6 Complex Fractions
The first step in simplifying complex fractions is to write the numerator as one
the denominator as one fraction.
Here are some to try. Answers are in the solution section.
For the function defined by f(x) = 3x - x2, we express in English that the
function takes a
number, multiplies it by 3 and subtracts its square.
For the function defined by g(x) = 1 - 3x2, we express in English that the
function takes a
number, squares it, multiplies the result by 3 and subtracts that result from 1.
Here are some for you to try. The answers are in the solution section.
1. If f(x) = 2x - x2, find f(x + h) and simplify completely.
2. If find f(x + h) and simplify completely.
3. If f(x) = 3 - 2x2, find f(x + h) and simplify completely.
8 Linear Equations
Rewrite the equation so that there are no variables in the denominator and no
Move all terms with the variable for which we are solving on one side of the equation and
move all terms without that variable on the other side. Factor out the variable and solve.
Example: Solve 3(x + 2) = 4x + 1 for x: We distribute to get 3x + 6 = 4x + 1.
x's on one side, we get 3x - 4x = 1 - 6 or -x = -5. So the answer is x = 5.
for y: Multiply both sides by 2x to get y = 2x(y + 1) or
y = 2xy + 2x. Isolate the terms with y to get y - 2xy = 2x or y(1- 2x) = 2x. The solution
Try the following problems. The answers are in the solution section.
2. Solve A(B + C) = BC + A for C.
9 Quadratic Equations
We can solve quadratic equations by factoring or by using the quadratic equation.
Example: Solve x2-x = 6 by factoring: We put all terms on one side, getting
x2-x-6 = 0.
Next, factor to get (x - 3)(x + 2) = 0. If a product ab equals 0, either a = 0 or b = 0. For
our problem, we conclude x - 3 = 0 or x + 2 = 0. The answer is x = 3 or x = -2.
Notice that, if ab = 6, we cannot conclude a = 6 or b = 6. Maybe a = 2 and b = 3
or a =1/2
and b = 12. So, if you factored without moving all terms to one side to get x(x - 1) = 6,
you cannot conclude x = 6 or x - 1 = 6.
Example: Solve x2 -x = 6 by using the quadratic formula: Again, we need to put
on one side, getting x2 - x - 6 = 0. To solve ax2 + bx + c = 0, the quadratic formula tells
In this example, a = 1, b = -1, and c = -6, so
. The solution is
Here are some problems to try. The answers are in the solution section.
1. Solve the equation for x: x2 - 2x = 8
2. Solve the equation for x:
10 Common Mistakes
Remember that For example,
Similarly, However, if x and y are nonnegative, and
In a similar vein, since
. However, since
Care needs to be taken when deciding if canceling is possible. We have that but
Recall that, for inequalities, if we multiply or divide both sides by a negative
inequality sign gets changed. We know that -3 ≤ 2 but, if we multiply by -2, we get
(-3)(-2) ≥ (2)(-2) since 6 ≥ -4.
Example: Solve 3 - 2x ≤ 9 for x: We get -2x ≤ 9 - 3 or -2x
≤ 6 or x ≥ -3. The
in interval notation is [-3;∞).
Example: Solve x + 2 > 4x - 1 for x: We get x - 4x >
-1 - 2 or -3x > -3 or
x < 1. The
answer in interval notation is (-∞; 1).
Decide if the following statements are true or false. Assume that all variable
Answers are in the solution section.
1. and the equation of the line is y - (-2) = -1(x - 4).
2. The points are (3, 0) and (0,-2). So
and the equation of the
3. To find the slope of the line 2x + 6y = 1, we solve for y: 6y = 1
- 2x so
The slope is and the equation of the parallel line is
1. Multiplying the equation by ABC on both sides, we get BC +AC = AB.
We isolate the terms with B on the left to get BC-AB = -AC. So B(C-A) = -AC
2. Distributing, we get AB +AC = BC +A. We isolate the terms with C on the left
get AC - BC = A - AB. So C(A - B) = A - AB and
1. x2 - 2x - 8 = 0. Factoring, we get (x - 4)(x + 2) = 0 so x
- 4 = 0 or x +
2 = 0. The
answers are x = 4,-2.
2. Multiplying both sides of the equation by x, we get 2+x = x2 +2x. So 0 = x2
+ x - 2
and 0 = (x + 2)(x - 1). The answer is x = -2, 1.
4. False: (2 + 3)3 = 53 = 125
5. False: If -2x < 4 then x > -2.
11. False: (3x)2 = 9x2