Question

HELP ME PLEASE ITS DUE TOMORROW AND IM ONLY A FRESHMAN PLSSS HELP

Answer 1

Answer:

No

Step-by-step explanation:

We could solve for c and see what the value of c should be to satisfy this equation.

However, in this case we are given a value for c already i.e. c=25 and all we need to do is plug it into the equation and see if it is consistent i.e. if the value on the LHS equals the value of RHS for that specific value of c

Using the value of c = 25 and plugging it into the LHS of the equation we have $\frac{1}{2} (25-13) =\frac{12}{2} = 6$

Doing the same with the RHS gives us 19-25 = -6

6 ≠ -6 (6 not equal to -6)

So Lance's solution of c = 25 is incorrect

If you actually wanted to solve for the correct value of c, proceed as follows
Multiply by 2 on both sides of the equation
$\frac{1}{2} (c-13) = 19 -2c\\\\\\text{This gives us } \\c-13 = 38-2c$

Adding 2c to both sides and adding 13 to both sides  (or, in other words moving -2c to the LHS and -13 to the RHS) gives us

c- (-2c) = 38 - (-13)

3c = 51

c = 17

Check this is correct by plugging it into the original equation and seeing that the LHS = RHS = 2

Answer 2

Answer:

No

Step-by-step explanation:

because c equals 17

hope this helps