6 pounds of apples and 3 pounds of oranges cost 24 dollars.5 pounds of apples and 4 pounds of oranges cost 23 dollars. Set up a system of equations and solve to find out the cost for each pound of apples and each pound of oranges.

cost of each pound of apple= $3

And cost of each pound of orange =$2

Step-by-step explanation:

Step 1

Let cost of apples = x

And cost of Oranges =y

Let 6 pounds of apples and 3 pounds of oranges cost 24 dollars be represented as

6 x + 3y= 24 equation 1

Also, Let 5 pounds of apples and 4 pounds of oranges cost 23 dollars be represented as

5x+ 4y= 23 equation 2

Step 2

6 x + 3y= 24 equation 1

5x+ 4y= 23 equation 2

Using substitution method to solve the equation

6 x + 3y= 24

24-6x=3y

y= 24-6x/3 = 8-2x

Y= 8-2x

Substituting the value of y= 8-2x  into equation 2

5x+ 4( 8-2x)= 23

5x+ 32 -8x= 23

32-23= 8x-5x

9=3x

x=9/3

x=3

Putting the value of x= 3 in equation 1 and solving to find y

6 x + 3y= 24

6(3) +3y= 24

18+3y=24

3y= 24-18

3y=6

y=6/3= 2

Therefore the cost of each pound of apple= $3

And cost of each pound of orange =$2

it’s supposed to be -4+45=41

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)