Help with ALGEBRA?

Question

I need help on two problems I couldn't figure out.

Two groups of students order burritos and tacos at a local restaurant. One order of 3 burritos and 4 tacos costs $11.33. The other order of 9 burritos and 5 tacos costs $23.56.
a. Write a system of equations that decribes this situation.
b. Solve by elimination to find the cost of a burrito and the cost of a taco.

a. A company sells brass and steel machine parts. One shipment contains 3 brass and 10 steel parts and costs $48. A second shipment contains 7 brass and 4 steel parts and costs $54. Find the cost of each type of machine part.
b. How much would a shipment containing 10 brass and 13 steel machine parts costs?

Answer 1

Problem One:
3 burritos and 4 tacos cost 11.33 3b + 4t = 11.33
9 burritos and 5 tacos cost 23.56 9b + 5t = 23.56

To solve: multiply top equation by 3.
3(3b + 4t = 11.33) becomes 9b + 12t = 33.99
9b + 5t = 23.56 stays as is 9b + 5t = 23.56

Now subtract the second equation from the first and you will get
7t = 10.43 which you can solve by dividing both sides by 7 to get t = 1.49 so a taco costs $1.49.

Now replace a t in either equation with 1.49 so 3b + 4t = 11.33 becomes 3b +4(1.49) = 11.33 or 3b + 5.96 = 11.33. Now subtract 5.96 from each side leaving 3b = 5.37 and finally divide both sides by 3 to get b = 1.79 so a burrito costs $1.79

Problem Two:
3 brass and 10 steel costs $48 3b + 10s = 48
7 brass and 4 steel costs $54 7b + 4s = 54

To solve: multiply top equation by 2 and the bottom equation by 5 so the number of s's will be equal

2(3b + 10s = 48) becomes 6b + 20s = 96
5(7b + 4s = 54) becomes 35 b + 20s = 270

Now subtract the top equation from the bottom equation and you will get 29b = 174. Divide both sides by 29 to get b = 6.

Now replace the b of either original equations with the 6 (say the top equation) to get 3(6) + 10S = 48 or written another way,
10s + 18 = 48. Now subtract 18 from both sides to get 10s = 30 and then divide both sides by 10 to get s = 3.

b=6 and s=3 so each brass part is $6 and each steel part is $3.

Answer 2

First problem:

Let b = burritos
let t = tacos

3b + 4t = 11.33 (3 burritos and four tacos is 11.33)
9b + 5t = 23.56 (9 burritos and 5 tacos is 23.56)

First you need to eliminate one variable by multiplying the euqations to get the coefficients to be opposites. If we multiply the top equation by -3, then we'll have -9b.

-9b - 12t = - 33.99
9b + 5t = 23.56

Add the colums.
- 7t = -10.43

Divide both sides by negative seven.
t = 1.49,

Substitute the value of t into one of the original equations to find b.
3b + 4t = 11.33
3b +4(1.49) = 11.33
3b + 5.96 = 11.33
3b = 5.34
b = 1.78

Then you should substitute both values into the second equation to check.

Second question, same basic game plan. Except that you might have to multiply both equations to get a column to eliminate.

Answer 3

1st problem:
Assign a variable for each item. i'll choose b for burritos and t for tacos.
Write an equation:
3 burritos and 4 tacos costs $11.33
9 burritos and 5 tacos costs $23.56
3b + 4t = 11.33
9b + 5t = 23.56
Pick a variable to eliminate. I'll choose b.
Multiply the top equation by -3 so that the b variable will be eliminated.
-9b - 12t = -33.99
9b + 5t = 23.56
the b is cancelled.
-7t = -10.43
divide each side by -7
tacos cost $1.49 each.
Plug 1.49 into one of the equations.
9b + 5(1.49) = 23.56
9b + 7.45 = 23.56
Solve for b
Burritos cost $1.79

Same idea for second problem, but i'm too lazy to do the work again.

Answer 4

the problems are pretty much the same, so lets do the second as it is easier to work with the simpler values.

first write our your equations:
3B + 10S = $48 B brass S steel
7B + 4S=$54
now you need to get to just one variable
from the second eq 4S=54-7B
divide by four s=(54/4) -(7/4)B

put that back into the first eq
3B + 10(54/4)-10(7/4)B=$48
simplify and solve

Answer 5

3Burritos + 4Tacos = $11.33
3B + 4T = 11.33
3B = 11.33 - 4T
B = (11.33 - 4T) / 3 .....EQN(1)

9Burritos + 5Tacos = $23.56
9B + 5T = 23.56 .....EQN(2)

substitute the value of B at eqn(2)

9[(11.33 - 4T)/3] + 5T = 23.56
33.99 -12T + 5T = 23.56
-7T = 23.56 - 33.99
-7T = -10.43
T = 1.49

substitute the value of T to eqn(1)
B = (11.33 - 4T) / 3
B = [11.33 - 4(1.49)] / 3
B = 1.79

Burritos = $1.79
Tacos = $1.49


3Brass + 10Steels = $48
3B + 10S = 48
B = (48 - 10S) / 3 .....EQN(1)

7Brass + 4Steels = $54
7B + 4S = 54 .... EQN(2)

substitue the value of B to eqn(2)
7[(48 - 10S) / 3] + 4S = 54
112 - 23.33S + 4S = 54
19.33S = 58
S = 3

substitute the value of S to eqn(1)
B = (48 - 10S) / 3
B = [48 - 10(3)] / 3
B = 6

Brass = $6
Steel = $3

10Brass + 13 Steel = 10(6) + 13(3) = $99

Answer 6

let B be the cost of Burritos
Let T be the cost of Tacos

Write two equations:

3B + 4T = 11.33
9B+ 5T = 23.56

Should be able to get it from there

Other Problem:

Let B = the cost of Brass parts
Let S = the cost of Steel parts

Write two equations:

3B + 10S = 48
7B + 4S = 54

Hope that gets you started.

Answer 7

Explanation of burrito and taco problem is below. The method for the brass and steel machine parts problem is similar, so you can work that out after understanding the first one...

Set:
x=burrito cost
y=taco cost

System of equations describing the situation:
3x+4y=11.33
9x+5y=23.56

Solving 1st equation for x:
3x=11.33-4y
x=[11.33-4y]/3

Substituting x expression into 2nd equation:
9 [11.33-4y]/3 + 5y = 23.56

Solving for y:
33.99 - 12y + 5y = 23.56

Combing similar expressions:
10.43 = 7y

Divide both sides by 7:

y = 1.49

Subsitute y=1.49 into equation solved for x:
x=[11.33-4*1.49]/3
x = [11.33 - 5.96] / 3
x = 5.37/3

x = 1.79

Answer 8

3b+4t=11.33
9b+5t=23.56

now multiply the 1st equation with 3.

9b+12t=33.99
9b+5t=23.56

now substract the 2nd from the 1st.

7t=10.43
now divide by 7
t= 1.49. now u should b able to find the price of b.
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
2. 3B+10S=48
7B+4S=54

solve by same process.

Answer 9

i wish i knew...sorry