Question

10 cookies and 3 brownies costs $11.25

Answer 1

c = cost of a cookie

b = cost of a brownie

(idk if u should put the variable in front of the number, but I'm putting it after)

10c + 3b = 11.25    

6c + 9b = 15.75

Isolate one of the variables of one of the equations, and substitute it into the other equation.

I will isolate "c" in the first equation

10c + 3b = 11.25       Subtract 3b on both sides

10c = 11.25 - 3b       Divide 10 on both sides

c = 1.125 - $\frac{3}{10}b$      or    c = 1.125 - 0.3b

Now substitute this into the second equation

6c + 9b = 15.75    Since c = 1.125-0.3b, u can plug in (1.125-0.3b) for c

6(1.125 - 0.3b) + 9b = 15.75   Distribute 6 into (1.125 - 0.3b)

6.75 - 1.8b + 9b = 15.75   Combine like terms

6.75 + 7.2b = 15.75    Subtract 6.75 on both sides

7.2b = 9       Divide 7.2 on both sides

b = 1.25     $1.25

You could continue to find  "c"

Since you know "b", you can plug it into one of the equations:

10c + 3b = 11.25

10c + 3(1.25) = 11.25

10c + 3.75 = 11.25

10c = 7.5

c = 0.75

PROOF

6(0.75) + 9(1.25) = 15.75

4.5 + 11.25 = 15.75

15.75 = 15.75

10(0.75) + 3(1.25) = 11.25

7.5 + 3.75 = 11.25

11.25 = 11.25