91Ó°ÊÓ

Understanding how to manipulate linear equations is crucial when solving for variables such as y. When presented with an equation like \(\frac{1}{5}(25-5 y)=4 x-9 y+13\), the goal is to rearrange it in a way that one can clearly see how y is influenced by x.

To do this, begin by distributing multiplication over addition or subtraction, a process known as the distributive property. For our case, applying this property to \(\frac{1}{5}(25-5 y)\) would result in \(5-y\), setting the stage for easier manipulation of the equation. It's akin to untangling a knot – the right moves will make the following steps significantly more straightforward.

Once untangled, relocate all the terms involving y to one side and the rest to the opposite to set a clear distinction between the variable of interest and the remaining elements. This part of the manipulation requires one to be comfortable with inverse operations – adding what's been subtracted and vice versa – in order to group like terms effectively. These strategies turn a daunting equation into a more manageable format that is primed for finding the value of y as a function of x.

To solve for a particular variable means to isolate it on one side of the equation. Imagine you are on a treasure hunt and y is the treasure, thus your goal is to remove all other elements away from y to claim it. Starting from a transformed equation like \(5 + 9y - y = 4x + 13\), you want to gather all the y terms on one side to clearly expose y’s relationship with x.

In our example, combining like terms by subtracting y from 9y gives us \(8y\), leaving us with \(8y = 4x + 8\). The balance of the equation must be maintained, so when you divide one side by 8, you must also divide the other side by 8. This final division is like opening the treasure chest – inside, you find that \(y = \frac{1}{2}x + 1\), which is y expressed explicitly as a function of x. Mastering isolation is an essential algebra skill, as it translates across a myriad of mathematical concepts.

Algebraic expressions are combinations of variables, numbers, and arithmetic operations. Think of them as sentences in the language of mathematics where variables like x and y can take on different values to make the sentence true. In our exercise, \(\frac{1}{2}x + 1\) is an algebraic expression where the variable x determines the value of y.

It’s important to understand each part of an algebraic expression: constants (like the number \(1\)), coefficients (the number \(\frac{1}{2}\) that multiplies the variable x), and the variables themselves. Grasping how to work with these components allows students to evaluate, simplify, and manipulate expressions to solve equations. This foundational knowledge is indispensable not just for linear equations but also for more complex mathematical problems involving polynomial and rational expressions.