System of equations for
-6=Ab^-3
-18=Ab^2
Ab^-3 = -6
Ab^2 = -18
divide to get rid of the A's and you have
b^-5 = 1/3
b^5 = 3
b = 3^(1/5)
Now use that to get A
I cant get A it keeps being nonreal
Ab^2 = -18
A(3^(1/5))^2 = -18
A(3(2/5)) = -18
A = -18 * 3^(-2/5)
= 18 / 9^(1/5)
= 2 * 9^(4/5)
It would be nice if you showed your work ...
To solve the system of equations -6 = Ab^(-3) and -18 = Ab^(2), we can use the method of substitution.
Let's begin by isolating one variable in either equation. Rearrange the first equation to isolate A:
-6 = Ab^(-3)
Divide both sides of the equation by b^(-3):
-6 / b^(-3) = A
Next, substitute this expression for A in the second equation:
-18 = Ab^(2)
-18 = (-6 / b^(-3)) * b^(2)
We can simplify this expression by multiplying -6 with b^2:
-18 = -6b^2 / b^(-3)
Now, let's simplify the expression further by multiplying -6 with b^2 and dividing by b^(-3):
-18 = -6b^2 * b^3
-18 = -6b^(2+3)
-18 = -6b^(5)
At this point, we can divide both sides of the equation by -6 to isolate b:
(-18 / -6) = b^(5)
3 = b^(5)
Now, take the fifth root of both sides of the equation to solve for b:
b = ∛3
Now that we have found the value of b, we can substitute it back into the first equation to find the corresponding value of A:
-6 = Ab^(-3)
-6 = A(∛3)^(-3)
-6 = A / (∛3)^3
-6 = A / 3
Multiply both sides of the equation by 3 to solve for A:
-18 = A
Therefore, the solution to the system of equations -6 = Ab^(-3) and -18 = Ab^(2) is A = -18 and b = ∛3.