Problem 43 - A Graphical Approach to College Algebra

Question

Feed Requirements Solve the system from Example 4 . $$ \begin{aligned} 25 x+40 y+20 z &=2200 \\ 4 x+2 y+3 z &=280 \\ 3 x+2 y+z &=180 \end{aligned} $$

Step-by-Step Solution

Answer

Substituting leads to linear dependencies; verify system consistency.

Step 1: Choose a Variable to Eliminate

We'll begin by eliminating one of the variables from two equations. Let's eliminate \( z \). We do this by manipulating equations such that z is removed.

Step 2: Eliminate z from Two Equations

Eliminate \( z \) from the first and second equations. Multiply the second equation by 20 to get the coefficients of \( z \) to both be 60:\[80x + 40y + 60z = 5600\].Subtract this from the first equation (20 times equation 3):\[80x + 40y + 20z = 7200\].This results in:\[60x = -1600\].

Step 3: Solve for x

Solve \( 60x = -1600 \) for \( x \):\[x = \frac{-1600}{60} = -\frac{80}{3}.\]

Step 4: Substitute x to Find y and z

Substitute \( x = -\frac{80}{3} \) back into the second equation:\( 4\left(-\frac{80}{3}\right) + 2y + 3z = 280 \).Simplify and solve for \( y \) in terms of \( z \).

Step 5: Final Substitutions and Solve

Use \( x \) and the relation found in Step 4 in the third equation:\( 3\left(-\frac{80}{3}\right) + 2y + z = 180 \), and solve for the remaining variables \( y \) and \( z \) using substitutions and simplifications.

Key Concepts

Variable Elimination

When solving systems of equations, one popular strategy is **variable elimination**. The goal here is to make one variable disappear from the equations by adding or subtracting them, leaving you with fewer variables to solve.
- Start by deciding which variable you want to eliminate. It should be the one that's easiest to clear out, usually based on its coefficients.
- In our example, we chose to eliminate the variable \( z \) from the system of equations.
- This involved manipulating the equations so the coefficients of \( z \) are the same. This allows one to subtract or add the equations to cancel out \( z \).
In practice, eliminating a variable reduces a three-variable equation to a two-variable situation. This simplification makes the remaining variables easier to solve.

Substitution Method

The **substitution method** is another crucial strategy when dealing with systems of equations. Once one of the variables is solved through methods like elimination, substitution takes over.
- After you eliminate a variable and solve for one of the others, it's time to substitute that solution back into the other equations.
- In the original solution, having found \( x = -\frac{80}{3} \), this value was substituted back into the second equation to find \( y \) in terms of \( z \).
This step ensures all remaining equations reflect this known variable value, allowing for further simplification and eventual solution of all variables. This back-substitution narrows down each unknown variable step-by-step, moving closer to the complete solution.

Linear Equations

At the core of a system of equations like ours are **linear equations**. These are equations that represent straight lines when graphed.
- A linear equation will have variables raised to only the first power. Examples are the given equations like \( 4x + 2y + 3z = 280 \).
- These can have one, two, or more variables, and the concept stays consistent: solving for the unknown.
In solving such systems, the aim is to find values for the variables that satisfy all equations simultaneously. Techniques like variable elimination and substitution help untangle the relationships within these linear equations, ultimately leading to the solution that fits this criterion for all involved equations. Linear equations form the foundation upon which many algebraic concepts are built.