Problem 26 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} 3 y+z=-1 \\ -x+2 z=-9+6 y \\ 9 y+3 z=-9+2 x \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (3, \frac{1}{3}, -2)\).

Step 1: Simplify Equations

Let's examine each equation in the system and seek ways to simplify them:\[ 3y + z = -1 \] is the first equation which is already simplified.For the second equation, \(-x + 2z = -9 + 6y\), we can rearrange it to isolate variables for substitution in later steps.The third equation, \(9y + 3z = -9 + 2x\) also can be rearranged if needed, simplifying it initially by dividing by 3 to get \(3y + z = -3 + \frac{2}{3}x\). We can recognize that \(3y + z\) appears in both the first and simplified third equations.

Step 2: Substitute First Equation into Third

From the first equation, we already know \(3y + z = -1\). Notice the similarity in the third equation: \(3y + z = -3 + \frac{2}{3}x\). Therefore, substitute \(-1\) for \(3y + z\) in the third equation:\[-1 = -3 + \frac{2}{3}x\].

Step 3: Solve for x

Now solve the equation from the previous step:\[-1 = -3 + \frac{2}{3}x\]Add \(3\) to both sides:\[2 = \frac{2}{3}x\]Multiply both sides by \(\frac{3}{2}\) to solve for \(x\):\[x = 3\]

Step 4: Substitute x into Second Equation

Now substitute the value of \(x = 3\) into the second equation:\(-x + 2z = -9 + 6y\)Substituting \(x = 3\):\[-3 + 2z = -9 + 6y\].

Step 5: Solve for z in terms of y

Rearrange the equation from the previous step:\[2z = -6 + 6y\]Divide each term by 2:\[z = -3 + 3y\].

Step 6: Substitute Value of z in First Equation

Now substitute \(z = -3 + 3y\) into the first equation:\[3y + z = -1\]\[3y + (-3 + 3y) = -1\]Combine like terms:\[6y - 3 = -1\].

Step 7: Solve for y

Add \(3\) to both sides:\[6y = 2\]Divide both sides by \(6\):\[y = \frac{1}{3}\].

Step 8: Substitute y back to find z

Now substitute \(y = \frac{1}{3}\) back into the expression for \(z\):\[z = -3 + 3y\]\[z = -3 + 3\left(\frac{1}{3}\right)\]Simplifying gives:\[z = -3 + 1 = -2\].

Key Concepts

Substitution Method

The substitution method is a powerful technique to solve systems of equations. This approach involves solving one of the equations for one variable and then substituting this expression into the other equations. This way, you reduce the number of variables, simplifying the system step by step until all variables are found.
Let's see how this works in practice. Assume we have two equations:

• Equation 1: \(3y + z = -1\)
• Equation 2: \(-x + 2z = -9 + 6y\)
• Equation 3: \(9y + 3z = -9 + 2x\)

First, simplify the third equation by dividing all terms by 3, giving us \(3y + z = -3 + \frac{2}{3}x\). Notice that the expressions for \(3y + z\) appear in both the first and third equations, allowing us to substitute them to find \(x\). This step-by-step replacement of variables and solving derived simpler equations eventually yields all the unknowns. This is one of the essence of the substitution method: reducing complexity through deliberate swapping and reducing the system.

Linear Equations

Understanding linear equations is essential for solving systems of equations efficiently. A linear equation describes a straight line on a graph and follows the standard form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are coefficients, and \(d\) is the constant term.
Linear equations have the following characteristics:

• They graph as straight lines.
• They do not have exponents on the variables.
• Each term is either a constant or the product of a constant and a single variable.

In our exercise, we deal with three linear equations: \(3y + z = -1\), \(-x + 2z = -9 + 6y\), and \(9y + 3z = -9 + 2x\). By rearranging and simplifying, such equations can often reveal natural opportunities for substitution or elimination, especially when multiple equations share similar terms like \(3y + z\) in this example.

Algebraic Solutions

Finding algebraic solutions to a system of equations means expressing variables in terms of numbers or other variables, eventually pinpointing their exact numerical values. This requires skillful manipulation of the equations, using algebraic rules and operations, such as addition, multiplication, and distribution.
The step-by-step solution process involves strategies like:

• Isolating one variable to substitute into another equation, reducing the number of variables step by step.
• Simplifying equations by combining like terms and using operations like dividing or multiplying through by constants to make solving more manageable.
• Solving the resulting simplified equations to find precise values for each unknown.

In our case, we began by determining \(x\) from a comparison of simplified equations and then found expressions for \(z\) and \(y\), ultimately substituting back to verify consistency. Mastering algebraic techniques like these can significantly simplify solving complex systems.