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Section 1.2 Variables and Solving

Subsection 1.2.1 Variables In Equations

Figure 1.2.1. Solving with Variables

When variables are first introduced, the are symbols that stand for some unknown or undetermined number. This is how they operate in equation. The most conventional variable is the letter \(x\text{.}\) The equation

\begin{equation*} 4 + x = 13 \end{equation*}

is a question. It asks: what number, when added to 4, produced 13? All equations with a single variable are implicitly questions: what value could the variable have to make the equation work? In this example, \(x = 9\) is the only possibility. Some equation have more than one possibility. The equation

\begin{equation*} p^2 - 7p + 10 = 0 \end{equation*}

asks what values of the variable \(p\) produce zero when put into this expression. This equation has two answers: \(p=2\) and \(p=5\text{,}\) since \(2^2 - 7(2) + 10 = 4 - 14 + 10 = 0\) and also \(5^2 - 7(5) + 10 = 25 - 35 + 10 = 0\text{.}\) We could even have the silly equation

\begin{equation*} h = h\text{.} \end{equation*}

This equation asks which numbers are equal to themselves. Obviously, all numbers are, so the variable \(h\) could be any number whatsoever.

Subsection 1.2.2 Solving Equation

If we have an equation with a variable

\begin{equation*} 5q - 6 = 25q + 9 \end{equation*}

and we want to know what value(s) of \(q\) satisfy the equation, we have to solve the equation. Solving here means making alterations to the equation to try to figure out what the variable might be, usually by isolating the variable. This leads us to a major, central idea in mathematics.

WE CAN PERFORM ANY OPERATIONS TO AN EQUATION AS LONG AS WE PERFORM THE SAME OPERATION TO BOTH SIDES OF THE EQUATION!
This is the principle we use to solve equation. Take the equation above. The first operation I'm going to do is add 6 to both sides of the equation.

\begin{align*} 5q - 6 + 6 \amp = 25q + 9 + 6 \\ 5q \amp = 25q + 15 \end{align*}

Then I'll divide both sides of the equation by 5.

\begin{align*} \frac{5q}{5} \amp = \frac{25q + 15}{5} \end{align*}

I can split up the fraction on the right (we are allowed to split up addition and subtraction in numerators of fractions; we'll talk more about manipulation of fraction in Chapter 5.

\begin{align*} \frac{5q}{5} \amp = \frac{25q}{5} + \frac{15}{5}\\ q \amp = 5q + 3 \end{align*}

Then I'll subtract \(5q\) from both sides of the equation.

\begin{align*} q - 5q \amp = 5q + 3 - 5q\\ -4q = 3 \end{align*}

Finally, I'll divide both sides of the equation by \(-4\text{.}\)

\begin{gather*} \frac{-4q}{-4} = \frac{3}{-4}\\ q = \frac{-3}{4} \end{gather*}

And I've solved the equation. The only value of the variable \(q\) that satisfies the original equation is \(\frac{-3}{4}\text{.}\) We could also write this as \(q = -0.75\) if we wished.

Solving can get quite complicated and there are many techniques. However, fundamentally the whole idea is that we preserve the equality of the situations by doing the same things to both sides of the equations; solving is the process of choosing the right operations to try to isolate the variable. (Sadly, there are many equations where no sequence of applying operations to both sides of the equation will produce a solution. The only such example that you'll need to be familiar with is quadratic equations, which are discussed in Chapter 13.)

\begin{align*} 5(x-7) \amp = \frac{x}{3} \\ \amp \text{Multiply both sides by 3.}\\ (3)(5)(x-7)) \amp = x \\ \amp \text{Multiply the 3 and 5.}\\ 15(x-7) \amp = x \\ \amp \text{Distribute the 15.}\\ 15x - 105 \amp = x \\ \amp \text{Subtract from both sides.}\\ 15x - 105 - x \amp = x - x \\ 14x - 105 \amp = 0\\ \amp \text{Add 35 to both sides.}\\ 14x \amp = 105 \\ \amp \text{Divide by 14 on both sides.}\\ x \amp = \frac{105}{14} \end{align*}
The solution to the original equation is \(x = \frac{35}{14}\text{.}\)
\begin{align*} 12t \amp = \frac{4t-7}{3} \\ \amp \text{Multiply both sides by 3.}\\ 36t \amp = 4t-7\\ \amp \text{Subtract from both sides.}\\ 36t - 4t \amp = 4t - 7 - 4t \\ 32t \amp = - 7 \\ \amp \text{Divide both sides by 32.}\\ t \amp = \frac{-7}{32} \end{align*}
This is the solution to the original equation.
\begin{align*} \frac{7k-2}{k} \amp = 4\\ \amp \text{Multiply both sides by }\\ 7k-2 \amp = 4k\\ \amp \text{Subtract from both sides.}\\ 3k - 2 \amp = 0 \\ \amp \text{Add 2 to both sides.}\\ 3k \amp = 2 \\ \amp \text{Divide both sides by 3.}\\ k \amp = \frac{2}{3} \end{align*}
These examples cover solving equations using the four basic arithmetic operations. Many more examples for solving with other functions and operations are found in subsequent sections.

Subsection 1.2.3 A Caution When Solving

As we saw in Example 1.2.4, there are some subtleties in solving. Many of these involve possible division by zero; we should always be aware of the possibility of division by zero by zero. This is particularly true when cancelling a variable. I can illustrate this with another example.

\begin{equation*} a(4a-6) = a(5a+1) \end{equation*}

In this equation, if we wanted to try to solve for \(a\text{,}\) it would be reasonable to divide by \(a\) to make both sides of the equation a little more approachable. That cancels off the \(a\) on both sides.

\begin{equation*} 4a-6 = 5a+1 \end{equation*}

From here, we could solve as before. Omitting the steps, the result is \(a = -7\text{.}\) However, if we look at the original equation, \(a = 0\) is also a solution, since the equation then result in the true statement \(0=0\text{.}\)

The point here is that whenever we divide by something on both sides of the equation, we have to be a bit aware of the possibility that the thing we wanted to divide by might have been zero. If it was zero, it can lead to extra solution which we might otherwise have missed.