36 ## WORD PROBLEMSTHAT LEAD TO ## SIMULTANEOUS EQUATIONSSection 1: Examples HERE ARE SOME EXAMPLES of problems that lead to simultaneous equations. Example 1. Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob. How much does each one actually have?
y be the .amount that Bob hasAlways let Now there are two unknowns. Therefore there must be two equations. (In general, the number of equations must equal the number of unknowns.) How can we get two equations out of the given information? We must translate each verbal sentence into the language of algebra. Here is the first sentence: "If Andre gave Bob $20, they would have the same amount." Algebraically: 1) (Andre -- Here is the second sentence: "While if Bob gave Andre $22, Andre would then have twice as much Algebraically: 2) (Andre has twice as much as Bob -- after Bob gives him $22.) To solve any system of two equations, we must reduce it to one equation in one of the unknowns. In this example, we can solve equation 1) for
-- and substitute it into equation 2):
Bob has $106. Therefore, according to the exression for 106 + 40 = $146. Example 2. 1000 tickets were sold. Adult tickets cost $8.50, children's cost $4.50, and a total of $7300 was collected. How many tickets of each kind were sold?
y be the .number of children's ticketsAgain, we have let
In equation 2), we will make the coefficients into whole numbers by multiplying both sides of the equation by 10:
We call the second equation 2' ("2 prime") to show that we obtained it from equation 2). These simultaneous equations are solved in the usual way. The solutions are:
To see the answer, pass your mouse over the colored area. Example 3. Mrs. B. invested $30,000; part at 5%, and part at 8%. The total interest on the investment was $2,100. How much did she invest at each rate?
(To change a percent to a decimal, see Skill in Arithmetic, Lesson 4.) Again, in equation 2) let us make the coefficients whole numbers by multiplying both sides of the equation by 100:
These are the simultaneous equations to solve. The solutions are:
Problem. Samantha has 30 coins, consisting of quarters and dimes, which total $5.70. How many of each does she have? To see the answer, pass your mouse from left to right Let The equations are:
To eliminate Multiply equation 1) by −10 and equation 2) by 100:
Therefore, Example 4. Mixture problem 1. First: "36 gallons of a 25% alcohol solution" means: 25%, or one quarter, of the solution is pure alcohol. One quarter of 36 is 9. That solution contains 9 gallons of pure alcohol. Here is the problem: How many gallons of 30% alcohol solution and how many of 60% alcohol solution must be mixed to produce 18 gallons of 50% solution? "18 gallons of 50% solution" means: 50%, or half, is pure alcohol. The final solution, then, will have 9 gallons of pure alcohol. Let y be the of 60% solution.number of gallons
Equations 1) and 2') are the two equations in the two unknowns. The solutions are:
Example 5. Mixture problem 2. A saline solution is 20% salt. How much water must you add to how much saline solution, in order to dilute it to 8 gallons of 15% solution? (This is more an arithmetic problem than an algebra problem.)
.2 That is, 2
Therefore, to 6 gallons of saline solution you must add 2 gallons of water. Example 6. Upstream/Downstream problem. It takes 3 hours for a boat to travel 27 miles upstream. The same boat can travel 30 miles downstream in 2 hours. Find the speeds of the boat and the current.
The student might review the meanings of "upstream" and "downstream," Lesson 25. We saw there that speed, or velocity, is
Therefore, according to the problem:
Here are the equations:
Enjoy! (The solutions are:
Please make a donation to keep TheMathPage online. Copyright © 2015 Lawrence Spector Questions or comments? E-mail: themathpage@nyc.rr.com |