Mathematics OpenStudy (krystal):

50000-x x is the amount of in \$ ann invested in the account that pays 2.5% annual interset OpenStudy (anonymous):

ok what are you trying to solve for? OpenStudy (krystal):

she invests 50,000 in two interst bearing accounts one pays 2.5 % interst and the second 3.75% at the end of one year the interest ann has earned on these is 1656.25 \$ how much she invest in each of the accounts OpenStudy (anonymous):

ok got it! OpenStudy (anonymous):

so what in an interest account, each year you earn the % of interest according to your original investment OpenStudy (anonymous):

so from the first account, she will earn 2.5% of 50,000 OpenStudy (anonymous):

from the second account, she will earn 3.75% of 50,000 OpenStudy (anonymous): OpenStudy (anonymous):

so we have two equations based on this information: OpenStudy (anonymous):

both involve amounts x and y OpenStudy (anonymous):

we know that .025x + .0375y = 1656.25 OpenStudy (anonymous):

that is, the interest earned between the two accounts adds up to 1656.25, and by definition .025 of X is earned, and .0375 of y is earned in that year OpenStudy (anonymous):

the other thing we know is that the total amount she invests is 50,000, so x + y = 50,000 OpenStudy (anonymous):

now since you have two variables, and two equations, you can use substitution to solve this OpenStudy (anonymous):

so using the second equation, we know that y = 50,000 - x OpenStudy (krystal):

i have to give a vebal descritpion of 50000-x OpenStudy (anonymous):

I'm not sure what you mean by verbal description OpenStudy (anonymous):

but we know that 50000 - x is the amount invested in one of the banks OpenStudy (krystal):

it says give a verbal description of what each of the follwoing expression represetn in the context of the problem letting x=2.5 %. 50,000 is the interest in the two accounts together OpenStudy (krystal):

she earned a total on the investements \$1656.25 OpenStudy (anonymous):

right. so what is the expression? 50,000 - x is the amount of money invested in the second bank, if we assume x is the amount invested in the 2.5% account OpenStudy (anonymous):

we know that because the total amounts are equal to 50000 OpenStudy (anonymous):

so x+y = 50000 OpenStudy (anonymous):

y = 50000 - x OpenStudy (anonymous):

and if x is the amount in the first account, 50000 - y is the amount in the 3.75% account OpenStudy (anonymous):

is that helping? OpenStudy (krystal):

i still dont get it but i think i understand ur saying if both x and y equally to 50000 than if you take away .025 away from that than it will give you the answer correct OpenStudy (krystal):

i need to know how to write it out OpenStudy (krystal):

if she has a total at the end of the year of 1656.25 it is asking how much did she invest in both accounts with one being 0.025 and one being 0.0375 annual interest OpenStudy (anonymous):

Since: 1. x + y = 50000 2. .025x + .0375y = 1656.25 We substitute for y in the second equation, using the first. i.e. y = 50000 -x Now substitute for y .025x + .0375 (50000 - x) = 1656.25. .025x + 1875 - .0375x = 1656.25 -.0125x = 1656.25 - 1875 -.0125x = -218.75 x = -218.75/-.0125 x = 17500 OpenStudy (anonymous):

so looking above, she invested 17,500 in the bank with 2.5% interest rate (also known as .025 interest) OpenStudy (anonymous):

now that you've solved for one of the values, you can solve for the other (y) by sticking this result back into either equation OpenStudy (anonymous):

so x+5 = 50000 OpenStudy (anonymous):

*x + y OpenStudy (anonymous):

since x = 17500, 17500 + y = 50000 OpenStudy (anonymous):

y = 32500 OpenStudy (anonymous):

so she invested 32500 in the second bank OpenStudy (anonymous):

(the one with .0375 interest) OpenStudy (anonymous):

now the final part is - check the answer. that is, at the end of the year, do those two values give you the correct total earned interest OpenStudy (anonymous):

so .025*17500 + 32500*.0375 = 437.50 + 1218.75 = 1656.25 OpenStudy (anonymous):

so we know it's right OpenStudy (anonymous):

make sense? OpenStudy (krystal):

kinda

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