Problem 45 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 45-48: Solve the system for \(a\) and \(b\). (Hint: Treat terms such as \(e^{3 x}, \cos x\), and \(\sin x\) as "constant coefficients.") $$ \left\\{\begin{array}{c} a e^{3 x}+b e^{-3 x}=0 \\ a\left(3 e^{3 x}\right)+b\left(-3 e^{-3 x}\right)=e^{3 x} \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(a = \frac{1}{6}\) and \(b = -\frac{1}{6}e^{6x}\).

Step 1: Solve the First Equation for One Variable

Given the system of equations, the first equation is \(a e^{3x} + b e^{-3x} = 0\). Rewriting terms, we can solve for \(a\) in terms of \(b\):\[a e^{3x} = -b e^{-3x}\]Divide both sides by \(e^{3x}\):\[a = -b e^{-6x}\]

Step 2: Substitute into the Second Equation

Take the expression for \(a\) from Step 1 and substitute it into the second equation of the system: \[a(3e^{3x}) + b(-3e^{-3x}) = e^{3x}\]Substituting the expression for \(a\) yields:\[-b e^{-6x}(3e^{3x}) + b(-3e^{-3x}) = e^{3x}\]

Step 3: Simplify the Substituted Equation

Simplify the expression from Step 2:\[-3b e^{-3x} + (-3b e^{-3x}) = e^{3x}\]Combine like terms:\[-6b e^{-3x} = e^{3x}\]

Step 4: Solve for Variable \(b\)

To solve for \(b\), divide both sides by \(-6e^{-3x}\):\[b = -\frac{1}{6} e^{6x}\]

Step 5: Solve for Variable \(a\)

Substitute the value of \(b\) back into the expression for \(a\) derived in Step 1:\[a = -(-\frac{1}{6} e^{6x}) e^{-6x}\]Simplify:\[a = \frac{1}{6}\]

Key Concepts

Exponential Equations

Exponential equations play an essential role in many fields of mathematics. In these types of equations, variables appear as exponents. The general form is something like \( a \cdot e^{bx} = c \). Exponential equations can model everything from population growth to radioactive decay. In our problem, terms like \( e^{3x} \) appear. These indicate that each solution involves variables in the exponent, making these equations more complex than linear ones.

Solving these requires different strategies compared to linear equations. For instance, we can treat the terms involving exponents as coefficients. In this case, \( e^{3x} \) acts like a constant factor that multiplies other variables. This perspective simplifies complex exponential interactions to resemble simpler problems often encountered in algebra.

Constant Coefficients

In many systems of equations, we deal with constant coefficients, which are just fixed numbers. However, in our particular problem, the constants are actually exponential terms like \( e^{3x} \). Although they might look complicated, under the rules of algebra, these can be treated like any numerical constant.

When we solve these types of systems, thinking of these exponential terms as fixed or constant can simplify the process. For example, if \( e^{3x} \) is treated as a constant, it remains unchanged across the equations. This approach helps to easily manipulate and substitute expressions, much like we would do with simple numbers or constants in regular system of equations.

• Recognizing exponential terms as constants simplifies complexities.
• Once viewed as constants, manipulation becomes similar to linear instances.

Solving Algebraically

Solving a system of equations algebraically means finding solutions using operations like addition, subtraction, multiplication, and division. In our exercise, this involves expressing one variable in terms of the other, and then substituting this expression into the second equation.

For instance, we started by solving the first equation for one variable, \( a \), in terms of another, \( b \). This manipulation allowed us to substitute and reduce the system to one equation with one variable. It highlights the power of algebraic techniques in solving systems efficiently.

• Step-by-step isolation of variables leads to simplification.
• Iterative substitution builds progression toward the final solution.

Substitution Method

The substitution method is a fantastic tool for solving systems of equations. It involves solving one equation for a variable and substituting that solution into another equation. Here we used the first equation to express \( a \) in terms of \( b \).

This method reduced the complexity of the system, allowing us to solve for \( b \) directly without needing to juggle both variables at once. After finding \( b \), we substituted back to find \( a \). This systematic approach applies to many kinds of systems, enabling clarity and precision.ul>

Turning multi-variable equations into single-variable problems is effective.

Substitution cuts through redundant complexity, focusing on core solving.