91Ó°ÊÓ

Linear equations are foundational in algebra and involve variables such as x, y, and z. These equations are called 'linear' because they graph as straight lines. Each term is either a constant or the product of a constant and a single variable. In the exercise, the linear equations are:
\[ x+y+z=10 \ x-y=-1 \ x+y=5 \]
Here’s a simple way to understand them: they form lines, and the solution is where these lines intersect. To solve them, we use methods like substitution or elimination. You can visualize this using a coordinate system where these lines will meet at a point representing the solution. Linear equations are easy to identify; look for equations where every term follows this form: \(ax + by + cz = d\).

The substitution method involves solving one equation for a specific variable and then substituting that solution into another equation. This process can simplify the system and make it easier to solve. Here’s how we can apply it:
1. Solve one equation for one of the variables. For example, using the third equation, \(x + y = 5\), we can solve for y:
\[ y = 5 - x \]
2. Substitute the value of y into another equation. Let’s substitute into the first equation:
\[ x + (5 - x) + z = 10 \]
This simplifies to:
\[ 5 + z = 10 \]
3. Solve for z:
\[ z = 5 \]
This method helps in breaking down complex systems into simpler, manageable parts.

The elimination method, also known as the addition method, involves adding or subtracting equations to eliminate a variable. This can be incredibly efficient for systems with more than two variables. Here’s the step-by-step process using our exercise:
1. The system of equations we have is:
\[ x + y + z = 10 \]
\[ x - y = -1 \]
\[ x + y = 5 \]

2. Add the second and third equations
\( (x - y) + (x + y) = -1 + 5 \)
This simplifies to:
\[ 2x = 4 \]
3. Solve for x:
\[ x = 2 \]
4. Substitute x into the third equation to find y:
\[ 2 + y = 5 \]
\[ y = 3 \]
5. Finally, substitute x and y back into the first equation to find z:
\[ 2 + 3 + z = 10 \]
\[ z = 5 \]
The elimination method is efficient when you can add or subtract equations to cancel out a variable, speeding up the solution process.