how do I solve this 8(c-9)=6(2c-12)-4c
\(8(c-9)=6(2c-12)-4c\\\\ 8c-72=12c-72-4c\\\\ 8c-72=8c-72\\\\0=0\\\\There\ are\ infinitely\ many\ solutions.\)
8(c-9) = 6(2c-12)-4c
Okay let us solve it together.
Multiply out and expand the brackets.
8c -89 = 62c -612 -4c
8c -72 = 12c -72 -4c Re arrange by taking the cs to one side
8c -12c +4c = -72 + 72
0 = 0
For this kind of statement it means C can be any number, and it would satisfy the equation.
Cheers.