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douald has investments totaling 4,000 in two accounts-one a savings account paying 3% interest and the other bond paying 5%. if the annual from the two investment was 180, how much did he have in invested at each rate?
Let funds deposited into account 1 be A and funds deposited into account 2 be B. A + B = 4000, since the total amount invested is $4,000. Since A earns interest at 3% and B earns interest at 5%, we can set up a second equation like this: A*.03 + B*.05 = 180
So you have A+B=4000 And .03A + .05B=180 Can you solve one of those for A or B? The first one, maybe?
no can u show me how
Let's solve for A by subtracting B from both sides: A+B=4000 - B -------------- A =4000-B
Now you can substitute that in for A in the second equation: .03(4000-B) + .05B = 180
i thought i subtract the like terms
I'm not sure what you mean.
o while ill do it your way
ok wat i do now
Distribute .03: .03*4000 - .03B + .05B = 180
ok i did it
Okay, what did you get? Then simplify it.
is the answer 120.02
Well, .03*4000 = 120.00. But that is not the answer to the problem. We need to solve for A and B.
ok is u workin it out to
I want you to understand how to get there, I'd rather not just give you the answer.
When you distribute .03: .03*4000 -.03B +.05B what do you get?
i know hw answer i just said was the wronge answer
.03*4000 -.03B+.05B = 120 + .02B = 180 Subtract 120 from both sides .02B=60 Divide 60/.02=B=3000 Plug back in to the original equation A+3000=4000 Therefore A = 1000 and B = 3000
want it be -0.35b=60
then wat do i do next
I worked it all out above.
how much did he have invested at each rate?
I already answered the question. .03*4000 -.03B+.05B = 120 + .02B = 180 Subtract 120 from both sides .02B=60 Divide 60/.02=B=3000 Plug back in to the original equation A+3000=4000 Therefore A = 1000 and B = 3000