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The substitution method is a popular technique for solving systems of linear equations. It's particularly useful when one equation is already solved for one variable, or can be easily solved. Here's a simplified approach:
1. Solve one equation for one of the variables.
2. Substitute this expression into the other equation.
3. Solve the resultant equation for the remaining variable.
4. Plug back the value to find the other variable.

Let's illustrate this method using the given system of equations:
1. First, from the equation \( 3y = 5x + 6 \), solve for y: \( y = \frac{5x + 6}{3} \). This is your substitution. 2. Replace y in the second equation \( x + y = 2 \) with \( \frac{5x + 6}{3} \): thus, \( x + \frac{5x + 6}{3} = 2 \). 3. Solve this equation for x (clearing fractions by multiplying each term by 3). 4. Finally, use the value of x to find y again.
This method avoids trial and error of guessing, making it algorithmic and systematic.

Linear equations play an essential role in algebra. They represent straight lines in coordinate geometry. Each term in a linear equation is either a constant or the product of a constant and a single variable.
• The standard form of a linear equation in two variables (x, y) is: \( Ax + By = C \).
Let's look at the given system.
\( 3y = 5x + 6 \) can be rewritten in standard form as \( 5x - 3y = -6 \).
\( x + y = 2 \) is already in standard form.
When solving systems of linear equations, you're essentially looking for points (x, y) that satisfy both equations at the same time.
Every step you take should get you closer to isolating x and y to see where these lines intersect.

Algebraic manipulation involves rearranging equations and expressions using algebraic properties. This is essential in solving equations, especially systems of linear equations.
• Common operations include adding, subtracting, multiplying, and dividing both sides of an equation by the same number or expression.
Consider Step 3 in our solution. After substituting \( y \) in \( x + y = 2 \), we clear the fraction:
Multiply everything by 3: \( 3x + (5x + 6) = 6 \)
Combine like terms: \( 8x + 6 = 6 \)
Subtract 6 from both sides: \( 8x = 0 \)
Divide by 8: \( x = 0 \)
This isolated the x term via sequential algebraic operations.