Problem 2 - Intermediate Algebra

Question

Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{rr} 2 x+3 y-4 z & =-10 \\ 2 y+3 z & =16 \\ 2 y-5 z & =-16 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (0, 2, 4)\).

Step 1: Identify the given system of equations

The given system of equations is:\[ \begin{align*} 2x + 3y - 4z &= -10 \ 0x + 2y + 3z &= 16 \ 0x + 2y - 5z &= -16 \end{align*} \] Notice the absence of the variable \(x\) in the second and third equations.

Step 2: Eliminate one variable from the second and third equations

Subtract the third equation from the second to eliminate \(y\):\[ (2y + 3z) - (2y - 5z) = 16 - (-16) \] This simplifies to:\[ 8z = 32 \] Divide by 8:\[ z = 4 \] So, we have found \(z = 4\).

Step 3: Substitute \(z = 4\) back into the second equation

Substitute \(z = 4\) into the second equation to solve for \(y\):\[ 2y + 3(4) = 16 \] Simplify and solve for \(y\):\[ 2y + 12 = 16 \ 2y = 4 \ y = 2 \] Thus, we have found \(y = 2\).

Step 4: Substitute \(y = 2\) and \(z = 4\) into the first equation

Now substitute \(y = 2\) and \(z = 4\) into the first equation to solve for \(x\):\[ 2x + 3(2) - 4(4) = -10 \] Simplify and solve for \(x\):\[ 2x + 6 - 16 = -10 \ 2x - 10 = -10 \ 2x = 0 \ x = 0 \] Thus, we have found \(x = 0\).

Step 5: Conclusion of the solution

After substituting the obtained values back into all equations to verify correctness, we find that our solution satisfies the entire system of equations. Thus, the solution to the system is \((x, y, z) = (0, 2, 4)\).

Key Concepts

Solution of Equations

Solving a system of linear equations means finding the values for the variables that make all the equations true. Think of it like a detective puzzle where the goal is to find the unique set of variable values that satisfy every equation in the system. Sometimes, there might be one solution, no solution, or even infinitely many solutions. Here's how to determine what kind of solution your system might have:

• If the system has a unique solution, the equations intersect at a single point, meaning there's exactly one set of values that works.
• If there's no solution, the equations are contradictory and will never meet, indicating parallel lines.
• If there are infinitely many solutions, the equations are actually the same line, just expressed differently.

In the example given, the system has a unique solution: \((x, y, z) = (0, 2, 4)\). This means these values satisfy each equation, and no other values will. Ensuring all equations are satisfied by the solution requires checking each equation with these variable values to verify the solution's accuracy.

Elimination Method

Elimination is a powerful technique to solve systems of equations. It's a bit like removing unwanted ingredients in a recipe to focus on the main flavor!Here's how you can use elimination:- Align your equations so all similar terms are vertically organized.- Decide which variable you want to eliminate (remove entirely from one or pairs of equations).- Add or subtract the equations to cancel out this variable.In our exercise, we eliminated the variable \(y\) by subtracting the third equation from the second, yielding an equation with only \(z\). This simplifies the problem, as smaller systems are easier to solve.Finding one variable such as \(z = 4\) then opens the path to solving for the other variables. It's by this systematic elimination process that we gradually narrow the solution down to those specific values of \((x, y, z)\).

Substitution Method

The substitution method is like solving a multi-step treasure map, where each clue leads to the next.Here's a step-by-step approach to using substitution:- Start with one equation and manipulate it to express one variable in terms of the others.- Replace, or substitute, this expression into another equation. This substitution reduces the number of variables in that equation.- Continue until you're left with a single equation with one unknown variable. Solve it.In our exercise, once we found \(z = 4\) using elimination, we substituted \(z = 4\) into the easier equation to find \(y = 2\). Afterwards, both \(y\) and \(z\) values were substituted into the initial equation to solve for \(x\).Substitution is highly effective in systems with equations that are already simplified, making it another favorite tool for tackling these mathematical challenges.