(6x^3/3x^4)^2 = (2/x)^2 = 4/x^2
Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.
x+2y=6
x-y=3
Multiply the bottom equation by two so then you have:
x+2y=6
2x-2y=6
The y's now cancel out:
x=6
2x=6
Add them together
3x=12
Divide
x=4.
To find y, plug x into either equation (*don't have to do both, but I will)
(4)+2y=6
(4)-y=3
Subtract four
2y=2
-y=-1
Divide each
2y/2 = 2/2
y=1
-y/-1 = -1/-1
y=1
The answer is:
x=4
y=1
I hope that helps!
Answer:
D. 4(3x-5) = 52
Step-by-step explanation:
52 is equal the sum of one side multiplied by 4
So
4(3x-5) = 52
Is the right answer
I think when you copied and pasted your question it re arranged itself
