Problem 31 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check your solutions. $$ \frac{x}{2}=\frac{5}{x+3} $$

Step-by-Step Solution

Answer

The solutions to the equation \(\frac{x}{2}=\frac{5}{x+3}\) are \(x=1\) and \(x=-5\).

Step 1: Clear the Fraction

To clear the fraction, multiply through by \(2(x+3)\), which is the product of denominators. By doing so, you get \(2x(x+3)=5*2\), which simplifies to \(2x^2+6x=10\).

Step 2: Re-arrange the equation

Re-arrange the equation to form a quadratic equation, that is, all terms in the left-side and zero on the right side. Subtract 10 from both sides, you get \(2x^2+6x-10=0\).

Step 3: Solve the quadratic equation

Factor the quadratic expression, \(2(x-1)(x+5)=0\). Setting each factor to zero and solving for x, we get \(x=1\) from \(x-1=0\) and \(x=-5\) from \(x+5=0\).

Step 4: Check the solutions

Substitute the found x-values individually on the original equation to ensure our original equation is defined. Substituting \(x=1\) and \(x=-5\) into the original equation, we find that both make the equation hold, thus \(x=1\) and \(x=-5\) are the solutions.

Key Concepts

Clearing Fractions

When faced with an equation involving fractions, such as \(\frac{x}{2} = \frac{5}{x+3}\), the first step to finding a solution is typically to eliminate the fractions. This process is called clearing fractions, and it helps simplify the equation to make it easier to solve. The key to clearing fractions is to find a common denominator for all the fractions in the equation. Once identified, each term in the equation is multiplied by this common denominator. This step clears the fractions and transforms the equation into a simpler, fraction-free form. In our exercise, the denominators were 2 and \(x + 3\). The product of these, \(2(x + 3)\), is used as the common denominator. Multiplying each term by this expression effectively removes the fractions:

• Multiply \(\frac{x}{2}\) by \(2(x+3)\) to get \(x(x+3)\)
• Multiply \(\frac{5}{x+3}\) by \(2(x+3)\) to get \(5 \times 2\)

By doing this, we transform the equation into \(2x^2 + 6x = 10\). From here, we can focus on solving a quadratic equation, as the fractions have been cleared.

Factoring Quadratics

In the revised equation, \(2x^2 + 6x - 10 = 0\), our task becomes solving the quadratic by factoring. Factoring quadratics involves rewriting the polynomial as a product of two simpler expressions. This technique is powerful because it breaks down the problem into smaller, more manageable pieces and often allows us to find solutions more easily.For the quadratic \(2x^2 + 6x - 10\), we look for factors of the whole expression. The most crucial observation is to find two numbers that multiply to give the product of the coefficient of \(x^2\) and the constant term (number without \(x\)), and that also add to give the coefficient of \(x\). Once identified, the equation can be expressed as:

• \(2(x-1)(x+5)=0\)

Solving by setting each factor equal to zero gives:

• \(x - 1 = 0\) → \(x = 1\)
• \(x + 5 = 0\) → \(x = -5\)

Factoring not only helps solve for \(x\), but also confirms that there may be multiple solutions, which is often the case when dealing with quadratics.

Checking Solutions

Once solutions have been determined by factoring, it is essential to verify that they satisfy the original equation. This step, known as checking solutions, ensures that no mistakes were made during the calculation and that the solutions are valid within the context of the original problem. For instance, from our example, we need to check if \(x = 1\) and \(x = -5\) satisfy the original equation \(\frac{x}{2} = \frac{5}{x+3}\). Substituting each value back into the original equation helps confirm their validity:

• For \(x = 1\), substitute into the left: \(\frac{1}{2}\); into the right: \(\frac{5}{4}\)
• For \(x = -5\), substitute into the left: \(-\frac{5}{2}\); into the right: \(-\frac{5}{-2}\)

Both calculations validate the solutions, confirming that both \(x = 1\) and \(x = -5\) are indeed correct. This step underscores the importance of substitution in checking the work done through algebraic manipulations.