B is the midpoint of AC and D is the midpoint of CE. solve for BD given that BD = 2y + 8 and AE = 6y + 12
Since B is the midpoint of AC, we can say that AB = BC.
Since D is the midpoint of CE, we can say that CD = DE.
We can also say that AC = AE + EC.
Substituting the given values:
AB = BC
2y + 8 = CD
AC = AE + EC
AC = 6y + 12 + EC
Since B is the midpoint of AC, we can say that AB = BC = (1/2)AC.
Substituting the values:
2y + 8 = (1/2)(6y + 12 + EC)
Multiply both sides by 2 to eliminate the fraction:
4y + 16 = 6y + 12 + EC
Subtract 6y from both sides:
4y - 6y + 16 = 12 + EC
-2y + 16 = 12 + EC
Subtract 12 from both sides:
-2y + 4 = EC
Since CD = DE:
2y + 8 = EC
Now we have a system of two equations:
-2y + 4 = 2y + 8
Combine like terms:
4 = 4y + 8
Subtract 8 from both sides:
-4 = 4y
Divide both sides by 4:
y = -1
Substituting this value back into BD:
BD = 2y + 8 = 2(-1) + 8 = -2 + 8 = 6
Therefore, BD = 6.