One of the most useful structures in mathematics is the proportion. You use it to parcel out amounts in fair and equal portions and to determine equivalent lengths, volumes, or amounts of money. A proportion is nothing more than two fractions set equal to one another, but it’s beautifully symmetric and functionally versatile
Solving Proportions for Missing Values
Proportions and their properties find their way into many financial and scientific applications. For instance, determining doses of medicine incorporates proportions involving a person’s weight. Vegetation and animal habitats are in a balance expressed with proportions as well. And figuring a seller’s share of the real estate taxes involves ratios and proportions.
Setting up and solving
The biggest challenge when solving a proportion for an unknown value is in setting the proportion up correctly. (I detail the correct way to set up a proportion in the earlier section, “Setting Up Proportions.”) When you have an unknown value, you let that value be represented by a variable, usually x, you cross-multiply, and then you solve the resulting equation for that variable. The following example gives you a chance to solve a proportion for a missing value.
Interpolating when necessary
Interpolation is basically inserting numbers between entries in a table. You assume that the change from one entry to another is pretty uniform and that you can find a number halfway between two numbers on the table by halving the difference. Proportions allow you to find numbers between the entries that are more or less than halfway between.
Handling Basic Linear Equations
When solving proportions for an unknown value, and when performing many mathematical computations involving measures, money, or time, you’ll likely find yourself working with an equation that has a variable in it. You solve for the value of the unknown by applying basic algebraic processes. These processes keep the integrity of the equation while they change its format so you can find the value of the unknown
Going the indirect route with indirect variation
When a 2 x 4 pine board is suspended between two buildings (with just the ends of the board at the building edges), the maximum weight that the board can support varies indirectly with the distance between the buildings. If the distance between the buildings is 10 feet, the board can support 480 pounds.
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