Problem 6 - College Algebra

Question

When \(80 \%\) of a number is added to the number, the result is 252 . What is the number? \(70 \%\) of what number is \(224 ?\)

Step-by-Step Solution

Answer

The answer to the first problem is 140. The answer to the second problem is 320.

Step 1: Setting up the first equation for the first problem

To solve the first problem, an equation can be set up. If the unknown number is represented as 'x', the equation would be: \(x + 0.8x = 252\).

Step 2: Solve the first equation

Combining like terms in the equation, we get \(1.8x = 252\). To find the value of 'x', divide 252 by 1.8. The result is \(x = 140\).

Step 3: Setting up the second equation for the second problem

To solve the second problem, another equation can be created. If the next unknown number is represented as 'y', then the equation would be \(0.7y = 224\).

Step 4: Solve the second equation

In order to find 'y', divide 224 by 0.7. The result is \(y = 320\).

Key Concepts

Percentages in Algebra

Understanding percentages in algebra is crucial for interpreting real-world scenarios. To convert a percentage into a decimal, which is needed for algebraic equations, you simply divide by 100. For example, 80% becomes 0.80 and 70% becomes 0.70. This allows us to use these values in equations to represent parts of a whole. In our exercise, when a problem states that '80% of a number is added to the number', we are dealing with the original number, which we can call 'x', and adding to it 80% of 'x' (which is 0.8x). Thus, the equation we constructed is:
\(x + 0.8x = 252\).
It's essential to view percentages as multipliers in the context of algebraic expressions. This perspective aids in setting up equations that accurately reflect the situations described in percentage-related word problems.

Solving Linear Equations

Linear equations form the foundation of algebra and are equations that can be written in the form \(ax + b = c\) with 'a', 'b', and 'c' representing constants and 'x' as the unknown variable we wish to solve for. The goal is to isolate the variable on one side of the equation.
For the problem given, \(x + 0.8x = 252\), we first combine like terms: this means adding up all the terms with the variable 'x'. Here, \(1x + 0.8x\) simplifies to \(1.8x\), so our equation becomes \(1.8x = 252\). To isolate 'x', we divide both sides of the equation by 1.8, which gives us the solution \(x = 140\). The process involves understanding how to combine like terms and manipulate the equation by performing the same operation on both sides until the variable 'x' is alone.

Equation Word Problems

Word problems can be nerve-wracking, but algebra provides the tools to translate them into solvable equations. The key is identifying the mathematical operations described by the words. In problems like our first example, phrases like 'is added to' suggest addition, while 'of' often indicates multiplication, especially in the context of percentages.
The solution begins by defining a variable to represent the unknown quantity in the problem. In the exercise given, we defined 'x' and 'y' for our unknown numbers. Once the equations are established, like \(0.7y = 224\), we can then use algebraic methods to isolate and solve for the variable. Remember to take each word problem step by step: interpret the language to form an equation, set up the proper algebraic structure, and then use algebraic rules to find the solution. With practice, even the most complex word problems become a series of routine, solvable steps.