Question

Solve and check each equation. Treat the constants in these equations as exact numbers. Leave your answers in fractional, rather than decimal, form. Simple Fractional Equations. $$5+\frac{7 x}{3}=\frac{x}{2}$$

Short answer

The solution to the equation is \( x = -\frac{30}{17} \).

Step-by-step solution

Clear Fractions

To clear fractions, find a common denominator for all the fractional terms in the equation. In this case, the common denominator is 6 (the least common multiple of 3 and 2). Multiply every term of the equation by 6 to eliminate the fractions.

Distribute and Simplify

After multiplying each term by the common denominator, distribute 6 to each term and simplify the resulting equation by combining any like terms.

Isolate x

Rearrange the resulting equation to isolate terms with an x on one side and the constant terms on the other. Do this by subtracting or adding terms from both sides of the equation, as necessary.

Solve for x

Solve the resulting linear equation for x by isolating x on one side.

Check the Solution

Substitute the value of x back into the original equation to ensure both sides of the equation are equal after the substitution.

Key concepts

Clearing Fractions in Equations

The process of clearing fractions in equations is a critical skill in algebra. This step involves eliminating all fractional terms from an equation to simplify solving. To accomplish this, we find a common denominator for all the fractions involved and multiply each term in the equation by this number.

For instance, in an equation like \(5+\frac{7x}{3}=\frac{x}{2}\), the common denominator would be 6, as it is the least common multiple of 3 and 2. By multiplying each term by 6, we get \(6\cdot 5 + 6\cdot \frac{7x}{3} = 6\cdot \frac{x}{2}\), which simplifies to \(30 + 14x = 3x\). This step transforms a fractional equation into a much simpler linear equation.

Finding Common Denominators

Finding common denominators is a fundamental step in both simplifying expressions and solving equations that involve fractions. The common denominator is the least common multiple (LCM) of the denominators in the equation. Determining the LCM allows us to combine fractions or clear them from an equation.

In our example, the denominators are 3 and 2. The LCM of 3 and 2 is 6, therefore, it's the common denominator. By focusing on this aspect, students can avoid mistakes and streamline the solving process. It's also a foundational concept that will aid in more advanced math, such as adding and subtracting fractions or working with complex rational expressions.

Linear Equation Solving

Linear equation solving is all about finding the value of the unknown variable that makes the equation true. Once we've cleared the fractions, solving the linear equation becomes a matter of isolating the variable on one side of the equation. This involves performing operations such as addition, subtraction, multiplication, or division on both sides of the equation to maintain equality.

In our cleared equation \(30 + 14x = 3x\), we can solve for \(x\) by first getting all the \(x\)-terms on one side, resulting in \(14x - 3x = -30\). Simplifying further, we get \(11x = -30\), and dividing both sides by 11 yields \(x = -\frac{30}{11}\). It is important to go through these steps methodically to avoid errors and ensure a thorough understanding of the algebraic principles involved.

Checking Equation Solutions

Checking equation solutions is just as important as solving the equation itself. This step verifies that our solution is correct and that it satisfies the original equation. To check, we substitute the found value of the variable back into the original equation and see if the expression balances.

For example, with the solution \(x = -\frac{30}{11}\), we substitute \(x\) in the original equation: \(5 + \frac{7(-\frac{30}{11})}{3} = \frac{-\frac{30}{11}}{2}\). Simplifying the left side yields \(5 - \frac{70}{33}\), and the right side simplifies to \( -\frac{15}{11}\). Once we find a common denominator and perform the arithmetic, we'll see if the two sides are equivalent and, thus, confirm the solution is correct.

This final check helps students reinforce their understanding of the concepts and ensures their confidence in the solution's accuracy.