form without fractions, making it easier to solve for \(x\). Cross multiplication works because it maintains the equality of the equation.

Verification of Solutions

Verification of solutions is an essential step after solving an equation to ensure that the solution is correct. This involves substituting the found value back into the original equation to see if it satisfies the equation. For the example provided, after calculating \(x = \frac{5}{4}\), we substitute this back into the original equation \(\frac{x+1}{6} = \frac{3x}{10}\) to verify:

• Calculate \(\frac{5/4 + 1}{6}\), which results in \(\frac{1}{2}\).
• Calculate \(\frac{3 \cdot 5/4}{10}\), which also results in \(\frac{1}{2}\).

Both sides equate to \(\frac{1}{2}\), confirming that \(x = \frac{5}{4}\) is indeed the correct solution. Verification is vital as it provides confidence in the accuracy of the solution.