l Use Substitution to solve c+p=25 and 4c + 3p = 84

Enter Equation 1

Enter Equation 2

    

Answer
c = 9 and p = 16

↓Steps Explained:↓



Use the substitution method to solve:

4c + 3p = 84

Check Format

Equation 2 is in the correct format.

Rearrange Equation 2 to solve for c:

4c + 3p = 84

Subtract 3p from both sides to isolate c:

4c + 3p - 3p = 84 - 3p

4c = 84 - 3p

Now divide by 4:

4c
4
=
  
84 - 3p
4

Plug Revised Equation 2 value into c:

1(c) + p = 25

1 * ((84 - 3p)/4) + p = 25

((84 - 3p)/4) + p = 25

Multiply equation 1 through by 4

4 * (((84 - 3p)/4) + p = 25)

4 * (((84 - 3p)/4) + p = 25)

84 - 3p + 4p = 100

Group like terms:

-3p + 4p = 100 - 84

1p = 16

Divide each side by 1

1p
1
=
  
16
1

p  =  16
  1

p = 16

Plug this answer into Equation 1

1c + 1(16) = 25

1c + 16 = 25

1c = 25 - 16

1c = 9

Divide each side by 1

1c
1
=
  
9
1

c  =  9
  1

c = 9



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