Solving a System of two Linear Equations.

Question

A coin purse contains dimes and quarters. The number of dimes is 5 less than twice the number of quarters. The total value is $4.00. Find the number od dimes and quarters.

 

Part A. Interpret the question and express the problem in Mathematical language.

 

Part B. Procedural Math. 

 

Answer 1

Total value = 4.00, let d = number of dimes, let q= number of quarters

 

We have also been given that d = 2q - 5

 

Total value = (0.10)(d) + (0.25)(q) = 4.00

 

Since d = 2q - 5, substitute for d

(0.10)(2q - 5) + 0.25q = 4.00

 

Distribute the 0.10 to get 0.20q -0.50 + 0.25q = 4.00

 

Combine like terms to get 0.45q - 0.50 = 4.00

 

                                      0.45q = 4.50

 

                                             q = 10           d = 2q - 5

                                                                 d = 15

 

There are 15 dimes and 10 quarters.

Answer 2

number of dimes (x) and number of quarters (y)

 

.10x + .25y = 4.00  This means that x amount of dimes plus y amount of quarters adds up to four dollars

 

x=2y-5 this expresses quarters in terms of dimes. It allows us to solve an equation with two variables.

 

we can put this in the equation in place of x

our new equation will look like this

 

 

.10(2y-5)+.25y=4.00

 

.20y - .50 + .25y = 4.00 now we add the like terms

 

.45y - .50 = 4.00 now we isolate y on one side of the equation

 

.45y = 4.50 Now we divide by .45

 

y = 10.  There are ten quarters.  we use this to find the number of dimes. we can put this number in our expression from earlier to find x

 

x=2y-5

x=2(10)-5

x= 15              there are fifteen dimes.  

 

Now we can put both of these numbers into our original equation to check our work

 

 

.10x + .25y = 4.00

.10(15) + .25(10)=4.00

1.50 + 2.50 = 4.00   This statement is true and therefore our math is correct.