Problem 35 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{3 a+2 b=27} \\ {5 a-7 b+c=5} \\ {-2 a+10 b+5 c=-29}\end{array} $$

Step-by-Step Solution

Answer

The solution is \(a = 7\), \(b = 3\), \(c = -9\).

Step 1: Set Up the Equations

We have three equations with three variables: \(3a + 2b = 27\), \(5a - 7b + c = 5\), and \(-2a + 10b + 5c = -29\). Our task is to find values for \(a\), \(b\), and \(c\) that satisfy all three equations.

Step 2: Solve for a Variable in One Equation

We will start by solving the first equation for \(a\) because this equation has only two variables. Solving \(3a + 2b = 27\) for \(a\), we subtract \(2b\) from both sides: \(3a = 27 - 2b\). Dividing by 3 gives \(a = 9 - \frac{2}{3}b\).

Step 3: Substitute for a in the Other Equations

We substitute \(a = 9 - \frac{2}{3}b\) into the second and third equations. For the second equation: \(5(9 - \frac{2}{3}b) - 7b + c = 5\). Simplifying gives \(45 - \frac{10}{3}b - 7b + c = 5\). For the third equation: \(-2(9 - \frac{2}{3}b) + 10b + 5c = -29\). Simplifying gives \(-18 + \frac{4}{3}b + 10b + 5c = -29\).

Step 4: Simplify Substituted Equations

Simplify the second equation: Combine the terms involving \(b\) to get \(-\frac{31}{3}b + c = 5 - 45 = -40\). Multiply the entire equation by 3 to clear the fraction: \(-31b + 3c = -120.\) Simplify the third equation: Combine like terms to get \(\frac{34}{3}b + 5c = -11\). Multiply by 3: \(34b + 15c = -33.\)

Step 5: Solve the System of Two Equations

We now have \(-31b + 3c = -120\) and \(34b + 15c = -33\). Solve these equations using elimination or substitution. Multiply the first equation by 5: \(-155b + 15c = -600\). Subtract the second from the new first: \(-155b + 15c - (34b + 15c) = -600 + 33\), giving \(-189b = -567\). Solving for \(b\), we get \(b = 3\).

Step 6: Back-Substitute to Find c

Substitute \(b = 3\) into \(-31b + 3c = -120\): \(-31(3) + 3c = -120\). This simplifies to \(-93 + 3c = -120\); thus, \(3c = -27\), and solving for \(c\), we find \(c = -9\).

Step 7: Back-Substitute to Find a

Use \(b = 3\) in the relationship we found for \(a\), \(a = 9 - \frac{2}{3}b\): \(a = 9 - \frac{2}{3}(3) = 9 - 2 = 7\).

Step 8: Verify the Solution

Plug \(a = 7\), \(b = 3\), \(c = -9\) back into the original equations to ensure they hold true: For \(3a + 2b = 27\), \(3(7) + 2(3) = 21 + 6 = 27\). For \(5a - 7b + c = 5\), \(5(7) - 7(3) - 9 = 35 - 21 - 9 = 5\). For \(-2a + 10b + 5c = -29\), \(-2(7) + 10(3) + 5(-9) = -14 + 30 - 45 = -29\). All equations check out, so the solution is correct.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when plotted on a graph. The general form of a linear equation is \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. In a system of linear equations, we have several such equations involving the same variables. Our task is to find the values of these variables that satisfy all the given equations simultaneously.
To illustrate, consider the system of equations in our problem:

• \(3a + 2b = 27\)
• \(5a - 7b + c = 5\)
• \(-2a + 10b + 5c = -29\)

These equations are linear because their highest power of variables is 1. Solving a system like this typically involves finding where the corresponding lines or planes intersect. Each intersection point represents a possible solution.

Substitution Method

The substitution method is a powerful tool for solving systems of equations, especially when one equation can easily be isolated for one variable. In this approach, we solve one of the equations for one variable, then substitute this expression into the other equations. This reduces the number of variables in those equations, making them easier to solve.
For example, in our solution, we first solved the equation \(3a + 2b = 27\) for \(a\), resulting in \(a = 9 - \frac{2}{3}b\). We then substituted this expression for \(a\) in the other two equations:

• Substituted in \(5a - 7b + c = 5\),
• Substituted in \(-2a + 10b + 5c = -29\).

This reduced the complexity of the system, making it easier to find the remaining variables.

Elimination Method

The elimination method, sometimes called the addition method, is another technique for solving systems of equations. The idea is to systematically eliminate one variable at a time by adding or subtracting equations from each other, until one variable can be directly solved.
In our solution, after simplifying the substituted equations, we used elimination by aligning the two simplified equations:

• \(-31b + 3c = -120\),
• \(34b + 15c = -33\).

We multiplied the first equation to parallel the coefficients of \(c\) and subtracted the aligned equations to eliminate \(c\), thereby solving for \(b\).
Once \(b\) was found, the elimination of variables becomes straightforward, leading us to find other unknowns.

Variables

Variables are symbols that denote unknown values in equations and systems. In linear equations, variables are typically letters such as \(a\), \(b\), and \(c\). They are placeholders for numbers that satisfy the equations within a system.
In our problem, there are three variables: \(a\), \(b\), and \(c\). Our goal is to determine their values such that all the provided equations hold true. Each equation gives a different relationship between these variables, allowing us, step by step, to hone in on their exact numerical values.
By solving the equations methodically, we determine that \(a = 7\), \(b = 3\), and \(c = -9\). These correspond to a specific point that satisfies all equations simultaneously.

Solution Verification

Verifying the solution of a system of equations is crucial to ensure accuracy. Solution verification involves substituting the found values of variables back into the original equations to check if they hold true.
In our example, once we found \(a = 7\), \(b = 3\), and \(c = -9\), we substituted these values back into each of the original equations:

• For \(3a + 2b = 27\), we found \(3(7) + 2(3) = 27\).
• For \(5a - 7b + c = 5\), the solution was verified with \(5(7) - 7(3) - 9 = 5\).
• For \(-2a + 10b + 5c = -29\), it held true as \(-2(7) + 10(3) + 5(-9) = -29\).

All equations were satisfied, confirming the correctness of our solution. This step is essential to validate that no calculation errors were made during the process.