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Word problems that lead to
simultaneous equations

Section 2

Back to Section 1:  Examples

Problem 1.   A woman is now 30 years older than her son.  15 years ago, she was twice as old.  What are the present ages of the woman and her son?

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Let x be the present age of the woman.

Let y be the present age of her son.

Here are the equations:

1)   Present age of the woman
in relation to her son:
x = y + 30.
2)   15 years ago: x − 15 = 2(y − 15).

To eliminate x, simply substitute equation 1) in equation 2), and solve for y.

You should find:  y = 45 years.

Therefore x, the age of the woman, is 75 years.

Problem 2.   A total of 925 tickets were sold for $5,925.  If adult tickets cost $7.50, and children's tickets cost $3.00, how many tickets of each kind were sold?   (Compare Example 2.)

Let x be the number of adult tickets.   Let y be the number of childeren's tickets.

Here are the equations:

1)   Total number of tickets: x + y = 925
2)   Total money collected: 7.5x + 3y = 5,925

In equation 2), make the coefficients into whole numbers by multiplying both sides of the equation by 10:

1)   x + y = 925
2')   75x + 30y = 59,250

To eliminate y, for example:

Multiply equation 1) by −30, and add.

The solution is:   x = 700,  y = 225.

Problem 3.   Mr. B. has $20,000 to invest.  He invests part at 6%, the rest at 7%, and he earns $1,280 interest.  How much did he invest at each rate?   (Compare Example 3.)

Let x be how much he inveted at 6%.   Let y be how much he inveted at 7%.

Here are the equations:

1)   Total investment: x + y = 20,000
2)   Total interest: .06x + .07y = 1,280
2')     6x + 7y = 128,000

To eliminate x, for example, from equations 1) and 2'):

Multiply equation 1) by −6, and add.

The solution is:   x = $12,000.  y = $8,000.

Problem 4.   Edgar has 20 dimes and nickels, which together total $1.40. How many of each does he have?   (Compare the Problem in Section 1.)

Let x be the number of dimes.   Let y be be the number of nickels.

Here are the equations:

1)   Total number of coins: x + y = 20
2)   Total value: .10x + .05y = 1.40
2')     10x + 5y = 140

To eliminate x, for example, from equations 1) and 2'), multiply equation 1) by −10, and add.

The solution is:   x = 8 dimes.  y = 12 nickels.

Problem 5.   How many gallons of 20% alcohol solution  and how many of 50% alcohol solution must be mixed  to produce 9 gallons of 30% alcohol solution?   (Compare Example 4.)

(9 gallons of 30% alcohol solution = .3 × 9 = 2.7 gallons of pure alcohol.)

Let x be the number of gallons of 20% solution.   Let y be the number of gallons of 50% solution.

Here are the equations:

1)   Total number of gallons: x + y = 9
2)   Total gallons of pure alcohol: .2x + .5y = 2.7
2')     2x + 5y = 27

To eliminate x, for example, from equations 1) and 2'), multiply equation 1) by −2, and add.

The solution is:   x = 6 gallons.  y = 3 gallons.

Problem 6.   15 gallons of 16% disenfectant solution is to be made from 20% and 14% solutions. How much of those solutions should be used?

(15 gallons of 16% solution = .16 × 15 = 2.4 gallons of pure disenfectant.)

Let x be the number of gallons of 20% solution.   Let y be the number of gallons of 14% solution.

Here are the equations:

1) Total number of gallons: x + y = 15
2) Total gallons of pure disenfectant:  .20x + .14y = 2.4
2')   20x + 14y = 240

To eliminate x, for example, from equations 1) and 2'), multiply equation 1) by −20, and add.

The solution is:   x = 5 gallons.  y = 10 gallons.

Problem 7.   It takes a boat 2 hours to travel 24 miles downstream  and 3 hours to travel 18 miles upstream. What is the speed of the boat in still water, and how fast is the current?   (Compare Example 6.)

Let x be the speed of the boat in still water.   Let y be the speed of the current.

Here are the equations:

1)   Downstream speed:  x + y = 24
 2
= 12
2)   Upstream speed: x y = 18
 3
= 6

To eliminate y, simply add the equations.

The solution is:   x = 9 mph.  y = 3 mph.

Problem 8.   An airplane covers a distance of 1500 miles in 3 hours
when it flies with the wind, and in 3 1
3
 hours when it flies against the

wind.  What is the speed of the plane in still air?   (Compare Example 6.)

Let x be the speed of the plane in still air.  Let y be the speed of the wind.

Here are the equations:

1)   Speed with the wind: x + y = 1500
   3
  =  500
2)   Speed against the wind : x y = word problem

To eliminate y, simply add the equations.

The solution is:   x = 475 mph.

Back to Section 1

end

Next Lesson:  Quadratic equations

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