Same college algebra college homework?

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Your Open Question: Same college algebra college question?

Your Open Question: College Algebra Math word problem?

"a man invests $2200 in three accounts that pays 6%, 8%, 9% in annual interest, respectively. He has three times as much invested at 9% as he does at 6%. If his total interest for th year is $178, how much is invested at each rate?" 
I know how to do 2 percentage and make an equation like X+Y=$2200 and like %+%=178. But now I have 3 percentage plus it saying something is three times as more. So how do I do this?

Asked by Gina 19 minutes ago 4 days left to answer. 
Answers (1) 
best is to use only one variable 
x @6%, 3x @9%, (2200-4x) @8%, so 
0.06x +0.09*3x + 0.08(2200-4x) = 178 
0.06x +0.27x -0.32x = 178 -0.08*2200 
0.01x = 2 
x = 200 
----------------- 
$200 @6% 
$600 @9% 
$1400 @8% 
-----------------

Someone answered this. But I still need explanation. The answers are right. But I have no idea how he got this or where the 4X came from. And in our class were required to make a chart and use X Y & Z. And also have 2 equations for the chart

Ex: dollars at 6%. Dollars at 8%. Dollars at 9% 
- X. -Y. -Z=2200 
And then .06X. .08Y. .09Z=178 And etc

Helppppppp :/

Asked by  Gina 2 minutes ago 4 days left to answer.

Anonymous · Accepted Answer

X = the number of dollars invested in 6% interest account
Y = the number of dollars invested in 8% interest account
Z = the number of dollars invested in 9% interest account

X + Y + Z = 2200 comes from the sentence "a man invests $2200 in three accounts that pays 6%, 8%, 9% in annual interest"

Z = 3X comes from the sentence "He has three times as much invested at 9% as he does at 6%."

0.06 X is the total interest for the year from the 6% account
0.08 Y is the total interest for the year from the 8% account
0.09 Z is the total interest for the year from the 9% account

0.06X + 0.08Y + 0.09Z = 178 comes from the sentence "his total interest for the year is $178"

They want to know what X, Y, and Z are ?

As you can see from the above, that we have three equations with three unknowns. This will permit us to solve for X, Y, and Z.

The three equations are:
X + Y + Z = 2200 
Z = 3X 
0.06X + 0.08Y + 0.09Z = 178 

Now we will substitute the z = 3X equation in for Z for the other two equations. This will reduce our problem to two equations and two unknowns. These will be the two equations for your chart.
The two equations are:
X + Y + 3X = 2200
0.06X + 0.08Y + 0.09(3X) = 178 

Combining the X terms for both equations you get:
4X + Y = 2200
0.33X + 0.08Y = 178

Now you can solve these two equations simultaneously like your teacher wants you to.
One way would be to solve the top equation for Y, and then substitute that into the Y in the bottom equation, like this:
4X + Y = 2200
Y = 2200 - 4X
Substituting the Y into 0.33X + 0.08Y = 178 you get:
0.33X + 0.08(2200 - 4X) = 178
0.33X + 0.08(2200) - 0.08(4X) = 178
0.33X + 176 - 0.32X = 178
0.01X = 2
X = 200

Since X = $200, from the Z = 3X equation we now know that Z = 3(200) = $600.
And since X + Y + Z = 2200 , then 200 + Y + 600 = 2200, so Y = $1400

Anonymous · Answer

4x-3y=22 -(4x+5y=6) _________ -8y=16 y= -2 since we now know y we can plug it into one of the equations and solve for x 4x - 3(-2) = 22 4x + 6 = 22 4x = 16 x = 4

Same college algebra college homework?

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