Question

Solve the equation by cross multiplying. Check your solution(s). $$\frac{x}{2 x+7}=\frac{x-5}{x-1}$$

Short answer

The solution for the given equation \(\frac{x}{2 x+7} = \frac{x-5}{x-1}\) is \(x = 7\). Other solutions are extraneous and do not satisfy the original equation.

Step-by-step solution

Cross Multiplication

Cross-multiply the fractions to get rid of the denominators. Multiply the denominator of the right-side fraction (\(x - 1\)) with the numerator of the left-side fraction (\(x\)) and the denominator of the left-side fraction (\(2x + 7\)) with the numerator of the right-side fraction (\(x - 5\)). This results in \((x)(x - 1) = (x -5)(2x + 7)\)

Simplifying

Simplify each side of the equation by carrying out the multiplication. This gives \(x^2 - x = 2x^2 + 7x - 10x - 35\) which simplifies to \(x^2 - x = 2x^2 - 3x - 35\)

Rearranging To Make A Quadratic Equation

Move all terms in the equation to one side to create a standard quadratic equation. We subtract \(x^2 - x\) from both sides resulting in \(0 = x^2 - 4x -35\)

Solving The Quadratic Equation

Now solve the quadratic equation. This equation can be factored to yield \((x -7)(x + 5) = 0\). Now, set each factor equal to zero and solve for \(x\). Hence, \(x = 7\) or \(x = -5\)

Checking The Solution

Replace \(x\) in the original equation by solutions obtained. If the left side equals the right side, then the solutions are valid. However, \(x = -5\) will make the denominator of the first fraction zero. Therefore, we discard \(x = -5\) and the only valid solution is \(x = 7\)

Key concepts

Cross Multiplication

Cross multiplication is a powerful technique that is often used to solve equations involving fractions. The basic idea is to eliminate the fractions by multiplying the denominator on one side of the equation by the numerator on the opposite side.
For example, with an equation like \( \frac{x}{2x+7} = \frac{x-5}{x-1} \), cross multiplication involves:

• Taking the numerator of the first fraction \(x\) and multiplying it by the denominator of the second fraction \(x-1\).
• Similarly, taking the numerator of the second fraction \(x-5\) and multiplying it by the denominator of the first fraction \(2x+7\).

This results in a new equation \( (x)(x - 1) = (x -5)(2x + 7) \), which is free from fractions. Now you can simplify and solve the equation more easily.
Cross multiplication is especially useful in algebra because it simplifies the manipulation of the equation. Remember, always check for extraneous solutions that might arise from values that make the denominator zero.

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). In our exercise, after simplifying and rearranging, we end up with the equation \( x^2 - 4x - 35 = 0 \).

This standard form helps us to apply different solving techniques such as factoring, completing the square, or using the quadratic formula. The key characteristic of a quadratic equation is that it involves an \( x^2 \) term, making the graph a parabola when plotted.

• The coefficient \(a\) influences the width and the direction of the parabola (upward or downward).
• The coefficient \(b\) affects the position of the axis of symmetry.
• The coefficient \(c\) is the y-intercept of the parabola.

Quadratic equations often have two solutions because they can be represented as the product of two binomials set to zero. These solutions are the points where the parabola intersects the x-axis.Make sure to verify any solutions, as some might not be valid due to restrictions like undefined fractions in the original equation.

Factoring

Factoring is a method used to simplify quadratic equations by expressing them as a product of simpler terms known as factors. Once you have a standard quadratic equation like \( x^2 - 4x - 35 = 0 \), the goal is to find two numbers that multiply to \( c \) (the constant term) and add to \( b \) (the coefficient of \(x\)).
In our example, we need to find two numbers that multiply to \(-35\) and add to \(-4\). These numbers are \(-7\) and \(+5\). Thus, the quadratic can be factored as \((x - 7)(x + 5) = 0\).

From here, you apply the zero product property, which states that if a product of two factors equals zero, then at least one of the factors must be zero. Thus, set each factor equal to zero:

• \(x - 7 = 0\), giving \(x = 7\)
• \(x + 5 = 0\), giving \(x = -5\)

These are the potential solutions to the quadratic equation. However, always substitute back into the original equation to check, as with this exercise, \(x = -5\) turns out to be invalid due to a zero denominator. Therefore, \(x = 7\) is the only valid solution.