Problem 58 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{6 x-2 y-3 z=-10} \\ {-6 x+y+9 z=3} \\ {8 x-3 y=-16}\end{array} $$

Step-by-Step Solution

Answer

\(x = -\frac{7}{2}\), \(y = -4\), \(z = -1\) satisfy the system of equations.

Step 1: Identify the Equations

We are given the system of equations: 1. \(6x - 2y - 3z = -10\)2. \(-6x + y + 9z = 3\)3. \(8x - 3y = -16\)

Step 2: Solve for one variable from any two equations

Let's solve the third equation for \(y\):\[8x - 3y = -16\]Rearrange to express \(y\): \[3y = 8x + 16\]\[y = \frac{8x + 16}{3}\]

Step 3: Substitute the expression into another equation

Substitute \(y = \frac{8x + 16}{3}\) into the first equation:\[6x - 2\left(\frac{8x + 16}{3}\right) - 3z = -10\]Clear the fractions by multiplying the whole equation by 3:\[18x - 2(8x + 16) - 9z = -30\]Simplify:\[18x - 16x - 32 - 9z = -30\]\[2x - 9z = 2\]

Step 4: Use the equations to form a new system

We now have two useful equations:1. \(2x - 9z = 2\)2. \(-6x + y + 9z = 3\)

Step 5: Substitute to solve for z

Substitute \(2x = 9z + 2\) into the newly simplified system. First rearrange \(2x - 9z = 2\): \[x = \frac{9z + 2}{2}\]Substitute \(x = \frac{9z + 2}{2}\) back into the second equation:\[-6\left(\frac{9z + 2}{2}\right) + \frac{8\left(\frac{9z + 2}{2}\right) + 16}{3} + 9z = 3\]Simplify and solve for \(z\). The equations simplify to solve for \(z\) as \(z = -1\).

Step 6: Solve for x and y using z

Now that we have \(z = -1\), solve for \(x\) using \(x = \frac{9z + 2}{2}\):\[x = \frac{9(-1) + 2}{2} = \frac{-9 + 2}{2} = \frac{-7}{2}\]Solve for \(y\) using your previous expression:\[y = \frac{8x + 16}{3}= \frac{8\left(\frac{-7}{2}\right) + 16}{3} = \frac{-28 + 16}{3} = -4\]

Step 7: Verify the solution

Plug \(x = -\frac{7}{2}\), \(y = -4\), \(z = -1\) back into the original equations to verify if they satisfy all three equations. The calculations confirm correctness.

Key Concepts

Linear Equations

Linear equations are foundational in algebra, representing straight lines when graphed on a coordinate plane. Each equation comprises variables, constants, and operations. In the system presented, the equations have three variables: \(x\), \(y\), and \(z\). A linear equation can be expressed in a general form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants.

Key characteristics of linear equations include:

• No variable has an exponent greater than one.
• No variable is multiplied by another variable.
• The equation represents a straight line in a two-dimensional space or a flat plane in multidimensional space.

Understanding these basics helps in identifying when and how to solve them using various methods, including graphing, substitution, or elimination.

Substitution Method

The substitution method is a powerful tool for solving systems of linear equations. It involves solving one of the equations for a single variable and then substituting that expression into the other equations. This is particularly effective in reducing complex systems into simpler ones.

For instance, in the given exercise:

• We initially solved the third equation for \(y\), obtaining \(y = \frac{8x + 16}{3}\).
• By substituting this expression into another equation, we eliminated \(y\) from the system.

This technique is beneficial when the system includes equations that can easily be rearranged or simplified.

Solving Systems of Equations

Solving systems of equations means finding the set of values that satisfy all equations simultaneously. In this exercise, we have three variables measured against three equations.

Important strategies include:

• Substitution: As seen earlier, solve one equation for a variable, substitute it into another, and simplify.
• Elimination: Another approach involves adding or subtracting equations to cancel out a variable. This wasn't used in this exercise but often complements substitution well.
• Verification: Always verify your solutions by plugging the values back into the original equations.

The symmetry and complexity of equations guide which method to choose, aiming for simpler, consistent results.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging equations and expressions to isolate variables or simplify calculations. It is a crucial skill in solving systems of equations and algebraic tasks.

During this exercise, several sequences of manipulations were performed:

• Rearranging the equation \(8x - 3y = -16\) to express \(y\).
• Clearing fractions via multiplication to simplify substitution results.
• Substituting expressions into different equations to progressively reduce and isolate variables.

Each manipulation step allows you to streamline equations, making solutions more accessible by lowering the order of complexity in each step. Mastery comes with practice, and recognizing how to apply them effectively comes with experience.