a@b = 3a + 2b
u@v = 3u + 2v = 20
v@u = 3v + 2u = 20
add them: 5u + 5v = 40
u + v = 8
then 2u@3v
= 6u + 6v
= 6(u+v) = 6(20) = 120
u@v = 3u + 2v = 20
v@u = 3v + 2u = 20
add them: 5u + 5v = 40
u + v = 8
then 2u@3v
= 6u + 6v
= 6(u+v) = 6(20) = 120
u@v=v@u=20 u=v
If u and v both = 4 then 3(4)+2(4)=12+8=20
u=4 v=4
2u@3v=2(4)@3(4)=6(4)+6(4)=24+24=48
Given that u@v = 20, we have:
3u + 2v = 20
Since v@u is also equal to 20, we have:
3v + 2u = 20
We can solve this system of equations to find the values of u and v.
First, multiply the first equation by 3, and multiply the second equation by 2:
9u + 6v = 60
6v + 4u = 40
Now, subtract the second equation from the first equation:
(9u + 6v) - (6v + 4u) = 60 - 40
9u + 6v - 6v - 4u = 20
(9u - 4u) + (6v - 6v) = 20
5u + 0 = 20
5u = 20
u = 4
Substitute the value of u into the second equation:
3v + 2(4) = 20
3v + 8 = 20
3v = 20 - 8
3v = 12
v = 4
Now we have the values of u and v.
To find the value of 2u@3v, substitute these values into the function:
2u@3v = 2(4) + 3(4)
2u@3v = 8 + 12
2u@3v = 20
Therefore, the value of 2u@3v is 20.
Let's start with the equation u@v = 20:
According to the definition of the function a@b, we have 3u + 2v = 20.
Similarly, for the equation v@u = 20:
Using the definition of the function a@b, we have 3v + 2u = 20.
Now we have a system of two linear equations:
3u + 2v = 20 (Equation 1)
3v + 2u = 20 (Equation 2)
To solve this system, we can use the method of substitution or elimination. Let's solve it using the method of elimination:
Multiply Equation 1 by 3:
9u + 6v = 60 (Equation 3)
Multiply Equation 2 by 2:
6v + 4u = 40 (Equation 4)
Now, subtract Equation 4 from Equation 3 to eliminate the variable v:
(9u + 6v) - (6v + 4u) = 60 - 40
Simplifying:
9u + 6v - 6v - 4u = 20
5u = 20
u = 4
Substituting the value of u = 4 into Equation 1:
3u + 2v = 20
3(4) + 2v = 20
12 + 2v = 20
2v = 8
v = 4
Now we have found the values of u = 4 and v = 4 that satisfy both equations u@v = 20 and v@u = 20.
To find the value of 2u@3v, we substitute the values of u = 4 and v = 4 into the definition of the function a@b:
2u@3v = 2(4) + 3(4) = 8 + 12 = 20
Therefore, the value of 2u@3v is 20.