Each problem that I solved became a rule which served afterwards to solve other problems.
[ expand all ]
This module provides an overview of methods for simplifying and evaluating expressions, and solving equations. You will first learn why each method works—this will allow you to use your reasoning and check your answers as you complete problems that require use of simplification or evaluation. If you only memorize laws and properties, you’ll find that you have no basis for checking your answers or figuring out a law or property that you are unable to recall. As one simple example, students who only memorize the distributive property, $$a(b + c) = ab + ac$$ often forget to distribute the factor across both terms, writing $$a(b+c) = ab+c$$. As another example, students who only memorize that $$x^{m}x^{n}=x^{m+n}$$ and $$(x^m)^n=x^{(m)(n)}$$ often don’t understand why the exponents are added in the first instance and multiplied in the second, thus using one rule when the other applies, e.g. adding exponents when they should multiply.
The idea of a simplified expression is that you start with one expression and produce an equivalent expression that is somehow “simpler”. For example, when applying the distributive property to the expression $$3x(x + 5) + {y^2}\left( {x + 5} \right)$$ we obtain the expression $$\left(3x+y^2\right)\left(x+5\right)$$. The two expressions are equivalent—this means that no matter what value x and y assume, the values of both expressions are the same—but the second is generally easier to work with, and therefore we would consider it simpler or more desirable than the first. As another example, when working with a quadratic function or formula, it is sometimes preferable to write the function in factored form, while in other instances it is preferable to write the formula in vertex form. This is because different forms of an expression convey different information about the quantities represented in the formula or function. In this case, the factored form of a quadratic function conveys information about the roots or solutions and the vertex form conveys information about the maximum or minimum value of the function.
This section provides an overview of the properties, laws, and conventions that you need to understand and be able to use when presented with problems that require changing the form of an expression.
To practice reasoning with these properties, laws and exponents, select the following links.
Multiplying and Dividing Fractions
Simplifying Rational Expressions
Operations on Rational Expressions
Rationalizing Denominators and Numerators
A variable is a letter that represents a quantity’s value (e.g., the length of a square’s side measured in centimeters, the number of inches burned from a 10-inch candle). The variable takes on different values as the quantity’s value varies.
Consider an expression such as 10-x that represents the length of a candle (that is originally 10 inches long) after the candle has burned x inches. Evaluating this expression when $$x = 2.3$$ involves substituting 2.3 in for x and performing the operations given in the expression. In this case 10-2.3 returns a value of 7.7. This value represents the length of the candle after 2.3 inches of the candle has burned off.
To practice evaluating expressions select the link:
One way to express the relationship between two quantities is by using formulas. To evaluate a formula that relates the values of two quantities we substitute a number for one of the variables and calculate to determine the value of the other variable. The formula $$p=4s$$, where the letter p represent a square’s perimeter and the letter s represents a square’s side length, defines the perimeter of a square in terms of its side length. We evaluate the expression 4s by substituting specific values of the side length into s and calculating (multiply the value for s by 4) to determine the perimeter. As the values of s vary so do the values of p. The formula is one way of indicating how p varies as s varies.
A formula that relates two varying quantities provides a concise and efficient way of determining the value of one quantity when the other is known. Determining a formula that relates two quantities sometimes requires use of advanced mathematics, but once the formula is known it can be committed to memory and used repeatedly.
Determining the relationship between the radius r and volume V of a sphere originally required integral calculus. But now the formula $$V=\frac{4}{3}\pi r^3$$ is known and can be used repeatedly to determine any sphere’s volume V when the sphere’s radius r is known.
What is the volume of a sphere that has a radius of 2 cm?
Think about it for a moment and then access this link 
to view answer.
To practice evaluating formulas select the following link:
Since many formulas are used frequently when solving word problems it is important that you remember these formulas after you understand why they are true.
Some common formulas to relate quantities in the following geometric figures follow:
Rectangles: $$\eqalign{
A &= lw \cr
P &= 2l + 2w \cr} $$
Squares: $$\eqalign{
A &= {s^2} \cr
P &= 4s \cr} $$
Circles: $$\eqalign{
A &= \pi {r^2} \cr
C &= 2\pi r = \pi d \cr
d &= 2r \cr} $$
Cubes: $$\eqalign{ \rm{Surface\;Area\;(SA)} &=6s^2 \\
V &= s^3 \\}$$
Spheres: $$\eqalign{
& SA = 4\pi {r^2} \cr
& V = \frac{4}
{3}\pi {r^3} \cr} $$
Solving equations and inequalities involves determining the value or values of the variable that makes the equation or inequality true. It is also the case that solving equations can be viewed as determining the value of the input variable of a formula when the value of the output variable is known. For example, take the formula above for determining the volume V of a sphere when the value of the radius r is known. Evaluating this formula when r = 4.7 inches involves substituting 4.7 for r and computing the value of V, giving $$V = \frac{4}{3}\pi {(4.7)^3} \approx 434.89$$ inches.
In another instance you may know a sphere’s volume and want to know the sphere’s radius. This would involve substituting a value for V and solving the equation for r. That is, you are given the value of the output variable V and you perform the operations needed to isolate r on one side of the equal sign. When you have done this, you have solved the equation for r.
If you need a spherical ball with a volume of 250 cubic inches, determine the radius of the sphere.
Think about it for a moment and then access this link to view answer.
To practice solving linear equations select the following link:
To practice solving inequalities select the following link:
To practice solving equations with absolute values select the following link:
To practice solving absolute value inequalities select the following link:
To practice solving equations with rational exponents select the following link: