Answer:
b = 3, which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0.
Step-by-step explanation:

The minus common multiply must be (b-3)*(b+3) = b² - 9

7b - 21 + 5b + 15 = 10b
2b = 6
b = 3
Which is an extraneous solution because if we replace b = 3 in the equation, we must be 5/0 and 30/0, and there is division by 0. So this solution is invalid.
System A:
6x + y = 2
-x - y = -3
System B:
2x - 3y = -10
-x-y = -3
Solve:
System A:
6x + y = 2
y = 2 - 6x
-x - (2-6x) = -3
-x - 2 + 6x = -3
5x = -3 + 2
5x = -1
x = -1/5
y = 2 - 6(-1/5)
y = 2 + 6/5
y = 2 + 1.2
y = 3.2 System A: x = -1/5 or -0.2 ; y = 3 1/5 or 3.2
System B:
2x - 3y = -10
2x = -10 + 3y
x = -5 + 1.5y
-x - y = -3
-(-5 + 1.5y) -y = -3
5 - 1.5y - y = -3
-2.5y = -3 - 5
-2.5y = -8
y = 3.2
x = -5 + 1.5(3.2)
x = -5 + 4.8
x = -0.2 System B: x = -0.2 ; y = 3.2
<span>B) They will have the same solution because the first equations of both the systems have the same graph.</span>
How tall are my class mates i think