Problem 9 - Algebra 2

Question

Solve each system of equations. \(2 x-y=2\) \(3 z=21\) \(4 x+z=19\)

Step-by-Step Solution

Answer

The solutions are \(x = 3\), \(y = 4\), and \(z = 7\).

Step 1: Solve the Equation for z

From the second equation, \(3z = 21\), divide both sides by 3 to solve for \(z\): \[ \frac{3z}{3} = \frac{21}{3} \] Thus, \(z = 7\).

Step 2: Substitute z into the Third Equation

Using the value of \(z\) from Step 1, substitute \(z = 7\) into the third equation: \(4x + z = 19\). This gives: \[ 4x + 7 = 19 \] Subtract 7 from both sides: \[ 4x = 12 \].

Step 3: Solve for x

Now, solve for \(x\) by dividing both sides by 4: \[ x = \frac{12}{4} = 3 \].

Step 4: Substitute x into the First Equation

Use the value for \(x\) found in Step 3 and substitute it into the first equation: \(2x - y = 2\). This becomes: \[ 2(3) - y = 2 \] Simplifying gives: \[ 6 - y = 2 \].

Step 5: Solve for y

Rearrange the equation from Step 4 to solve for \(y\) by subtracting 6 from both sides: \[ -y = 2 - 6 \] Therefore: \[ -y = -4 \] Finally, multiply both sides by -1: \( y = 4 \).

Key Concepts

Solving equations

Solving equations involves finding the values of variables that make an equation true. In the given problem, there are three equations that form a system, and our task is to find values for variables \(x\), \(y\), and \(z\) that satisfy all these equations simultaneously. By solving each equation one by one and using the solutions in subsequent equations, we can unravel the values of all variables. This approach is systematic and ensures accuracy, as we check consistency across all given equations.

Substitution method

The substitution method is a powerful technique for solving systems of equations. It involves expressing one variable in terms of another, then substituting this expression into other equations. In our exercise, we started by solving for \(z\) from the equation \(3z = 21\). Once we determined that \(z = 7\), we substituted \(z\) into the third equation \(4x + z = 19\). This effectively reduced the number of variables, simplifying the process of finding \(x\). The substitution method helps break down complex multi-variable situations into simpler, single-variable ones.

Linear equations

Linear equations are mathematical expressions where each term is either a constant or the product of a constant and a single variable. They represent straight lines on a graph. In our system:

• \(2x - y = 2\)
• \(3z = 21\)
• \(4x + z = 19\)

Each equation features variables with an exponent of one, characteristic of linear equations. Solving linear equations of a system often involves combining equations to eliminate variables, ultimately finding a consistent solution point that satisfies all given equations.

Algebraic manipulation

Algebraic manipulation allows us to rearrange and simplify equations to isolate variables. It involves operations like addition, subtraction, multiplication, and division. For instance, in the exercise, we divided by 3 to solve for \(z\) in \(3z = 21\). Later, subtraction was used to isolate \(4x\) in the equation \(4x + 7 = 19\), and division again to solve for \(x\). Finally, subtraction and multiplication by \(-1\) were used to solve for \(y\) in the last steps. These manipulations are essential skills for balancing and solving equations efficiently.