Let a = price of an apple

Let p = price of a pear

Given: 4a+6p=5 and 8a+2p=4

Find: a=?

4a+6p = 5

Subtract 4a from both sides:

4a+6p-4a = 5-4a

6p = 5-4a

Divide both sides by 6:

6p/6 = (5-4a)/6

p = (5-4a)/6

8a+2p = 4

Plug in (5-4a)/6 for p:

8a+2[(5-4a)/6] = 4

8a+[(5-4a)/3] = 4

Change 8a to a fraction with a common denominator, 3:

(24a/3)+[(5-4a)/3] = 4

On the left side, do the operation to the numerator but keep the denominator the same:

(24a+5-4a)/3 = 4

(20a+5)/3 = 4

Multiply both sides by 3:

[(20a+5)/3](3) = 4(3)

20a+5 = 12

Subtract 5 from both sides:

20a+5-5 = 12-5

20a = 7

Divide both sides by 20:

20a/20 = 7/20

a = 0.35

Answer: $0.35

*By the way, I solved by the substitution method. There is also the elimination method but I personally find this way easier. If in the first part, you happened to get "a" .equals something in terms of "p", then you could have also solved for "p" first, then substituted that in, and solved for "a".