Question 641683
Let {{{x}}} be the amount (in $) of money the woman has invested.
Let {{{y}}} be the rate at which she invested her money, written as a decimal (so 1% would be 0.01, 2% would be 0.02, and so on).
The interest she gets every year (her yearly income) is calculated as
${{{x*y}}}
An interest rate 1% lower than {{{y}}} would be {{{y-0.01}}} .
If the woman had invested $6000 at that lower rate, her yearly income would be
${{{6000(y-0.01)}}} , and that is the same as ${{{x*y}}} , so
{{{6000(y-0.01)=xy}}} .
An interest rate 1% higher than {{{y}}} would be {{{y+0.01}}} .
If the woman had invested $4500 at that lower rate, her yearly income would be
${{{4500(y+0.01)}}} , and that is the same as ${{{x*y}}} , so
{{{4500(y+0.01)=xy}}} .
We have a system of non-linear equations:
{{{system(6000(y-0.01)=xy,4500(y+0.01)=xy)}}}
From there we get
{{{6000(y-0.01)=4500(y+0.01)}}} --> {{{6000y-60=4500y+45}}} --> {{{6000y-60-4500y=4500y+45-4500y}}} --> {{{1500y-60=45}}} --> {{{1500y-60+60=45+60}}} -->  {{{1500y=105}}} --> {{{1500y/1500=105/1500}}} --> {{{highlight(y=0.07)}}} ,
The interest rate is 7%.
Substituting into {{{6000(y-0.01)=xy}}} ,
{{{6000(0.07-0.01)=0.07x}}} --> {{{6000(0.06)=0.07x}}} --> {{{360=0.07x}}} --> {{{360/0.07=0.07x/0.07}}} --> {{{360/0.07=x}}} --> {{{highlight(x=5142.86)}}} (rounded).
The woman had $5142.86 invested.