THAT LEAD TO
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?
Solution. Let x be the amount of money that Andre has. Let y be the amount that Bob has.
Always let x and y answer the question -- and be perfectly clear about what they represent!
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."
1) x − 20 = y + 20.
(Andre -- x -- has the same amount as Bob, after he gives him $20.)
Here is the second sentence:
"While if Bob gave Andre $22, Andre would then have twice as much
2) x + 22 = 2(y − 22).
(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 x --
-- and substitute it into equation 2):
Bob has $106. Therefore, according to the exression for x, Andre has
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?
Solution. Let x be the number of adult tickets. Let y be the number of children's tickets.
Again, we have let x and y answer the question. And again we must get two equations out of the given information. Here they are:
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: x = 700, y = 300.
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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: x = $10,000, y = $20,000.
Problem. Samantha has 30 coins, consisting of quarters and dimes, which total $5.70. How many of each does she have?
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Let x be the number of quarters. Let y be the number of dimes.
The equations are:
To eliminate y:
Multiply equation 1) by −10 and equation 2) by 100:
Therefore, y = 30 − 18 = 12.
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 x be the number of gallons of 30% solution. Let y be the number of gallons of 60% solution.
Equations 1) and 2') are the two equations in the two unknowns.
The solutions are: x = 6 gallons, y = 12 gallons.
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.)
Solution. Let s be the number of gallons of saline solution. Now all the salt will come from those s gallons. So the question is, What is s so that 20% of s -- the salt -- will be 15% of 8 gallons?
.2s = .15 × 8 = 1.2
2s = 12.
s = 6.
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.
Solution. Let x be the speed of the boat (without a current). Let y be the speed of the current.
The student might review the meanings of "upstream" and "downstream," Lesson 25. We saw there that speed, or velocity, is distance divided by time:
Therefore, according to the problem:
Here are the equations:
(The solutions are: x = 12 mph, y = 3 mph.)
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