Problem 17 - Algebra 1

Question

Solve the equation by cross multiplying. $$\frac{5}{x+4}=\frac{5}{3(x+1)}$$

Step-by-Step Solution

Answer

The solution to the equation is x = 0.5.

Step 1: Cross Multiplication

Cross multiplying would be applied on the given proportion. Cross Multiplication is the process of cancelling out denominators with other denominators by multiplying them. The equation would look like this after cross-multiplication: \(5 * 3(x+1) = 5 * (x+4)\)

Step 2: Distribute and Simplification

Distribute the 5 into both terms of the binomial on the right-hand side and the 5*3 (which is 15) into both terms of the binomial on the left-hand side. This results in: 15x + 15 = 5x + 20.

Step 3: Form a Simpler Equation

Make the equation simpler by moving the 5x term on the right side to the left side. As a result, the equation now reads: 15x - 5x = 20 - 15.

Step 4: Solving the Equation

Solving the equation gives: 10x = 5. Therefore, x = 5/10 = 0.5

Key Concepts

Solving Equations

When faced with an equation, the ultimate goal is to find the value of the unknown variable that makes the equation true. Solving equations effectively is a fundamental skill in algebra. In the context of this exercise, you must isolate the variable 'x' to determine its value. Cross multiplication is a method often used to solve equations where the variable of interest is part of a fraction.

To navigate through the solving process, intermediate steps such as simplification and reorganizing terms are needed. For instance, after cross multiplying the given equation, simplification requires combining like terms and isolating the variable, which can sometimes necessitate moving terms from one side of the equation to the other. This reorganization respects the fundamental principle that whatever operation is performed on one side of an equation must also be applied to the other side, to maintain the equality.

Distributive Property

The distributive property is a critical algebraic property that allows us to multiply a single term by each term within a parenthesis. For example, when you have an expression like \(5(x + 4)\), applying the distributive property would involve multiplying the number 5 by each term inside the parenthesis, resulting in \(5x + 20\).

In the equation from the exercise, you distribute \(5\) into \((x+4)\) and \(15\) into \((x+1)\) after cross multiplying. This property is instrumental in simplification, which makes equations easier to solve. Without this property, algebraic expressions would be much more difficult, if not impossible, to streamline into solvable equations.

Linear Equations

Linear equations are the most basic form of algebraic equations and they describe a straight line when plotted on a graph. The standard form of a linear equation is \(Ax + By = C\), where A, B, and C are constants, and 'x' and 'y' are variables. In our exercise, the equation simplifies to a single variable linear equation of the form \(10x = 5\), a straightforward case in which only one variable exists, and therefore, no graph plotting is required.

Once simplified, this linear equation can then easily be solved by isolating 'x', which results in the variable equaling a constant. The solution to our provided exercise, \(x = 0.5\), falls exactly into this category. Linear equations are fundamental as they are the foundation upon which more complex algebraic concepts build.