Problem 51 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 0.5 a+0.3 b=2.2 \\ 1.2 c-8.5 b=-24.4 \\ 3.3 c+1.3 a=29 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(a = 2\), \(b = 4\), \(c = 8\).

Step 1: Analyze the System

First, we look at the given system of equations: \[ \begin{align*} 0.5a + 0.3b &= 2.2 \ (1) \1.2c - 8.5b &= -24.4 \ (2) \3.3c + 1.3a &= 29 \ (3) \end{align*} \] We identify that this system involves three equations with three variables, namely \(a\), \(b\), and \(c\). The goal is to find values for these variables that satisfy all three equations.

Step 2: Simplify Equation (1)

Simplify equation (1) by clearing the decimal by multiplying through by 10: \[ 5a + 3b = 22 \] This gives us a simpler equation to work with.

Step 3: Simplify Equation (2)

Similarly, simplify equation (2) by multiplying through by 10 to eliminate decimals: \[ 12c - 85b = -244 \] We now have another simplified equation ready for use.

Step 4: Simplify Equation (3)

For equation (3), multiply through by 10: \[ 33c + 13a = 290 \] All equations are now simplified and ready for further manipulation.

Step 5: Solve for One Variable

Choose one of the equations to express one variable in terms of the others. From equation (1), solve for \(a\): \[ 5a = 22 - 3b \] \[ a = \frac{22 - 3b}{5} \]

Step 6: Substitute for \(a\) in Equation (3)

Substitute \(a = \frac{22 - 3b}{5}\) into equation (3): \[ 33c + 13\left(\frac{22 - 3b}{5}\right) = 290 \] This gives: \[ 33c + \frac{286 - 39b}{5} = 290 \] Simplify and get \[ 165c + 286 - 39b = 1450 \] \[ 165c - 39b = 1164 \] (new equation (3'))

Step 7: Use Substitution to Solve Equations (2) and (3') for One Variable

Now we have:\[ 12c - 85b = -244 \] \[ 165c - 39b = 1164 \] Let's solve these two equations. Multiply the first by 39 to align the coefficients of \(b\): \[ 468c - 3315b = -9516 \] Multiply the second equation by 85: \[ 14025c - 3315b = 98940 \] Subtract these equations to solve for \(c\): \[ 13557c = 108456 \] Therefore, \[ c = 8 \]

Step 8: Find \(b\) Using \(c\) Value

Substitute \(c = 8\) into \(12c - 85b = -244\) to find \(b\): \[ 12(8) - 85b = -244 \] \[ 96 - 85b = -244 \] \[ -85b = -340 \] \[ b = 4 \]

Step 9: Find \(a\) Using \(b\) Value

Substitute \(b = 4\) into the expression for \(a\) to find its value: \[ a = \frac{22 - 3(4)}{5} \] \[ a = \frac{22 - 12}{5} \] \[ a = 2 \]

Step 10: Verify the Solution

Substitute \(a = 2\), \(b = 4\), and \(c = 8\) back into the original equations to verify: 1. \(0.5(2) + 0.3(4) = 2.2\)2. \(1.2(8) - 8.5(4) = -24.4\)3. \(3.3(8) + 1.3(2) = 29\)Each equation holds true, verifying the solution is accurate.

Key Concepts

Solving systems of equations

Systems of linear equations are a set of two or more equations with the same set of unknowns or variables. Solving these systems involves finding the values of the variables that make all equations true simultaneously. This is a fundamental concept in linear algebra and is critical for modeling and solving real-life problems.
The given exercise features three equations involving three variables: \(a\), \(b\), and \(c\). These are linear equations, meaning they represent straight lines when graphed. Solving them typically involves looking for a single set of values that satisfy all the equations at the same time.
Each equation can provide unique information about the relationship between the variables. For example, expressing one variable in terms of others and using substitution or elimination to reduce the system to fewer equations with fewer variables is a common approach.
Solving systems of equations is essential not just for understanding algebra but also for applications across different fields like engineering, economics, and computer science.

Substitution method

The substitution method is a straightforward way to solve a system of equations, especially helpful when one of the equations is easily solved for one variable. Here's how it generally works:

• First, solve one of the equations for one variable. For instance, solve \(5a + 3b = 22\) for \(a\), yielding \(a = \frac{22 - 3b}{5}\).
• Replace this expression for \(a\) in the other equations to eliminate \(a\) from those equations. This allows the focus to shift to the other variables.
• In this exercise, substituting for \(a\) in equation (3) simplifies the problem, reducing the number of variables in each equation.

This method is particularly suited when one of the variables with its coefficient makes the algebraic manipulation easy, and it helps in breaking down complex systems into simpler parts. The key advantage is that it reduces the number of variables, allowing the solver to tackle a more straightforward system.

Linear algebra

While the step-by-step approach presented combines both the substitution method and equation simplification, these are rooted in concepts central to linear algebra. Linear algebra studies vectors, vector spaces, and systems of linear equations—foundational tools for solving complex problems.

• The solution of this exercise exemplifies the translation of real numbers systems into simpler equations to offer insight into the relationships of different variables.
• Understanding matrix representations of systems and the methods to manipulate them for solutions illustrates a broader skill in linear algebra beyond just computation; it's about interpreting and modeling relationships.
• Critical techniques, like simplifying systems and using properties of equality, underline how linear algebra helps express practical problems abstractly, allowing exact and consistent solutions.

In essence, mastering linear algebra allows one to tackle systems systematically and symbolically, and it offers methods to handle mathematical modeling and problem-solving tasks efficiently.