<span>Nope. A consistent system is either dependent (both equations represent the same line) or independent. So in an independent system, the lines intersect in exactly one point, which means there is exactly one solution. A dependent system has an infinite number of solutions. An inconsistent system has no solution, so no system of linear equations can have exactly two solutions.</span>
Step-by-step explanation:
Start by creating expressions
t= A tray of tomato plants
3t = $16.47
Solve the equation above
3t/3= 16.47/3
t= 5.49
A tray of tomato plants costs $5.47
Now create the ratios
b= A tray of basil plants
t:b = 3:2
Substitute t
5.49:b = 3:2
Rewrite as fractions
= ![\frac{3}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B2%7D)
Cross multiply
5.49(2) = 3(b)
10.98 = 3b
10.98/3 = 3b/3
3.66 = b
A tray of basil plants costs $3.66
In total she bought 3 trays of tomato plants, which cost $16.47, and 10 trays of basil plants
T0 find out the cost of the 10 basil plants, multiply 3.66 by 10
3.66 × 10 = 36.6
She bought 10 trays of basil plants for $36.60 and 3 trays of tomato plants for $16.47
Add the 2 values together
36.60+16.47 = 53.07
$53.07 is the total
I hope this helps!!!
We can explicitly find the inverse. If
is the inverse of
, then
![f\left(f^{-1}(x)\right) = \dfrac{f^{-1}(x)+2}{f^{-1}(x)+6} = x](https://tex.z-dn.net/?f=f%5Cleft%28f%5E%7B-1%7D%28x%29%5Cright%29%20%3D%20%5Cdfrac%7Bf%5E%7B-1%7D%28x%29%2B2%7D%7Bf%5E%7B-1%7D%28x%29%2B6%7D%20%3D%20x)
Solve for the inverse :
![\dfrac{f^{-1}(x) + 2}{f^{-1}(x) + 6} = x](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%5E%7B-1%7D%28x%29%20%2B%202%7D%7Bf%5E%7B-1%7D%28x%29%20%2B%206%7D%20%3D%20x)
![\dfrac{f^{-1}(x) + 6 - 4}{f^{-1}(x) + 6} = x](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%5E%7B-1%7D%28x%29%20%2B%206%20-%204%7D%7Bf%5E%7B-1%7D%28x%29%20%2B%206%7D%20%3D%20x)
![1 - \dfrac4{f^{-1}(x) + 6} = x](https://tex.z-dn.net/?f=1%20-%20%5Cdfrac4%7Bf%5E%7B-1%7D%28x%29%20%2B%206%7D%20%3D%20x)
![1 - x = \dfrac4{f^{-1}(x) + 6}](https://tex.z-dn.net/?f=1%20-%20x%20%3D%20%5Cdfrac4%7Bf%5E%7B-1%7D%28x%29%20%2B%206%7D)
![f^{-1}(x) + 6 = \dfrac4{1-x}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%20%2B%206%20%3D%20%5Cdfrac4%7B1-x%7D)
![\implies f^{-1}(x) = \dfrac4{1-x} - 6](https://tex.z-dn.net/?f=%5Cimplies%20f%5E%7B-1%7D%28x%29%20%3D%20%5Cdfrac4%7B1-x%7D%20-%206)
Then when x = -6, we have
![f^{-1}(-6) = \dfrac4{1-(-6)} - 6 = \dfrac47-6 = \boxed{-\dfrac{38}7}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28-6%29%20%3D%20%5Cdfrac4%7B1-%28-6%29%7D%20-%206%20%3D%20%5Cdfrac47-6%20%3D%20%5Cboxed%7B-%5Cdfrac%7B38%7D7%7D)
Alternatively, we can first solve for x such that
. Then taking the inverse of both sides,
. (The difference in this method is that we don't compute the inverse for all x.)
We have
![\dfrac{x+2}{x+6} = -6](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B2%7D%7Bx%2B6%7D%20%3D%20-6)
![x + 2 = -6 (x + 6)](https://tex.z-dn.net/?f=x%20%2B%202%20%3D%20-6%20%28x%20%2B%206%29)
![x + 2 = -6x - 36](https://tex.z-dn.net/?f=x%20%2B%202%20%3D%20-6x%20-%2036)
![7x = -38](https://tex.z-dn.net/?f=7x%20%3D%20-38)
![\implies x = \boxed{-\dfrac{38}7}](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%20%5Cboxed%7B-%5Cdfrac%7B38%7D7%7D)
Answer:
12
Step-by-step explanation: