1/(b+2)+1/(b+2)=3/(b+1)
Is this right?
(b+1)+(b+1)=b+3
2b+2=b+3
b+2=3
b=1
1/(b+2)+1/(b+2)=3/(b+1)
Combine.
2/(b+2) = 3/(b+1)
Multiply both sides by (b+2) and (b+1).
2(b+1) = 3(b+2)
2b + 2 = 3b + 6
-4 = b
To solve this equation, we need to find the value of "b" that satisfies the equation:
1/(b+2) + 1/(b+2) = 3/(b+1)
To start, let's simplify the equation by combining the fractions on the left-hand side:
2/(b+2) = 3/(b+1)
Next, we can cross-multiply to eliminate the denominators:
2(b+1) = 3(b+2)
Now, distribute the terms on both sides of the equation:
2b + 2 = 3b + 6
Next, let's isolate the "b" term by moving all the "b" terms to one side of the equation and the constant terms to the other side:
2 - 6 = 3b - 2b
-4 = b
Therefore, the value of "b" that satisfies the equation is -4.