Problem 34 - Algebra 2

Question

Solve each system of equations. $$ \begin{array}{l}{3 x-3 y=6} \\ {-6 y=-30} \\ {5 z-2 x=6}\end{array} $$

Step-by-Step Solution

Answer

Solution: \((x, y, z) = (7, 5, 4)\)

Step 1: Solve the Second Equation for y

The second equation is \(-6y = -30\). To solve for \(y\), divide both sides by \(-6\) to isolate \(y\):\[y = \frac{-30}{-6} = 5\]

Step 2: Substitute y into the First Equation to Find x

Now that we have \(y = 5\), substitute this value into the first equation: \[3x - 3(5) = 6\]Simplify and solve for \(x\):\[3x - 15 = 6\]Add 15 to both sides:\[3x = 21\]Divide by 3:\[x = 7\]

Step 3: Substitute x into the Third Equation to Solve for z

Use the value \(x = 7\) in the third equation: \[5z - 2(7) = 6\]Solve for \(z\):\[5z - 14 = 6\]Add 14 to both sides:\[5z = 20\]Divide by 5:\[z = 4\]

Step 4: Verify the Solution

Substitute \(x = 7\), \(y = 5\), and \(z = 4\) back into the original system to ensure all equations are satisfied:1. \(3x - 3y = 6\): \(3(7) - 3(5) = 21 - 15 = 6\), correct.2. \(-6y = -30\): \(-6(5) = -30\), correct.3. \(5z - 2x = 6\): \(5(4) - 2(7) = 20 - 14 = 6\), correct.All equations are satisfied.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that describe a straight line when graphed on a coordinate plane. In simpler terms, they're equations where each variable is raised only to the power of one. You won't find any squares, roots, or higher powers involved. This makes linear equations straightforward to work with in terms of algebra.

A typical form of a linear equation in two variables is expressed as:

• \( ax + by = c \)

Here, \( a \), \( b \), and \( c \) are constants, and \( x \) and \( y \) are variables.

When solving systems of equations like those in the exercise, we are working with more than one linear equation at a time. In such cases, we aim to find a set of values for the variables that satisfy all the given equations simultaneously. This means the solution is the point where the lines represented by these equations intersect.

Substitution Method

The substitution method is a popular technique for solving systems of equations. It's especially useful when one of the equations is already solved for one variable, or can easily be solved for one variable.

Here's how the substitution method typically works:

• First, solve one of the equations for one of the variables.
• Next, substitute this expression into the other equation(s). This gives you an equation with only one variable.
• Solve for that variable.
• Once you have found the value of one variable, substitute it back into one of the original equations to find the value of the other variable(s).

In the exercise, this method was applied by first solving for \( y \) from the second equation and then substituting \( y \) into the first equation to find \( x \). Finally, \( x \) was substituted into the third equation to find \( z \). This step-by-step approach makes the process much clearer and manageable.

Algebraic Solutions

Algebraic solutions involve using algebraic techniques to find unknown values in equations. They are based on logical manipulations of equations to isolate the variables and solve for their values.

Several key algebraic operations are used when finding solutions:

• Addition or subtraction to both sides of the equation to cancel out terms.
• Multiplication or division across the equation to simplify expressions.
• Substitution to replace one variable with another equivalent expression.

In the original exercise, these techniques were used to isolate and compute the values of \( x \), \( y \), and \( z \).

Understanding these operations is crucial for solving more complex algebraic systems as they provide the foundational skills needed to manipulate and solve almost any algebraic expression. The careful application of these principles ensures that all solutions are accurate and align with the original equations.