# Difference between revisions of "Algebraic manipulation"

Algebraic manipulation involves doing opposite operations (undoing) to any equation to solve for a certain variable (often by isolation).

## Properties of Equality

Any time we add, subtract, multiply, divide, square, square root, etc. to one side, we must do it to the other side to maintain equality.

### Example

\begin{align*} x^2 + 18 &= 43 \\ x^2 &= 25 \\ x &= \boxed{\pm 5} \end{align*} In the example, we first subtract 18 from the left side to isolate the $x^2$. However, we also have to subtract 18 from the right side to maintain equality. The right hand side becomes $43 - 18 = 25$.

We then square root the left side to get the $x$ by itself. However, we also have to square root the right side to maintain equality.

## Cross Multiplication

Cross multiplication is a common method of solving proportions. Essentially, one is multiplying both sides by the denominators of both sides.

### Example

\begin{align*} \frac{x}{3} &= \frac{10}{5} \\ 5 \cdot x &= 3 \cdot 10 \\ 5x &= 30 \\ x &= 6 \end{align*} The above method of solving is an example of cross-multiplication. During the cross-multiplication, we actually multiplied both sides by $3 \cdot 5$. On the LHS, the $3$ gets divided out (leaving a $5$ for multiplication), and on the RHS, the $5$ gets divided out (leaving a $3$ for multiplication).

## Potential Extraneous Solutions

In some methods of isolation like squaring both sides or multiplying both sides by a non-constant, we could introduce additional "solutions" that aren't really solutions to the original equation. Such solutions are called "extraneous solutions". Thus, when doing such methods, it is a good idea to check the solutions by plugging it back into the original equation.

Additionally, when dividing both sides by a non-constant, we could eliminate perfectly valid solutions. Thus, when possible, it's often better to factor than to divide both sides.

### Example

\begin{align*} \frac{x^2 - 2x}{x-5} &= \frac{15}{x-5} \\ x^2 - 2x &= 15 \\ x^2 - 2x - 15 &= 0 \\ (x-5)(x+3) &= 0 \\ x &= 5, -3 \end{align*} When solving, we start by multiplying both sides by $x-5$. Then we used algebraic manipulation to get all terms to one side to form a quadratic. After factoring, we get $x = 5, -3$.

Plugging in $x = -3$ satisfies the original equation because $\frac{9+6}{-8} = \frac{15}{-8}$. However, plugging in $x = 5$ does not satisfy the original equation because $\frac{15}{0}$ on both sides is undefined. Thus, $x = -3$ is the only valid solution, and $x = 5$ is an extraneous solution.

The extraneous solution appeared when we multiplied both sides by $x - 5$. The original equation banned $x = 5$ as a possible solution because it would make the denominator zero. However, after multiplying both sides by $x-5$, there is no denominator that eliminates $x = 5$ as a possible solution.