Solve value problems by setting up a system of equations

One application of system of equations are known as value problems. Value problems are ones where each variable has a value attached to it. For example, if our variable is the number of nickels in a person’s pocket, those nickels would have a value of five cents each. We will use a table to help us set up and solve value problems. The basic structure of the table is shown below.

Also Read : Solve systems of equations using the addition/elimination method

Also Read:

The first column in the table is used for the number of things we have. Quite often, this will be our variables. The second column is used for the that value each item has. The third column is used for the total value which we calculate by multiplying the number by the value. For example, if we have 7 dimes, each with a value of 10 cents, the total value is 7 · 10 = 70 cents. The last row of the table is for totals. We only will use the third row (also marked total) for the totals that are given to use. This means sometimes this row may have some blanks in it. Once the table is filled in we can easily make equations by adding each column, setting it equal to the total at the bottom of the column. This is shown in the following example.

Also Read : Solve systems of equations using substitution

Example 1:

In a child’s bank are 11 coins that have a value of S1.85. The coins are either quarters or dimes. How many coins each does child have?

Solution:

Using value table, use q for quarters, d for dimes Each quarter’s value is 25 cents, dime’s is 10 cents

Multiply number by value to get totals

We have 11 coins total. This is the number total.

We have 1.85 for the final total,

Write final total in cents (185)

Because 25 and 10 are cents

First and last columns are our equations by adding

q + d =11

25q + 10d =185

Solve by either addition or substitution.

Using addition, multiply first equation by -10

-10 (q + d) =(11)( -10)

-10q -10d =-110

Add together equations

Divide both sides by 15

q = 5

We have our q, number of quarters is 5

Plug into one of original equations

(5) + d =11

Subtract 5 from both sides

d = 6

We have our d, number of dimes is 6.

World View Note: American coins are the only coins that do not state the value of the coin. On the back of the dime it says “one dime” (not 10 cents). On the back of the quarter it says “one quarter” (not 25 cents). On the penny it says “one cent” (not 1 cent). The rest of the world (Euros, Yen, Pesos, etc) all write the value as a number so people who don’t speak the language can easily use the coins.

Ticket sales also have a value. Often different types of tickets sell for different prices (values). These problems can be solve in much the same way.

Example 2:

There were 41 tickets sold for an event. Tickets for children cost S1.50 and tickets for adults cost S2.00. Total receipts for the event were S73.50. How many of each type of ticket were sold?

Solution:

Using our value table, c for child, a for adult Child tickets have value 1.50, adult value is 2.00 (we can drop the zeros after the decimal point)

Multiply number by value to get totals

We have 41 tickets sold. This is our number total

The final total was 73.50

Write in dollars as 1.5 and 2 are also dollars

First and last columns are our equations by adding

c + a = 41

1.5c + 2a =73.5

We can solve by either addition or substitution

We will solve by substitution.

c + a = 41

Solve for a by subtracting c

c – c + a = 41 – c

a = 41 – c

Substitute into untouched equation

1.5c + 2(41 – c) = 73.5

Distribute

1.5c + 82 – 2c = 73.5

Combine like terms

-0.5c + 82 = 73.5

Subtract 82 from both sides

-0.5c = -8.5

Divide both sides by -0.5

c = 17

We have c, number of child tickets is 17

Plug into a = equation to find a

a = 41 – (17)

a = 24

We have our a, number of adult tickets is 24.

Our solution is 17 child tickets and 24 adult tickets.

Some problems will not give us the total number of items we have. Instead they will give a relationship between the items. Here we will have statements such as “There are twice as many dimes as nickels”. While it is clear that we need to multiply one variable by 2, it may not be clear which variable gets multiplied by 2. Generally the equations are backwards from the English sentence. If there are twice as many dimes, than we multiply the other variable (nickels) by two. So the equation would be d = 2n. This type of problem is in the next example.

Example 3:

A man has a collection of stamps made up of 5 cent stamps and 8 cent stamps. There are three times as many 8 cent stamps as 5 cent stamps. The total value of all the stamps is S3.48. How many of each stamp does he have?

Solution:

Use value table, f for five cent stamp, and e for eight Also list value of each stamp under value column

Multiply number by value to get total

The final total was 338(written in cents)

We do not know the total number, this is left blank.

3 times as many 8cent stamps as 5cent stamps

e = 3f

Total column gives second equation

5f + 8e = 348

Substitution, substitute first equation in second

5f + 8(3f) =348

Multiply first

5f +24f =348

Combine like terms

29f = 348

Divide both sides by 29

f =12

We have f. There are 12 five cent stamps

Plug into first equation

e =3(12)

e = 36

We have e, There are 36 eight cent stamps

Our solution is 12 five cent, 36 eight cent stamps.

The same process for solving value problems can be applied to solving interest

problems. Our table titles will be adjusted slightly as we do so.

Our first column is for the amount invested in each account. The second column is the interest rate earned (written as a decimal – move decimal point twice left), and the last column is for the amount of inter set earned. Just as before, we multiply the investment amount by the rate to find the final column, the interest earned. This is shown in the following example.

Example 4:

A woman invests S4000 in two accounts, one at 6% interest, the other at 9% interest for one year. At the end of the year she had earned S270 in interest. How much did she have invested in each account?

Solution:

Use our investment table, x and y for accounts

Fill in interest rates as decimals

Multiply across to find interest earned.

Total investment is 4000,

Total interest was 276

First and last column give our two equations

x + y = 4000

0.06x + 0.09y = 270

Solve by either substitution or addition

Use Addition, multiply first equation by -0.06

-0.06(x + y) =(4000)( -0.06)

-0.06x -0.06y =-240

Add equations together

Divide both sides by 0.03

y =1000

We have y, S1000 invested at 9%

Plug into original equation

x + 1000 = 4000

Subtract 1000 from both sides

x = 3000

We have x, S3000 invested at 6%

And our solution is S1000 at 9%and S3000 at 6%.

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