If he used 34 bottles for every batch just divide 214 bottles he used by 34 it takes for a batch
he made 6.3 batches
This is Bridge Algebra 2 stuff. Well you see that 2x+y=8 and x-1=y. To slove for y you have to slove for x first to find y. So 2x+x-1=8 ( the reason I did that is because there is a y in that equation and I placed it on that equation) now lets slove this equation for x. So 3x-1=8( I add 2x with x) 3x=9 ( I made -1 a positive one and add it to 8) x=3 ( I divided 9/3 to get 3). Now that you have slove x we will use x to slove for y. So 3-1=y, 2=y. Your answer is y=2.
I’d say that it’s definitely random choice for the squares, pentagons, and hexagons. But, if we are looking at how common it would be between the 3 of them, then I would say that it would be one of the hexagons, because there are more hexagons than any of the other shapes in the bag.
Answer:
Ok, we have a system of equations:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
First, we want to isolate one of the variables,
As we have almost the same expression (x*y) in the right side of both equations, we can see the quotient between the two equations:
(6*x + 3*y)/(2*x + 4*y) = 6/5
now we isolate one off the variables:
6*x + 3*y = (6/5)*(2*x + 4*y) = (12/5)*x + (24/5)*y
x*(6 - 12/5) = y*(24/5 - 3)
x = y*(24/5 - 3)/(6 - 12/5) = 0.5*y
Now we can replace it in the first equation:
6*x + 3*y = 6*x*y
6*(0.5*y) + 3*y = 6*(0.5*y)*y
3*y + 3*y = 3*y^2
3*y^2 - 6*y = 0
Now we can find the solutions of that quadratic equation as:

So we have two solutions
y = 0
y = 2.
Suppose that we select the solution y = 0
Then, using one of the equations we can find the value of x:
2*x + 4*0 = 5*x*0
2*x = 0
x = 0
(0, 0) is a solution
if we select the other solution, y = 2.
2*x + 4*2 = 5*x*2
2*x + 8 = 10*x
8 = (10 - 2)*x = 8x
x = 1.
(1, 2) is other solution