Example 1.   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.

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?

In equation 2), let us 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).

Example 2.   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?

Solution.   

Problem 1.   Samantha has 30 coins, 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 3.   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 4.   Mixture problem 2.   A saline soution 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

That is,

2s = 12.

s = 6.

Therefore, to 6 gallons of saline solution you must add 2 gallons of water.

Example 5.   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.