Problem 10 - Elementary Algebra

Question

For the following problems, solve the rational equations. $$ \frac{x+1}{4}=\frac{x-3}{2} $$

Step-by-Step Solution

Answer

Answer: 7

Step 1: Find the common denominator

The given equation has two denominators: 4 and 2. The least common multiple of these two numbers is 4. So, the common denominator is 4.

Step 2: Cross-multiply

Now, we will cross-multiply both sides of the equation to get rid of the fractions. $$ \frac{x+1}{4}=\frac{x-3}{2} $$ Multiply both sides by 4: $$ 4 \cdot \frac{x+1}{4} = 4 \cdot \frac{x-3}{2} $$ On the left side, the 4's cancel out: $$ (x+1) = 4 \cdot \frac{x-3}{2} $$ On the right side, 4 is divided by 2: $$ (x+1) = 2(x-3) $$

Step 3: Distribute and simplify

Now, distribute the 2 on the right side of the equation and simplify. $$ (x+1) = 2x - 6 $$

Step 4: Solve for x

Our next step is to isolate the variable x. First, subtract x from both sides of the equation: $$ 1 = x - 6 $$ Finally, add 6 to both sides of the equation to solve for x: $$ x = 7 $$ So, the solution to the given equation is x = 7.

Key Concepts

Common Denominator

The quest to solve rational equations often begins with finding a common denominator, which is crucial when dealing with fractions. The common denominator is essentially a common ground for all the fractions involved in the equation, allowing us to combine them and move forward in solving the problem. Think of it like this: If you're trying to compare slices of pizza from different sized pizzas, you need to think of them in terms of slices from a pizza of the same size to make a fair comparison.

In the given exercise, the denominators are 4 and 2. To find the least common denominator, you look for the smallest multiple that both denominators share. Since 4 is a multiple of 2, it’s the least common multiple and serves as our common denominator. Working with a common denominator simplifies the equation, turning it into an easier puzzle to solve.

Cross-Multiplication

Cross-multiplication is a method often used to solve equations involving two fractions. It's like creating a bridge between two fractions, allowing them to cross over and eliminate the denominators. To apply cross-multiplication, you multiply the numerator of one fraction by the denominator of the other, and vice versa, thus 'crossing' them.

In our exercise, cross-multiplication comes into play after identifying our common denominator. By multiplying both sides of the equation by the common denominator, we effectively cancel out the denominators on each side, leaving us with a simpler equation without fractions. This process is akin to leveling the playing field so that all the variables and constants can play nicely without the complication of fractions in between.

Distributive Property

Once we've cleared the fractions, the distributive property becomes our next step. It's a property that allows you to 'distribute' or multiply a single term across terms inside parentheses. Imagine handing out apples to a group of friends; you're distributing the apples evenly. Mathematically speaking, if you have 2(x-3), you give the 2 to both x and -3 separately.

In this case, after using cross-multiplication, we get to a point where we need to distribute 2 across the terms within the parentheses on one side of our equation. To do this, you multiply 2 by x, and then 2 by -3, effectively simplifying the equation even further. This is a fundamental step in breaking down an equation into basic components, making it easier to understand and solve.

Isolate the Variable

The final act of our equation-solving drama is to isolate the variable, which means to get the variable on one side of the equation all by itself. In other words, you want to have your x (or any variable you're solving for) stand alone, like a lone tree in a field, so you can clearly see what it's worth.

To achieve this, you perform operations that 'undo' anything attached to the variable. In our exercise, we have x on both sides of the equation after distributing. To isolate x, we subtract it from both sides to get it alone on one side. Then, we deal with any constants attached to the variable – in this case, adding 6 to both sides to fully isolate x on one side of the equation. With the variable isolated, we now have our solution: the value of x that makes the equation true.