Algebra II For Dummies. Sterling Mary Jane

      In this equation, the one variable is the x. The a, b, and c are coefficients and constants. (If you go to Chapter 12, you can see linear equations with two or three variables.) But, no matter how many variables you see, the common theme to linear equations is that each variable has only one solution or value that works in the equation.

      The graph of the single solution of a linear equation, if you really want to graph it, is one point on the number line – the answer to the equation. When you up the ante to two variables in a linear equation, the graph of all the solutions (there are infinitely many) is a straight line. Any point on the line is a solution. Three variable solutions means you have a plane – a flat surface.

      

Generally, algebra uses the letters at the end of the alphabet for variables; the letters at the beginning of the alphabet are reserved for coefficients and constants.

      

To solve a linear equation, you isolate the variable on one side of the equation. You do so by adding the same number to both sides – or you can subtract, multiply, or divide the same number on both sides.

      For example, you solve the equation 4x – 7 = 21 by adding 7 to each side of the equation, to isolate the variable and the multiplier, and then dividing each side by 4, to leave the variable on its own:

      When a linear equation has grouping symbols such as parentheses, brackets, or braces, you deal with any distributing across and simplifying within the grouping symbols before you isolate the variable. For instance, to solve the equation 5x – [3(x + 2) – 4(5 – 2x) + 6] = 20, you first distribute the 3 and –4 inside the brackets:

      You then combine the terms that combine and distribute the negative sign (–) in front of the bracket; it’s like multiplying through by –1:

      Simplify again, and you can solve for x:

      

When distributing a number or negative sign over terms within a grouping symbol, make sure you multiply every term by that value or sign. If you don’t multiply each and every term, the new expression won’t be equivalent to the original.

      To check your answer from the previous example problem, replace every x in the original equation with –2. If you do so, you get a true statement. In this case, you get 20 = 20. The solution –2 is the only answer that works – focusing your work on just one answer is what’s nice about linear equations.

Clearing out fractions

      The 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:

Extraneous (false) solutions can occur when you alter the original format of an equation. When working with fractions and changing the form of an equation to a more easily solved form, always check your answer in the original equation. For the previous example problem, you insert x = 3 into

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.

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