91Ó°ÊÓ

A linear equation is an equation of the form \(ax + by + cz = d\). Each term is either a constant or the product of a constant and a single variable.
Linear equations graph as straight lines and are foundational in algebra.
Consider the system: \begin{array}{l} 3x - z = -3 5y + 2z = -6 4x + 3y = -8 \right.\right. Breaking it down into simpler pieces allows easier management.
When working with linear equations, following a structured approach helps maintain clarity.
Label the equations Step 1: Label the system for easy referencing: let Equation 1: \(3x - z = -3\), Equation 2: \(5y + 2z = -6\), Equation 3: \(4x + 3y = -8\).

The substitution method solves a system of equations by replacing one variable with an equivalent expression.
Let's dive into an example: Step 2: Solve one equation for one of the variables. Using Equation 1 (\(3x - z = -3\)), solve for \(z\): \(z = 3x + 3\).
Step 3: Replace the found variable in another equation.
Substitute \(z = 3x + 3\) into Equation 2 to find \(y\):
This becomes \(5y + 2(3x + 3) = -6\).
Simplify: \(5y + 6x + 6 = -6\) leads to \(5y + 6x = -12\). Now solve for \(y\): \(y = \frac{-12 - 6x}{5}\).
This substitution technique helps eliminate variables, making the problem easier.

To solve a system of linear equations, find one variable at a time, then back-substitute until all variables are found.
Here's how it works in practice: Step 4: Substitute calculated values into other equations.
Starting with \(y = \frac{-12 - 6x}{5}\), substitute it into Equation 3: \(4x + 3(\frac{-12 - 6x}{5}) = -8\).
Multiply through by 5 to clear the fractions: \(20x + (-36 - 18x) = -40\) simplifies to: \(2x - 36 = -40\), so \(2x = -4\), leading to \(x = -2\).
Step 5: Use the found variable to determine others. Back-substitute \(x = -2\) into \(z = 3x + 3\) to find \(z\): \(z = 3(-2) + 3 = -3\).
Next, substitute \(x = -2\) into \(y = \frac{-12 - 6(-2)}{5}\) to find \(y\): \(y = \frac{-12 + 12}{5} = 0\).
Always verify the solution by substituting back into the original equations.