Please help me with algebra question

Total invested: T=$9,500
Amount invested at 10%: x
Amount invested at 7%: y

(1) x+y=9,500

The total yield is $842.00
10%x=10x/100→10%x=0.1x
7%y=7y/100→7%y=0.07y
(2) 10%x+7%y=842.00
0.1x+0.07y=842.00

We have a system of 2 equations ans 2 unknowns:
(1) x+y=9500
(2) 0.1x+0.07y=842

Using the method of substitution. Isolating y from the first equation:
(1) x+y=9500
x+y-x=9500-x
y=9500-x   (3)

Replacing y by 9500-x in the second equation:
(2) 0.1x+0.07y=842
0.1x+0.07(9500-x)=842
0.1x+0.07(9500)-0.07x=842
0.1x+665-0.07x=842
0.03x+665=842

Solving for x:
0.03x+665-665=842-665
0.03x=177

Dividing both sides of the equation by 0.03:
0.03x/0.03=177/0.03
x=5900

Replacing x by 5,900 in the equation (3):
(3) y=9500-x
y=9500-5900
y=3600

Answers:
How much was invested at 10%?   $5,900.00
How much was invested at 7%?     $3,600.00