306 algebra De mystif ieD
Summary
In this chapter, we learned how to:• Solve percent problems. If we increase x by p × 100% (with p written as a
decimal number), the new amount is x + px. If we decrease x by p × 100%,
the new amount is x – px.
• Solve problems with a known formula. If a problem involves a known for-
mula, such as a formula from geometry, we use information in the prob-
lem to eliminate all but one variable, giving us an equation to solve.
• Solve number sense problems having two unknowns. We are given two sets
of information about a pair of numbers. One of these sets of information
involves a sum. We use the information in the other set to eliminate one
of the variables in the sum, giving us an equation to solve.
• Solve number sense problems having three unknowns. We are told the sum
of the three numbers. The numbers are compared to each other. We use
these comparisons to write the sum using a single variable, giving us an
equation to solve. If two variables are compared to a third, we let the vari-
able represent this number and write the other two numbers in terms of
this number.
• Solve problems involving coins. Coin problems are similar to number sense
problems. Instead of the being told the sum of the numbers, we are told a
total amount of money. We let the variable represent the number of one
of the coins and write the number of the other coins in terms of the same
variable. We then multiply the number of each coin by the value of the
coin. The sum of these values is the total amount of money.
• Solve investment problems. Money is divided into two different investments,
paying different interest rates. We are told the interest earned. We represent the
amount of one of the investments with the variable. If A and B are the interest
rates, written as decimal numbers and x the amount invested at interest A,
then we solve the following equation: Ax+−Bx()TotalI=nterestearned.^
• Solve mixture problems. Mixture problems involve mixing together two
different concentrations of a solution to produce the final concentration.
We let x represent the amount of one of the concentrations. We know the
amount of either the other concentration or the final concentration. If we
know the amount of the second concentration, the amount of the final
concentration is “x + Amount of the second concentration.” If we know