91Ó°ÊÓ

Rearranging equations is a fundamental skill in algebra that helps you isolate one variable to make it easier to solve for it.
Consider the equation given in the exercise: \(5x - y = 10\). We need to solve for \(y\). To rearrange the equation:
* First, add \(y\) to both sides to move it away from the negative side: \(5x = 10 + y\)
* Next, subtract 10 from both sides to isolate \(y\): \(y = 5x - 10\).
Rearranging makes it simpler to perform substitutions later. Take your time with each step and verify your changes to avoid mistakes.

The substitution method is an excellent way to find specific solutions to a linear equation once it's rearranged. In our exercise, \(y\) is isolated as \(y = 5x - 10\).
* First, choose simple values for \(x\), such as -1, 0, and 2—easy to manage numbers.
* Substitute each chosen \(x\) value back into the equation to find \(y\). For example:
- When \(x = -1\):\(y = 5(-1) - 10\) leads to \(-5 - 10 = -15\). Thus, \((-1, -15)\).
- When \(x = 0\):\(y = 5(0) - 10\) results in \(0 - 10 = -10\). Giving us \((0, -10)\).
- When \(x = 2\):\(y = 5(2) - 10\) gives us \(10 - 10 = 0\). Providing \((2, 0)\).
Substitution breaks the problem into manageable pieces, making it easier and precise.

Finding solutions for linear equations involves substituting chosen values back into the rearranged equation. Each substitution provides a pair of \((x, y)\) that satisfies the equation.
* Choosing values for \(x\) ensures you get different solutions. In the example: - \((-1, -15)\) when \(x = -1\).
- \((0, -10)\) when \(x = 0\).
- \((2, 0)\) when \(x = 2\).
Ensuring there are three distinct pairs: each solution represents a point where the line described by the equation cross the coordinate plane.
Practice finding solutions by selecting varied values and calculating the corresponding \(y\). This strengthens understanding and prepares you for more complex linear systems.