# WORD PROBLEMS ON SYSTEMS OF EQUATIONS

Problem 1 :

A person has \$8.80 in pennies and nickels. If there are twice as many nickels as pennies, how many pennies does Calvin have ?, How many nickels ?

Solution :

In this kind of problem, it’s good to do everything in cents to avoid having to work with decimals. So the person has 880 cents total.

Let p be the number of pennies.

Then, the number of nickels is 2p.

Now, we can arrange the information in a table as shown below.

The total value of the coins (880) is the value of the pennies plus the value of the nickels.

From the last column of the table, we have

p + 10p  =  880

11p  =  880

Divide each side by 11.

p  =  80

2p  =  2(80)  =  160

So, the person has 80 pennies and 160 nickels.

Problem 2 :

A total of 78 seats for a concert are sold, producing a total revenue of \$483. If seats cost either \$2.50 or \$10.50, how many \$2.50 seats and how many \$10.50 seats were sold ?

Solution :

Let x be the number \$2.50 seats and y be the number of \$10.50 seats.

Now, we can arrange the information in a table as shown below.

From the second and last columns of the table, we have

x + y  =  78  and  2.5x + 10.5y  =  483

Multiply the second equation by 10 to clear decimals :

x + y  =  78  and  25x + 105y  =  4830

Divide the second equation by 5 and simplify :

x + y  =  78  and  5x + 21y  =  966

Multiply the first equation by -5 :

-5x - 5y  =  -390  and  5x + 21y  =  966

Add the above two equations :

16y  =  576

Divide each side by 16 :

y  =  36

Substitute 36 for y in the first equation :

x + 36  =  78

Subtract 36 from each side.

x  =  42

So, 42 of the \$2.50 seats and 36 of the \$10.50 seats were sold.

Problem 3 :

Jacob has some 32-cent stamps, some 29-cent stamps, and some 3-cent stamps. The number of 3-cent stamps is 5 less than the number of 29-cent stamps, while the number of 29-cent stamps is 10 less than the number of 32-cent stamps. The total value of the stamps is \$9.45. How many of each stamp does he have ?

Solution :

Let x be the number of 32-cent stamps, y be the number of 29-cent stamps and z be the number of 3-cent stamps.

Now, we can arrange the information in a table as shown below.

From the last column of the table, we have

32x + 29y + 3z  =  945 ----(1)

The number of 3-cent stamps is 5 less than the number of 29-cent stamps, so

z  =  y − 5 ----(2)

The number of 29-cent stamps is 10 less than the number of 32-cent stamps, so

y  =  x − 10 ----> y + 10  =  x

x  =  y + 10 ----(3)

Substitute (y - 5) for z and (y + 10) for x in (1).

(1)---->  32(y + 10) + 29y + 3(y - 5)  =  945

32y + 320 + 29y + 3y - 15  =  945

64y + 305  =  945

Subtract 305 from each side.

64y  =  640

Divide each side by 64.

y  =  10

Substitute 10 for y in (2).

(2)---->  z  =  10 - 5

z  =  5

Substitute 10 for y in (3).

(3)---->  x  =  10 + 10

x  =  20

So, Jacob has 20 32-cent stamps, 10 29-cent stamps, and 5 3-cent stamps.

Problem 4 :

Peterson mixes candy that sells for \$2.00 per pound with candy that costs \$3.60 per pound to make 50 pounds of candy selling for \$2.16 per pound. How many pounds of each kind of candy did he use in the mix ?

Solution :

Let x be the number of pounds of \$2 candy and y be the number of pounds of \$3.60 candy.

Now, we can arrange the information in a table as shown below.

From the second and last columns of the table, we have

x + y  =  50  and  2x + 3.6y  =  108

Multiply the second equation by 10 to clear decimals :

x + y  =  50  and  20x + 36y  =  1080

Divide the second equation by 4 and simplify :

x + y  =  50  and  5x + 9y  =  270

Multiply the first equation by -5 :

-5x - 5y  =  -250  and  5x + 9y  =  270

Add the above two equations :

4y  =  20

Divide each side by 4 :

y  =  5

Substitute 5 for y in the first equation :

x + 5  =  50

Subtract 5 from each side.

x  =  45

So, Peterson mixed 45 pounds of the \$2 candy and 5 pounds of the \$3.60 candy.

Problem 5 :

Jonathan mixes an alloy containing 14% silver with an alloy containing 24% silver to make 100 pounds of an alloy with 18% silver. How many pounds of each kind of alloy did he use ?

Solution :

Let x be the number of pounds of 14% alloy and y be the number of pounds of 24% alloy.

Now, we can arrange the information in a table as shown below.

From the second and last columns of the table, we have

x + y  =  100  and  0.14x + 0.24y  =  18

Multiply the second equation by 100 to clear decimals :

x + y  =  100  and  14x + 24y  =  1800

Divide the second equation by 2 and simplify :

x + y  =  100  and  7x + 12y  =  900

Multiply the first equation by -7 :

-7x - 7y  =  -700  and  7x + 12y  =  900

Add the above two equations :

5y  =  200

Divide each side by 5 :

y  =  40

Substitute 40 for y in the first equation :

x + 40  =  100

Subtract 40 from each side.

x  =  60

So, Jonathan mixed 60 pounds of the 14% alloy and 40 pounds of the 24% alloy.

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