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Difference between revisions of "Algebraic manipulation"

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<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{x}{3} &= \frac{10}{5} \\
 
\frac{x}{3} &= \frac{10}{5} \\
−
5x &= 3 \cdot 10 \\
+
5 \cdot x &= 3 \cdot 10 \\
 
5x &= 30 \\
 
5x &= 30 \\
 
x &= 6
 
x &= 6

Latest revision as of 17:49, 8 November 2020

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.

Problems

Introductory

Intermediate

See Also

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