Solving Word Problems with Linear Equations

1. The Art of Translation: From English to Math

The biggest challenge in word problems is often not the math itself, but correctly translating the problem's language into an equation. Your background in computer science involves translating logic into code; this is a very similar process of translating human language into the precise language of mathematics.

A systematic approach is key. A great resource from Purplemath outlines four essential steps.

The four steps are:

1. Read the entire problem first to get the context.
2. Work in an organized manner, clearly defining what your variables represent (e.g., "let x = the cost of a pen").
3. Look for keywords that indicate mathematical operations.
4. Apply your "real-world" knowledge.

Step 3 is where most of the translation work happens. Certain words and phrases are direct signals for specific operations.

To see this in action, let's watch a video that demonstrates how to convert simple sentences into one-variable equations.

As both the video and the Purplemath article point out, be very careful with the phrase "less than". For example, "6 less than 5 times a number" is written as $5x-6$, not $6-5x$. The order is reversed from how we read it in English. A good way to remember this is to think of a real-world example: if your salary is "₹10,000 less than your manager's," you calculate your salary by taking the manager's salary and subtracting 10,000 from it.

2. Solving Systems of Two Linear Equations

Sometimes, a problem involves two unknown quantities and gives you two distinct pieces of information. This leads to a system of two linear equations with two variables (commonly $x$ and $y$). To solve for the variables, you need to use both equations together.

For the CAT, the two most practical and time-efficient methods for solving these systems are Substitution and Elimination.

Let's quickly review the core ideas:

• Substitution Method:

  1. Solve one equation for one variable (e.g., solve for $x$ in terms of $y$).
  2. Substitute this expression into the other equation.
  3. You now have one equation with one variable. Solve it.
  4. Plug the result back into your expression from step 1 to find the second variable.

• Elimination Method:

  1. Multiply one or both equations by numbers so that the coefficient of one variable is the same (or opposite) in both equations.
  2. Add or subtract the equations to eliminate that variable.
  3. You now have one equation with one variable. Solve it.
  4. Plug the result back into either original equation to find the second variable.

The elimination method is often faster, especially when the equations are neatly aligned.

3. Putting It All Together: CAT-Style Word Problems

Now, let's apply this entire process to a typical word problem you might encounter in the CAT. We'll use an example from the EduRev resource you just reviewed.

Problem:
In 2010, a library contained a total of 11500 books in two categories – fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was a 10% increase in the fiction category while there was a 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

Let's break this down systematically:

1. Identify Unknowns & Assign Variables:
   The core unknowns are the number of books in each category in the base year, 2010.

   • Let $f$ be the number of fiction books in 2010.
   • Let $n$ be the number of non-fiction books in 2010.

2. Translate Sentences into Equations:

   • "In 2010, a library contained a total of 11500 books...":
     $f+n=11500\text{(Equation 1)}$
   • "a 10% increase in the fiction category... a 12% increase in the non-fiction category... a total of 12760 books":
     A 10% increase means the new amount is 110% of the original, or $1.10f$. A 12% increase is $1.12n$.
     $1.10f+1.12n=12760\text{(Equation 2)}$

3. Solve the System of Equations:
   Let's use the substitution method. From Equation 1, we can easily express $n$:
   $n=11500-f$
   Now, substitute this into Equation 2:
   $1.10f+1.12(11500-f)=12760$
   Solve for $f$:
   $1.10f+12880-1.12f=12760$
   $-0.02f=12760-12880$
   $-0.02f=-120$
   $f=\frac{-120}{-0.02}=6000$
   So, there were 6000 fiction books in 2010.

4. Answer the Final Question:
   The question asks for the number of fiction books in 2015. This is after the 10% increase.
   Number of fiction books in 2015 = $1.10\times f=1.10\times 6000=6600$.

This step-by-step process—Define, Translate, Solve, and Answer—is your blueprint for success with these problems.

Conclusion

In this lesson, you've built a robust framework for converting word problems into linear equations and solving them. This skill is not just one topic among many; it's a foundational technique that unlocks solutions to problems in percentages, geometry, and many other areas of the CAT syllabus.

Key Takeaways:

• Always follow a structured process: Define your variables clearly, Translate sentences using keywords, Solve the resulting equation(s), and ensure you Answer the specific question asked.
• Be cautious with tricky phrases like "less than," where the mathematical order is reversed.
• For two-variable problems, the Substitution and Elimination methods are your most efficient tools for the CAT.
• When in doubt, you can test your translated expression with simple numbers to see if it makes logical sense, as suggested by the Purplemath resource.

Preview of the Next Lesson:
We will continue our journey through algebra by moving from linear equations (where the variable's highest power is 1) to quadratic equations (where the highest power is 2). You will learn how to solve these equations and use their properties to tackle a new class of problems.