91Ó°ÊÓ

Linear equations are fundamental in algebra, forming the basis for understanding higher-level math concepts. A linear equation is typically in the form \(ax + b = c\), where \(a\), \(b\), and \(c\) are constants. To solve these equations, the goal is to find the variable's value that makes the equation true.
In the problem given, we have the equation \(x + y = 3\), which is a linear equation in two variables, \(x\) and \(y\). Solving it requires knowing the value of one of the variables (in this case \(x = \frac{1}{2}\)) to find the other variable, \(y\). By substituting the known value into the equation, we convert it into a simple linear equation with one unknown, making it easier to solve.

The substitution method is a technique used to solve systems of equations or a single equation when you have some known values. It's especially useful when you have equations with two variables and need to find one by knowing the value of the other.
In our exercise, we started with the equation \(x + y = 3\) and the given \(x = \frac{1}{2}\). By substituting \(x = \frac{1}{2}\) into the equation, the equation becomes \(\frac{1}{2} + y = 3\). This substitution simplifies the problem to one with a single variable, \(y\), which can then be easily isolated and solved.

Isolating a variable is a crucial step in solving equations, as it allows you to express the unknown variable on one side of the equation. The primary goal is to manipulate the equation until you have the variable by itself.
To isolate \(y\) in the equation \(\frac{1}{2} + y = 3\), you subtract \(\frac{1}{2}\) from both sides. This subtraction step gives you:

• \(y = 3 - \frac{1}{2}\)

Now, to simplify the right side, convert 3 into \(\frac{6}{2}\), so the subtraction becomes

• \(y = \frac{6}{2} - \frac{1}{2} = \frac{5}{2}\)

This simplification clearly isolates \(y\), resulting in a solution of \(y = \frac{5}{2}\). Recognizing how to isolate variables is key to solving linear equations efficiently.