ALgebra help please??

3(8r - 5) = -4(7 - 6r)
3*8r - 3*5 = -4*7 + 4*6r
24r - 15 = -28 + 24r
24r - 24r = -28 + 15
0 = -13
(no solutions)

multiply and distribute 24r-15=28+24r
combine terms 24r+13=24r
get r on one side and numbers on other side so then subtract 24r and then it looks like this 13=0
there is no real number that makes this solution true

24r - 15 = -28 + 24r
no real number so no solution

24r - 15 = -28 +24r

There is no unknown because when you collect like terms the r terms cancel each other out and you are left with numbers.

multiply out both sides

24r - 15 = -28 + 24r

0 = -13

Yes, it is No real number.

Then 24r-15=-28+24r,it is impossible to have -15=-28

distributive property of multiplication

24r-15= -28+24r
24r-24r= -13

what?

it can't be solved..

i dont know

3(8r - 5) = -4(7 - 6r)
24r - 15 = -28 + 24r
24r - 24r = -28 + 15

therefore...
it is no real solution for the algebra... because there is no unknown(r)

3(8r - 5) = -4(7-6r)

3 x (8r - 5) = -4 x( 7 - 6r)

3 x (8r - 5 + 6r) = -4 x 7

3 x (14r - 5) = -28

42r - 15 = -28

42r = -13

42r divided by 42 = -13 divided by 42

r = -0.31
= - 31/100

i know my answer is right and it was fun working it out! thanx. it is a real number! just a really irrational one.