I decided to create a matrix to solve this.
So, I let x = Amount for Hannah
y = Amount for Francine
z = Amount for Peter
This gave me the following information:
x + y = 120
x + z = 230
6y = z; therefore: -6y + z = 0
I then set up the following matrix:
1 1 0 = 120
1 0 1 = 230
0 -6 1 = 0
I then did: R1 - R2 = new R2; which gave me the new matrix iteration of:
1 1 0 = 120
0 1 -1 = -110
0 -6 0 = 0
I then did: 6R2 + R3 = new R3; which gave me the new matrix iteration of:
1 1 0 = 120
0 1 -1 = -110
0 0 -5 = -660
I then did a row reduction on R3 by: -(1/5)R3 = new R3; which gave the new matrix iteration of:
1 1 0 = 120
0 1 -1 = -110
0 0 1 = 132
I then did: R3 + R2 = new R2; which gave the new matrix iteration of:
1 1 0 = 120
0 1 0 = 22
0 0 1 = 132
I then did: -R2 + R1 = new R1; which gave the new matrix iteration of:
1 0 0 = 98
0 1 0 = 22
0 0 1 = 132
This gave me: (x, y, z) of: (98, 22, 132)
Check # 1: 98 + 22 = 120 [Check]
Check # 2: 98 + 132 = 230 [Check]
check # 3: 22 x 6 = 132 [Check]
To answer the question, Hannah has $98
