supose you bought eight oranges and one grapefruit for a total of $4.60. later day, you bought six oranges and three grapefruits for a total of $4.80. what is the price of each type of fruit?

To find the price of each type of fruit, we can set up a system of linear equations. Let's represent the price of an orange as "x" and the price of a grapefruit as "y".

From the first statement, we can create the equation:
8x + 1y = 4.60 ------ Equation (1)

And from the second statement, we can create another equation:
6x + 3y = 4.80 ------ Equation (2)

We now have a system of linear equations. We can solve this system to find the values of x and y.

Multiplying Equation (2) by 2 will help us eliminate the y variable:
2(6x + 3y) = 2(4.80)
12x + 6y = 9.60

Now, let's subtract the modified Equation (2) from Equation (1):

(8x + 1y) - (12x + 6y) = 4.60 - 9.60
8x + y - 12x - 6y = -5.00

Simplifying the left side:
-4x - 5y = -5.00 ------ Equation (3)

Now we have a system of two equations:
-4x - 5y = -5.00 ------ Equation (3)
12x + 6y = 9.60 ------ Equation (2)

To eliminate the y variable, we can multiply Equation (3) by 6:
6(-4x - 5y) = 6(-5.00)
-24x - 30y = -30.00

Adding this modified Equation (3) to Equation (2):

(12x + 6y) + (-24x - 30y) = 9.60 - 30.00
-12x - 24y = -20.40

Simplifying the left side:
-12x - 24y = -20.40 ------ Equation (4)

We now have a new system of two equations:
-12x - 24y = -20.40 ------ Equation (4)
12x + 6y = 9.60 ------ Equation (2)

When we add Equation (4) to Equation (2), the x variable cancels out:
(-12x - 24y) + (12x + 6y) = (-20.40) + (9.60)
-18y = -10.80

Dividing both sides by -18:
y = -10.80 / -18
y = 0.60

So, the price of one grapefruit is $0.60.

Now, substitute this value of y into Equation (1):
8x + 1(0.60) = 4.60
8x + 0.60 = 4.60
8x = 4.60 - 0.60
8x = 4.00

Dividing both sides by 8:
x = 4.00 / 8
x = 0.50

Therefore, the price of one orange is $0.50.

To find the price of each type of fruit, we can set up a system of equations based on the given information.

Let's denote the price of an orange as 'O' and the price of a grapefruit as 'G'.

From the first purchase, we can write the equation:
8O + 1G = $4.60 ...(Equation 1)

From the second purchase, we can write the equation:
6O + 3G = $4.80 ...(Equation 2)

Now, we can solve this system of equations to find the values of 'O' and 'G'.

Multiply Equation 1 by 3 to eliminate G:
24O + 3G = $13.80 ...(Equation 3)

Subtract Equation 2 from Equation 3:
(24O + 3G) - (6O + 3G) = $13.80 - $4.80
18O = $9

Divide both sides of the equation by 18:
O = $9 / 18
O = $0.50

Now, substitute the value of O into Equation 1 to find G:
8(0.50) + G = $4.60
4 + G = $4.60

Subtract 4 from both sides of the equation:
G = $4.60 - $4
G = $0.60

So, the price of each orange is $0.50 and the price of each grapefruit is $0.60.

assuming the prices did not change during the day,

8o+g=460
6o+3g=480

18o=900
o=50
g=60

so, oranges cost $0.50, grapefruit cost $0.60