Problem 39 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 2 a+3 b-2 c=18 \\ 5 a-6 b+c=21 \\ 4 b-2 c-6=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( a = 8 \), \( b = 4 \), and \( c = 5 \).

Step 1: Rewrite third equation

Rearrange the third equation: \( 4b - 2c - 6 = 0 \). Add 6 to both sides to get: \( 4b - 2c = 6 \).

Step 2: Solve for c in terms of b

Divide the third equation by 2: \( 2b - c = 3 \). Solve for \( c \): \( c = 2b - 3 \).

Step 3: Substitute c into the first and second equations

Substitute \( c = 2b - 3 \) into the first equation: \( 2a + 3b - 2(2b - 3) = 18 \) and simplify to find \( a \) in terms of \( b \): \( 2a + 3b - 4b + 6 = 18 \rightarrow 2a - b = 12 \rightarrow 2a = b + 12 \).

Step 4: Solve for a in terms of b

From the simplified equation, solve for \( a \): \( a = \frac{b + 12}{2} \).

Step 5: Substitute c and a into the second equation

Substitute \( c = 2b - 3 \) and \( a = \frac{b + 12}{2} \) into the second equation: \( 5\left(\frac{b + 12}{2}\right) - 6b + (2b - 3) = 21 \).

Step 6: Simplify and solve for b

Simplify the substituted equation: \( \frac{5(b + 12)}{2} + (2b - 3) - 6b = 21 \rightarrow \frac{5b + 60}{2} + 2b - 3 - 6b = 21 \rightarrow \frac{5b + 60}{2} - 4b - 3 = 21 \).

Step 7: Clear fractions and solve for b

Multiply the entire equation by 2 to clear fractions: \( 5b + 60 - 8b - 6 = 42 \rightarrow -3b + 54 = 42 \).

Step 8: Complete solving for b

Subtract 54 from both sides: \( -3b + 54 - 54 = 42 - 54 \rightarrow -3b = -12 \). Divide by -3: \( b = 4 \).

Step 9: Find corresponding values of a and c

Use \( b = 4 \) in \( a = \frac{b + 12}{2} \): \( a = \frac{4 + 12}{2} \rightarrow a = 8 \). Use \( b = 4 \) in \( c = 2b - 3 \): \( c = 2(4) - 3 = 5 \).

Step 10: Solution summary

The solution to the system of equations is \( a = 8 \), \( b = 4 \), and \( c = 5 \).

Key Concepts

Substitution Method

The Substitution Method is a way to solve systems of linear equations by substituting one equation into another. This technique turns a system of equations into a single equation with just one variable. Once that variable is solved for, the result is substituted back to find the other variables. This method works best when one of the equations is easy to solve for one variable.
In the given problem, we started by rearranging the third equation to solve for one variable in terms of another: solving the third equation, we found that \( c = 2b - 3 \). This expression for \( c \) was then substituted into the other equations.

This substitution simplified the system, allowing us to express \( a \) in terms of \( b \), and eventually solve for all variables step by step. The step-by-step substitution enabled the transformation of a complex problem into an easier one.

Dependent Equations

Dependent Equations arise in a system when the equations represent the same geometric line. When this occurs, it means that any solution that satisfies one equation will satisfy the other. In essence, these systems have an infinite number of solutions.
In the present problem, we did not encounter dependent equations. Instead, the solution was unique at the end, indicating independent lines. However, understanding dependent systems is crucial. If during substitution, you end up with an identity such as \( 0 = 0 \), then consider that the equations might be dependent.

• Dependent equations usually arise when one equation is a multiple or a linear combination of another.
• Graphically, they represent the same line, thus overlapping.

Being aware of this will help you recognize systems with infinite solutions.

Inconsistent Systems

An Inconsistent System consists of equations that have no solution. Graphically, these lines are parallel, meaning they never intersect. When solving, if you encounter a contradiction such as \( 0 = 5 \), the system is inconsistent.
Although the given problem ended with a solution, understanding inconsistent systems is essential in equations solving. They indicate that there is no common solution that satisfies all equations simultaneously.

• A contradiction during solving (e.g. contradictive statement) suggests the system is inconsistent.
• It implies lines are parallel and never meet in any point in the coordinate plane.
• This awareness can save time during problem-solving, ensuring you identify unsolvable systems early.

Recognizing these systems helps anticipate no solution outcomes in algebraic systems.