Question 155060
Let x = amount invested at 9% and y = amount invested at 8%

Since Walt only invested $9,000, this means that the total (ie sum) of both accounts add to $9,000. Algebraically, this means that {{{x+y=9000}}}


{{{I=Prt}}} Start with the simple interest formula. Note: "P" is the amount invested, "r" is the interest rate, and "t" is the time.


{{{I=(x)(0.09)}}} Plug in {{{P=x}}} and {{{r=0.09}}} and {{{t=1}}} (since this only occured last year)



So the interest formula for the 9% account is {{{0.09x}}}


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{{{I=Prt}}} Go back to the simple interest formula. 



{{{I=(y)(0.08)}}} Plug in {{{P=y}}} and {{{r=0.08}}} and {{{t=1}}} 



So the interest formula for the 8% account is {{{0.08y}}}



Since we know that the total interest adds to $770, this means that {{{0.09x+0.08y=770}}}



{{{9x+8y=77000}}} Multiply <b>every</b> term by 100 to make every number a whole number




So we have the system of equations:


{{{system(x+y=9000,9x+8y=77000)}}}



{{{x+y=9000}}} Start with the first equation



{{{y=9000-x}}} Subtract "x" from both sides



{{{9x+8y=77000}}} Move onto the second equation



{{{9x+8(9000-x)=77000}}} Plug in {{{y=9000-x}}}



{{{9x+72000-8x=77000}}} Distribute



{{{x+72000=77000}}} Combine like terms.



{{{x=5000}}} Subtract 72,000 from both sides



So Walt invested $5,000 at 9%. 



{{{y=9000-x}}} Go back to the isolated equation



{{{y=9000-5000}}} Plug in {{{x=5000}}}



{{{y=4000}}} Subtract



So Walt invested $4,000 at 8%



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Answer:


So Walt invested $5,000 at 9% and $4,000 at 8%