Algebra II For Dummies. Sterling Mary Jane

problem with fractions, like cats, is that they aren’t particularly easy to deal with. They always insist on having their own way – in the form of common denominators before you can add or subtract. (Or, with cats, they get hissy.) And division? Don’t get me started!

      

Seriously, though, the best way to deal with linear equations that involve variables tangled up with fractions is to get rid of the fractions. Your game plan is to multiply both sides of the equation by the least common denominator of all the fractions in the equation.

      To solve

, for example, you multiply each term in the equation by 70, which is the least common denominator (also known as the least common multiple) for fractions with the denominators 5, 7, and 2:

      Now you distribute the reduced numbers over each parenthesis, combine the like terms, and solve for x:

and get 1 + 2 = 3 or 3 = 3. This one checks.

Isolating different unknowns

      When you see only one variable in an equation, you have a pretty clear idea what you’re solving for. When you have an equation like 4x + 2 = 11 or 5(3z – 11) + 4z = 15(8 + z), you identify the one variable and start solving for it.

      Life isn’t always as easy as one-variable equations, however. Being able to solve an equation for some variable when it contains more than one unknown can be helpful in many situations. If you’re repeating a task over and over – such as trying different widths of gardens or diameters of pools to find the best size – you can solve for one of the variables in the equation in terms of the others.

      The equation

for example, is the formula you use to find the area of a trapezoid. The letter A represents area, h stands for height (the distance between the two parallel bases), and the two b’s are the two parallel sides called the bases of the trapezoid.

      If you want to construct a trapezoid that has a set area, you need to figure out what dimensions give you that area. You’ll find it easier to do the many computations if you solve for one of the components of the formula first – for h, b₁, or b₂.

      To solve for h in terms of the rest of the unknowns or letters, you multiply each side by two, which clears out the fraction, and then divide by the entire expression in the parentheses:

      You can also solve for b₂, the measure of the second base of the trapezoid. To do so, you multiply each side of the equation by two, and then divide each side by h.

      

      Next, subtract b₁ from each side of the equation.

      

or

Paying off your mortgage with algebra

      A few years ago, one of my mathematically challenged friends asked me if I could help her figure out what would happen to her house payments if she paid $100 more each month on her mortgage. She knew that she’d pay off her house faster, and she’d pay less in interest. But how long would it take and how much would she save? I created a spreadsheet and used the formula for an amortized loan (mortgage). I made different columns showing the principal balance that remained (solved for P) and the amount of the payment going toward interest (solved for the difference), and I extended the spreadsheet down for the number of months of the loan. We put the different payment amounts into the original formula to see how they changed the total number of payments and the total amount paid. She was amazed. I was even amazed! She’s paying off her mortgage much sooner than expected!

      

When you rewrite a formula aimed at solving for a particular unknown, you can put the formula into a graphing calculator or spreadsheet to do some investigating into how changes in the individual values change the variable that you solve for (see a spreadsheet example of this in the “Paying off your mortgage with algebra” sidebar).

      Linear Inequalities: Algebraic Relationship Therapy

      Equations – statements with equal signs – are one type of relationship or comparison between things; they say that terms, expressions, or other entities are exactly the same. An inequality is a bit less precise. Algebraic inequalities show relationships between two numbers, a number and an expression, or between two expressions. In other words, you use inequalities for comparisons.

      Inequalities in algebra are less than (<), greater than (>), less than or equal to (< ), and greater than or equal to (> ). A linear equation has only one solution, but a linear inequality has an infinite number of solutions. When you write

, for example, you can replace x with 6, 5, 4, –3, –100, and so on, including all the fractions that fall between the integers that work in the inequality.

      

Here are the rules for operating on inequalities (you can replace the < symbol with any of the inequality symbols, and the rule will still hold):

      ✔ If a < b, then a + c < b + c (adding any number c).

      ✔ If a < b, then a – c < b – c (subtracting any number c).

      ✔ If a < b, then

(multiplying by any positive number c).

      ✔ If a < b, then

(multiplying by any negative number c).

      ✔ If a < b, then

(dividing by any positive number c).

      ✔ If a < b, then

(dividing by any negative number c).

      ✔ If

, then

(reciprocating fractions).

      Notice that the direction of the inequality changes only when multiplying or dividing by a negative number or when reciprocating (flipping) fractions.

      

You must not multiply or divide each side of an inequality by zero. If you do so, you create an incorrect statement. Multiplying each side of 3 < 4 by 0, you get 0 < 0, which is clearly a false statement. You can’t divide each side by 0, because you can never divide anything by 0 – no such number with 0 in the denominator exists.

Solving linear inequalities

      To solve a basic linear inequality, you first move all the variable terms to one side of the inequality and the numbers to the other. After you simplify the inequality down to a variable and a number, you can find out what values of the variable will make the inequality into a true statement. For example, to solve 3x + 4 > 11 – 4x, you add 4x