Question

and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question. A jar contains 50 nickels and dimes worth $4.20. How many of each kind of coin are in the jar? x = nickels y = dimes
7. Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- and y-variables. Solve the system by the elimination method. Be sure to state your final answer in terms of the original question.


The concession stand at an ice hockey rink had receipts of $7000 from selling a total of 3000 sodas and hot dogs. If each soda sold for $2 and each hot dog sold for $3, how many of each were sold?


x = sodas

y = hot dogs

8. Carry out the row operation on the matrix.


R1 â R2

Answer 1

Answer:

A. x = 16 , y = 34

B x = 2000, y = 1000

Step-by-step explanation:

Solution:-

- We will first define our variables ( x and y ) as follows:

                     x: The number of "nickels" in the jar

                     y: The number of "dimes" in the jar.

- Nest we will write down the rates of nickel and dime in dollar equivalent as ( P_n and P_d, respectively )  as follows:

                    P_n = $0.05  ... ( 5 cents )

                    P_d = $0.10   ... ( 10 cents )

- We are told that the jar contains a total of "50" coins comprised of nickel and dimes. Since, we don't know the exact amount of nickel and dimes in the jar. We will express the statement mathematically using the previously defined variables as follows:

                     $x + y = 50$     ... Eq1

- Secondly, the total worth of the jar is given to be " $4.2 ". The respectively value of each coin was iterated above. We will compute the total worth of the jar by expressing in terms of x and y as follows:

                    $P_n*x + P_d*y = 4.2\\\\0.05x + 0.1*y = 4.2$  ... Eq 2

- We have two equations [ Eq1 and Eq2 ] comprising of two variables. We can solve them simultaneously for a unique solution ( x and y ).

- To solve by elimination. We will first multiply the [ Eq1 ] by "- 0.1 " throughout as follows:

                    $-0.1x - 0.1y = -5\\\\0.05x + 0.1y = 4.2$

- Now we will add the two equations and eliminate the variable " y " and solve for " x ":

                    $-0.05x = -0.8\\\\x = 16$

- Now plug the value of " x " in either of the derived equations and solve for "y":

                    $y = 50 - 16\\y = 34$

Answer: There are 16 nickels and 34 dimes in the jar of total worth $4.2.

- We will first define our variables ( x and y ) as follows:

                     x: The number of "sodas" sold

                     y: The number of "hot dogs" sold.

- Nest we will write down the rate charged for soda and hot-dogs equivalent as ( P_s and P_h, respectively )  as follows:

                    P_s = $2 / soda

                    P_h = $3 / hot-dog

- We are told that " 3000 " sodas and hot-dogs were sold at the concession stand . Since, we don't know the exact amount of sodas and hot-dogs sold. We will express the statement mathematically using the previously defined variables as follows:

                     $x + y = 3000$   ... Eq1

- Secondly, the total amount expressed on the receipts after selling "x" many sodas and " y " many hot--dogs was " $7000 ". The respectively value of each commodity sold was iterated above. We will compute the total value of items sold by expressing in terms of x and y as follows:

                    $P_s*x + P_h*y = 7000\\\\2x + 3y = 7000$ ... Eq 2

- We have two equations [ Eq1 and Eq2 ] comprising of two variables. We can solve them simultaneously for a unique solution ( x and y ).

- To solve by elimination. We will first multiply the [ Eq1 ] by "-2 " throughout as follows:

                    $-2x - 2y = -6000\\\\2x + 3y = 7000$

- Now we will add the two equations and eliminate the variable " x " and solve for " y ":

                    $y = 1000$

- Now plug the value of " y " in either of the derived equations and solve for "x":

                    $x = 3000 - 1000\\\\x = 2000$

Answer: The concession sold 2000 sodas and 1000 hot-dogs of total worth $7000.