Ok, I asked this one already and received 2 answers but some of the work in the answers left me confused, help

Question

Homer has three investments totaling $110,000.these earn interest at 4%,6% & 8%. His total income from these investments is $7200. the income from the 8% investment exceeds the tot income form the other two investments by $800. find how much he has invested at 4%

the answer given for this was $20,000.

thanks for the help from the two people who responded but some of the answer I didn't understand where it was coming from, I really suck at word problems and need every bit explained. Thanks to anyone who can help me.

Answer 1

Hi,

Make one equation adding the money in each of the 3 accounts together to equal the total of $110,000. If x = money at 4%, y = money at 6%, and z = money at 8%, then this equation is

x + y + z = 110,000

Next the total income from these investments was $7,200.
They got 4% of $x from the first account. 4% of $x is .04x.
They got 6% of $y from the first account. 6% of $x is .06y.
They got 8% of $z from the first account. 8% of $x is .08z.

This equation is:

.04x + .06y + .08z = 7,200

If the income from the 8% investment exceeds the total income form the other two investments by $800, then the 8% income - (4% income + 6% income) = 800. That equation is:

.08z - (.04x + .06y) = 800

Eliminating parentheses and putting terms in the x,y,z order gives the 3rd equation as:

-.04x - .06y + .08z = 800


So, your 3 equations are:

x + y + z = 110,000
.04x + .06y + .08z = 7,200
-.04x - .06y + .08z = 800

The easiest way to start solving this system is to add the last 2 equations together. When you do that, the x and y terms drop out and you can solve for z.

.04x + .06y + .08z = 7,200
-.04x - .06y + .08z = 800
------------------------------------------
__________.16z = 8,000

z = $50,000 That means $50,000 was invested at 8% interest.

Putting 80,000 into the first 2 equations for z changes them as follows:

x + y + z = 110,000
x + y + 50,000 = 110,000
x + y = 60,000 <== new first equation


.04x + .06y + .08z = 7,200
.04x + .06y + .08(50,000) = 7,200
.04x + .06y + 4,000 = 7,200
.04x + .06y = 3,200 <== new second equation

Here are the equations we want to solve for x and y.

x + y = 60,000 <== new first equation
.04x + .06y = 3,200 <== new second equation

Let's solve the first equation for x:
x + y = 60,000
x = 60,000 - y

Now substitute 60,000 - y into the second equation in place of x. Then solve for y.

.04x + .06y = 3,200
.04(60,000 - y) + .06y = 3,200
2,400 - .04y + .06y = 3,200
.02y = 800
y = 40,000 That means $40,000 was invested at 6% interest.

Since x + y + z = 110,000 if we fill in the "y" and "z" values, we get:

x + 40,000 + 50,000 = 110,000
x + 90,000 = 110,000
x = 20,000 That means $20,000 was invested at 4% interest.

I hope that is clear enough to help you!! :-)

Answer 2

Ok, let's say that homer investments are x, y, and z (in $).
So we know:

x + y + z = 110000 -- EQ1

We know that the interest for them are 4%, 6% and 8%. Their combined income is 7200. We can put that in an equation:

4x/100 + 6y/100 + 8z/100 =7200 ---EQ2

And the income from the 8% investment (in our equatin system, z) exceeds the total of the other two by 800.

8z/100 - (4x/100 + 6y/100) = 800 --EQ3

to find out how much he invested at 4% means to find x (using the variables in our system).

We are going to use EQ2 and EQ3 to start:

EQ2 4x/100 + 6y/100 + 8z/100 = 7200
EQ3 8z/100 -4x/100 - 6y/100 = 800

we add up both EQ2 + EQ3

8z/100 + 8z/100 = 8000

multiply both sides by 100, then divide both sides by 8

z + z = 100000
2z= 100000
z= 50000
one down, two more to go

from EQ1

x + y + z =110000

x + y + 50000 = 110000

x + y = 60000 --- EQ4

now we are going to use EQ2 and EQ4 to find x and y

EQ2 4x/100 + 6y/100 + (8 * 50000)/100 = 7200
EQ4 x + y = 60000

but first, lets simplify EQ2
4x/100 + 6y/100 + 4000=7200
4x/100 + 6y/100 = 3200
4x + 6y = 320000
2x + 3y = 160000
y = (160000 - 2x)/3

so from EQ1:

x + 160000/3 - 2x/3 = 60000
3x + 160000 -2x = 180000
x = 20000

and that's how you get your answer.

Answer 3

I answered this before. Sorry if I did not make things clear.
Let x = amount invested at 4%
Lety= amount invested at 6 %
Let z= amount invested at 8%
Total amount invested = x+y+z = $110,000 <-- Eq 1
.04x +.05y +.08z = $7200 = interest from the three investments <-- Eq 2
.08z = .04x + .06 y +800 <-- Eq 3 the interest from 8% investment = the sum of the 4% and 6% investments + $800.
Summarizing, we have 3 equations and 3 unknowns:
x+y+z + $100000 <-- Eq 1
.04x+.06y+.08z = $7200<-- Eq 2
.08z = .04x+.06y +$800 <-- Eq 3
-.04x -.06y + .08z = 800 <-- Eq 4 = Eq 3 rearranged.
Now add Eq 4 to Eq2 getting:
.16z = $8000 --> z = $50,000. {Nice how x and y canceled}
Now replace z with $50,000 in EQ1 and Eq4 getting:
x+y+50,000 = 110,000, or
x+y = 60,000 <-- EQ 5, and
-.04x - .06y +.08(50,000) = 800, or
-.04x-.06y = -3200 <-- Eq 6
Now divide Eq 6 by .04 getting:
-x -1.5y = -80,000<-- Eq 7
Add Eq 7 to Eq 5 getting:
-.5y = -20,000
y = $40,000
Therefore x = 110,000 - 50,000-40,000 = $20,000

Check: x+y+z = 20,000+40,000+50,000 = $110,000
.04(20,000) +.06(40,000) +.08(50,000) = $7200
$4000 = $800 + $2400 +$800

Answer 4

I'll give the equations:

x + y + z = 110,000
.04x + .06y + .08z = 7,200
.08z = .04x + .06y + 800

To simplify, multiply the next 2 equations by 50 to remove the decimals:
x + y + z = 110,000
2x + 3y + 4z = 360,000
4z = 2x + 3y + 40,000.


x + y + z = 110,000 ......... equation (1)
2x + 3y + 4z = 360,000 .. equation (2)
2x + 3y - 4z = -40,000. ... equation (3)

The best way to solve this is by eliminating variables.
Say, to eliminate z.
4 * eqn (1) + eqn (3) ...... equation (4) : 6x + 7y = 400,000
eqn(2) + eqn (3) ............. equation (5) : 4x + 6y = 320,000

To solve for x:
6*eqn(4) - 7* eqn(5).

The answer is really 20,000.
(x = 20000, y = 40000, z = 50000)

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