Problem 4 - Calculus for Biology and Medicine

Question

Solve the following equations: (a) \(|2 x+4|=|5 x-2|\) (b) \(|1+2 u|=|5-u|\) (c) \(\left|4+\frac{1}{2}\right|=\left|\frac{3}{2} t-2\right|\) (d) \(|2 s-6|=|3-s|\)

Step-by-Step Solution

Answer

(a) \(x = 2, -\frac{2}{7}\); (b) \(u = \frac{4}{3}, -6\); (c) \(t = \frac{13}{3}, -\frac{5}{3}\); (d) \(s = 3\).

Step 1: Understanding Absolute Values

Absolute values represent the magnitude of a number, disregarding its sign. For two absolute expressions to be equal, either they both are equal or their negatives are equal.

Step 2: Equation (a) \\(|2x + 4| = |5x - 2|)

Set the equation in two cases. First, solve for \(2x + 4 = 5x - 2\). Then solve for \(2x + 4 = -(5x - 2)\).

Step 3: Case 1: \(2x + 4 = 5x - 2\)

Rearrange to isolate \(x\):\[2x + 4 = 5x - 2\]\[4 + 2 = 5x - 2x\]\[6 = 3x\]\[x = 2\].

Step 4: Case 2: \(2x + 4 = -(5x - 2)\)

Distribute the negative and solve for \(x\):\[2x + 4 = -5x + 2\]\[2x + 5x = 2 - 4\]\[7x = -2\]\[x = -\frac{2}{7}\].

Step 5: Solution for (a)

The solutions are \(x = 2\) and \(x = -\frac{2}{7}\).

Step 6: Equation (b) \\(|1+2u| = |5-u|)

Set the equation in two cases:1) \(1 + 2u = 5 - u\)2) \(1 + 2u = -(5 - u)\).

Step 7: Case 1: \(1 + 2u = 5 - u\)

Rearrange to isolate \(u\):\[2u + u = 5 - 1\]\[3u = 4\]\[u = \frac{4}{3}\].

Step 8: Case 2: \(1 + 2u = -(5 - u)\)

Distribute the negative and solve for \(u\):\[1 + 2u = -5 + u\]\[2u - u = -5 - 1\]\[u = -6\].

Step 9: Solution for (b)

The solutions are \(u = \frac{4}{3}\) and \(u = -6\).

Step 10: Equation (c) \\(|4 + \frac{1}{2}| = |\frac{3}{2}t - 2|)

Simplify and consider the two cases:1) \(4.5 = \frac{3}{2}t - 2\)2) \(4.5 = - (\frac{3}{2}t - 2)\).

Step 11: Case 1: \(4.5 = \frac{3}{2}t - 2\)

Solve by isolating \(t\):\[4.5 + 2 = \frac{3}{2}t\]\[6.5 = \frac{3}{2}t\]\[t = \frac{6.5 \times 2}{3} = \frac{13}{3}\].

Step 12: Case 2: \(4.5 = - \left(\frac{3}{2}t - 2\right)\)

Simplify and solve for \(t\):\[4.5 = -\frac{3}{2}t + 2\]\[4.5 - 2 = -\frac{3}{2}t\]\[2.5 = -\frac{3}{2}t\]\[t = \frac{2.5 \times -2}{3} = -\frac{5}{3}\].

Step 13: Solution for (c)

The solutions are \(t = \frac{13}{3}\) and \(t = -\frac{5}{3}\).

Step 14: Equation (d) \\(|2s - 6| = |3 - s|)

Follow similar steps to solve:1) \(2s - 6 = 3 - s\)2) \(2s - 6 = -(3 - s)\).

Step 15: Case 1: \(2s - 6 = 3 - s\)

Solve by isolating \(s\):\[2s + s = 3 + 6\]\[3s = 9\]\[s = 3\].

Step 16: Case 2: \(2s - 6 = -(3 - s)\)

Distribute and solve:\[2s - 6 = -3 + s\]\[2s - s = -3 + 6\]\[s = 3\].

Step 17: Solution for (d)

The solution is \(s = 3\).

Key Concepts

Algebraic Solutions

Algebraic solutions are a method of solving equations using algebraic manipulation to find the value(s) of the variable(s). In the context of absolute-value equations, algebraic solutions involve setting up different possible scenarios (cases) due to the nature of absolute values. When we have an equation like \(|A| = |B|\), it indicates that the expressions within the absolute values can be equal directly, or they can be negatives of each other. Thus, algebraic solutions require solving for both of these cases. We'll:

• Set up Case 1: The expressions inside the absolute values are equal, i.e., \(A = B\).
• Set up Case 2: One expression is the negative of the other, i.e., \(A = -B\).

This dual approach is necessary because the absolute value represents a non-directional distance from zero, which means the input could be on either side of the number line. For each case, you'll rearrange the equation to isolate the variable, and solve accordingly to find potential solutions.

Isolation of Variables

The isolation of variables is a critical step in solving any algebraic equation, including absolute-value equations. It involves rearranging the equation so that the variable (like \(x, u, t, \) or \(s\)) stands alone on one side of the equation. This process allows us to directly find the value of the variable. Here's how it works:

• Combine like terms to simplify the equation.
• Add or subtract terms to get all instances of the variable on one side of the equation.
• Factor or divide to solve for the variable, if necessary.

For example, in the equation \(2x + 4 = 5x - 2\), you would first move the term with the variable \(5x\) from one side to another by subtracting it from both sides. This eventually results in the variable being isolated. By performing such operations, each of these absolute-value equations can be solved systematically, leading to the identification of the unknown values. By isolating the variables step by step, students can achieve a deeper understanding of the equation-solving process.

Educational Math Problems

Educational math problems often aim to enhance learning by engaging students with practical scenarios that require the application of mathematical concepts. Absolute-value equations like those presented in the exercise are an example of such problems. They help students develop their analytical skills and improve their understanding of basic algebraic principles. Here are some ways these problems are educational:

• They challenge students to identify different cases and think logically about each.
• They reinforce the understanding of absolute values as expressing magnitude irrespective of sign.
• They require a grasp of algebraic techniques such as distribution, factoring, and simplifying expressions.

By working through problems involving multiple steps and cases, students learn to organize their work and think critically. Moreover, these math problems lay the foundation for more advanced algebra topics, like solving systems of equations or quadratic equations. This progressive learning helps build both confidence and competence in mathematical problem-solving.