Word Problem - System of Linear Equations Type 4 (ALEKS)

[Denda]

Hi there! I'm new to the forum. Please be kind. Learning Algebra through ALEKS at college, which doesn't go into too much detail, teacher not much help. I'm trying to understand how to work this problem.

A chef is using 2 brands of italian dressing. If she combines 200 mL of the first brand with 100 mL of the second brand, she gets 300 mL of a mixture that is 11% vinegar. If she combines 90 mL of the first brand with 180 mL of the second brand, she gets 270 mL of a mixture that is 12% vinegar.
What are the %'s of vinegar in the two brands of dressing?
First Brand:
Second Brand:

I've got it to:
200x + 100y = 11(300)
90x + 180y = 12(270)

For some stupid reason, I can't figure out how to get the answer. I have the answer, but am unable to work my way back on this one. Any help would be greatly appreciated.

First Brand: 10%
Second Brand: 13%

Thank you

 

[Denis]

Hi there! I'm new to the forum.
200x + 100y = 11(300)
90x + 180y = 12(270)

Welcome, Denda.

200x + 100y = 3300 [1]
90x + 180y = 3240 [2]

[1]*18: 3600x + 1800y = 59400
[2]*10: 900x + 1800y = 32400

[1]-[2]: 2700x = 27000

OK?

 

[Denda]

Thank you Denis!

What clued you in on to use the *18 & *10, by dividing the y variable by 10?

I apologize, I try to understand the clues of these darn word problems and then it helps me with the different types of wording. Thank you again for your help. It is very much appreciated!

 

[Mrspi]

Thank you Denis!

What clued you in on to use the *18 & *10, by dividing the y variable by 10?

I apologize, I try to understand the clues of these darn word problems and then it helps me with the different types of wording. Thank you again for your help. It is very much appreciated!

You want to get the coefficients of one of the variables, either x or y, to be the same (or to be opposites....works equally well). Then when you subtract (or add) the two equations together, you can eliminate one of the variables. Multiplying the first equation by 18 and the second by 10 turns both y terms into 1800y.

Suppose I decide that I want the coefficients of the x terms to be opposites? I note that if I multiply the first equation by -9, and the second by 20, the coefficients of the x terms will become -1800 and + 1800:

-9(200x) + (-9)(100y) = -9(11)(300)
20(90x) + 20(180y) = 20(12)(270)

-1800x - 900y = -29700
1800x + 3600y = 64800

Now, add the two equations together:
-1800x - 900y = -29700
1800x + 3600y = 64800
-----------------------------
0x + 2700y = 35100
Divide both sides of the equation by 2700:
y = 13

You choose your multipliers so that you get the coefficients of one of the variables to add or subtract to 0, eliminating that variable.

I hope this helps you.

 

[Denda]

Thank you for the explaination! It helps quite a bit. I'm so glad I found this website!

 

[Denis]

Denda, that was one way to solve; there are others; here's another:

200x + 100y = 3300 [1]
90x + 180y = 3240 [2]

[1]: x = (3300 - 100y) / 200
[2]: x = (3240 - 180y) / 90

[1][2]: (3300 - 100y) / 200 = (3240 - 180y) / 90
Solve for y...then for x

Another would be to substitute [1] x = (3300 - 100y) / 200 in [2]:
90[(3300 - 100y) / 200] + 180y = 3240

Whatever turns you on :wink: