Solving Linear Equations And Age Problems A Step-by-Step Guide

This comprehensive guide will walk you through the process of solving linear equations, covering a range of examples to solidify your understanding. We'll break down each step, ensuring clarity and accuracy in your problem-solving approach. This guide covers the solution of three mathematical problems. The first one is finding the truth set of a given equation. The second is solving a given equation, and the third is solving a word problem involving ages. Let's dive in!

Question 50: Unveiling the Truth Set of a Linear Equation

To find the truth set of the equation $\frac{2}{3}(3 y-1)-(y+2)=\frac{1}{3}$, our primary goal is to isolate the variable y on one side of the equation. This involves a series of algebraic manipulations, each designed to simplify the equation while maintaining its balance. Let's break down the process step-by-step:

1. Eliminate the Fractions: The presence of fractions often complicates the solving process. To eliminate them, we multiply both sides of the equation by the least common multiple (LCM) of the denominators. In this case, the LCM of 3 and 1 (the implicit denominator of the terms without fractions) is 3. Multiplying both sides by 3, we get:

   $$3 \times \left[\frac{2}{3}(3 y-1)-(y+2)\right]=3 \times \frac{1}{3}$$

   This simplifies to:

   $$2(3y - 1) - 3(y + 2) = 1$$

2. Distribute: Next, we distribute the constants outside the parentheses to the terms inside:

   $$2(3y - 1) - 3(y + 2) = 1$$

   $$6y - 2 - 3y - 6 = 1$$

3. Combine Like Terms: Now, we combine the terms with the variable y and the constant terms separately:

   $$(6y - 3y) + (-2 - 6) = 1$$

   $$3y - 8 = 1$$

4. Isolate the Variable Term: To isolate the term with y, we add 8 to both sides of the equation:

   $$3y - 8 + 8 = 1 + 8$$

   $$3y = 9$$

5. Solve for y: Finally, we divide both sides by 3 to solve for y:

   $$\frac{3y}{3} = \frac{9}{3}$$

   $$y = 3$$

Therefore, the truth set of the equation is {3}. This means that when y is equal to 3, the equation holds true. To verify this, we can substitute y = 3 back into the original equation:

$$\frac{2}{3}(3(3)-1)-(3+2)=\frac{1}{3}$$

$$\frac{2}{3}(9-1)-5=\frac{1}{3}$$

$$\frac{2}{3}(8)-5=\frac{1}{3}$$

$$\frac{16}{3}-5=\frac{1}{3}$$

$$\frac{16}{3}-\frac{15}{3}=\frac{1}{3}$$

$$\frac{1}{3}=\frac{1}{3}$$

The equation holds true, confirming our solution. Finding the truth set of an equation is a fundamental concept in algebra, and mastering these steps is crucial for solving more complex problems.

Question 51: Solving Linear Equations with Multiple Steps

Our next challenge is to solve the equation $5(a-5)-\frac{1}{2}(2 a+6)=4$. This equation involves parentheses and fractions, requiring us to apply the distributive property and eliminate fractions to isolate the variable a. Let's break down the solution process:

1. Distribute: Begin by distributing the constants outside the parentheses to the terms inside:

   $$5(a - 5) - \frac{1}{2}(2a + 6) = 4$$

   $$5a - 25 - a - 3 = 4$$

2. Combine Like Terms: Combine the terms with the variable a and the constant terms separately:

   $$(5a - a) + (-25 - 3) = 4$$

   $$4a - 28 = 4$$

3. Isolate the Variable Term: To isolate the term with a, add 28 to both sides of the equation:

   $$4a - 28 + 28 = 4 + 28$$

   $$4a = 32$$

4. Solve for a: Divide both sides by 4 to solve for a:

   $$\frac{4a}{4} = \frac{32}{4}$$

   $$a = 8$$

Therefore, the solution to the equation is a = 8. We can verify this by substituting a = 8 back into the original equation:

$$5(8-5)-\frac{1}{2}(2 (8)+6)=4$$

$$5(3)-\frac{1}{2}(16+6)=4$$

$$15-\frac{1}{2}(22)=4$$

$$15-11=4$$

$$4=4$$

The equation holds true, confirming our solution. Solving linear equations like this requires careful attention to detail and a systematic approach. Mastering these steps will enable you to tackle a wide range of algebraic problems.

Question 52: Tackling Word Problems Involving Ages

Now, let's tackle a word problem. This requires translating the given information into a mathematical equation and then solving it. The problem states: "The sum of the ages of two brothers Kofi and Kweku is 35. Kofi's age is..." The problem is incomplete, but let's assume the full question is: "The sum of the ages of two brothers Kofi and Kweku is 35. Kofi's age is 4 years more than twice Kweku's age. Find their ages."

To solve this problem, we'll use variables to represent the unknown ages and then form equations based on the given information:

1. Define Variables: Let Kofi's age be k and Kweku's age be q.

2. Formulate Equations: Translate the given information into equations:

   • "The sum of the ages of two brothers Kofi and Kweku is 35":

     $$k + q = 35$$

   • "Kofi's age is 4 years more than twice Kweku's age":

     $$k = 2q + 4$$

3. Solve the System of Equations: We now have a system of two equations with two variables. We can use substitution or elimination to solve for k and q. Let's use substitution. Since we have an expression for k in terms of q, we can substitute this into the first equation:

   $$(2q + 4) + q = 35$$

4. Solve for q: Combine like terms and solve for q:

   $$3q + 4 = 35$$

   $$3q = 31$$

   $$q = \frac{31}{3}$$

5. Solve for k: Substitute the value of q back into either equation to solve for k. Let's use the second equation:

   $$k = 2\left(\frac{31}{3}\right) + 4$$

   $$k = \frac{62}{3} + \frac{12}{3}$$

   $$k = \frac{74}{3}$$

So, Kweku's age is $\frac{31}{3}$ years, and Kofi's age is $\frac{74}{3}$ years. However, ages are usually whole numbers. Let's consider a slightly different version of the problem to make the answer more realistic. Assume Kofi's age is 4 more than Kweku's age. So, the equations would be:

• $$k + q = 35$$

• $$k = q + 4$$

Substituting the second equation into the first:

$$(q + 4) + q = 35$$

$$2q + 4 = 35$$

$$2q = 31$$

$$q = 15.5$$

And,

$$k = 15.5 + 4 = 19.5$$

Again, we get non-integer ages. Let's try a final variation: Suppose Kofi's age is 4 years more than twice Kweku's age:

• $$k + q = 35$$

• $$k = 2q + 4$$

Substituting the second equation into the first:

$$(2q + 4) + q = 35$$

$$3q + 4 = 35$$

$$3q = 31$$

$$q = \frac{31}{3}$$

This still leads to non-integer solutions. The original problem statement seems to have some inconsistencies leading to non-integer solutions which are not practical for age problems. However, solving word problems requires careful reading, translating the information into equations, and then using algebraic techniques to find the solution.

Conclusion: Mastering Linear Equations

In this guide, we've explored various aspects of solving linear equations, from finding truth sets to tackling word problems. Each example highlights the importance of a systematic approach, careful algebraic manipulation, and verification of the solution. By mastering these techniques, you'll be well-equipped to solve a wide range of mathematical problems. Remember, practice is key to success in mathematics, so continue to challenge yourself with new problems and solidify your understanding.