Answer:
There is one possible solution
Step-by-step explanation:
Here, we want to get the number of solutions
We can proceed to solve the equation
we have two y values so we can directly equate the x parts
-2x - 4 = 3x + 3
3x + 2x = -3-4
5x = -7
x = -7/5
To get y, we substitute
We can only get one value of y too
So we have a point (x,y) as the solution to the system of equations
Answer:
B) -16
Step-by-step explanation:
-4-3-1-8
subtract them all lol
7+9= 16
Then add the negative symbol
-16
Therefore
(4y+6) +y = 46
We can combine the like terms:
5y + 6 = 46
Subtract 6 on both sides:
5y+6-6=46-6
Therefore 5y =40
Divide by 5 on both sides
5y/5 = 40/5
Therefore
y= 8
We found y but to find the value x we use y.
X= 4y+6
Therefore
x= (4*8)+ 6
Therefore
x= 32 + 6
x= 38.
Answer:
D. ![xy\sqrt[3]{9y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9y%7D)
Step-by-step explanation:
![\sqrt[3]{9x^3y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9x%5E3y%5E4%7D)
![\sqrt[3]{9}\sqrt[3]{x^3}\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7D%5Csqrt%5B3%5D%7Bx%5E3%7D%5Csqrt%5B3%5D%7By%5E4%7D)
The ![\sqrt[3]{x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D)
cancels out to become x:
![\sqrt[3]{9}x\sqrt[3]{y^4}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E4%7D)
Split the 
![\sqrt[3]{9}x\sqrt[3]{y^3}\sqrt[3]{y^1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B9%7Dx%5Csqrt%5B3%5D%7By%5E3%7D%5Csqrt%5B3%5D%7By%5E1%7D)
![\sqrt[3]{y^3} =y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E3%7D%20%3Dy)
![xy\sqrt[3]{9} \sqrt[3]{y}](https://tex.z-dn.net/?f=xy%5Csqrt%5B3%5D%7B9%7D%20%5Csqrt%5B3%5D%7By%7D)
Put the cube root of y and cube root of 9 together:
The answer you're looking for my good lad is C
