Solving Systems of Linear Equations in Two Variables?

Question

Using the Substitution method, how would I solve my problem? What am I doing wrong? I'm getting a fraction. I dont think I need one.

3m -5r = 8
m=-2r-1

Answer 1

OK
substitute the value of m in the first equation
So first equation becomes
3( -2r - 1 ) - 5r = 8
-6r -3 - 5r = 8
-11r - 3 = 8
-11r = 11
r = -1

and substitue this value in second equation
m = -2( -1) - 1
m = 2 -1
m = 1

Answer 2

3m-5r=8..............1
m=-2r-1................2

substituting value of m from equation 2 into equation 1
3(-2r-1) -5r =8
-6r-3-5r=8
-11r=8+3
-11r = 11
r=11/-11
r= -1
putting this value of r in equation 2
m= -2(-1) -1
m=2-1
m=1
so answer is r= -1, m= 1

Answer 3

3 (-2r-1)-5r=8
-6r-3-5r=8
-11r=8+3
-11r=11
r=-1

Answer 4

3(-2r-1)-5r=8
-6r-3-5r=8
-11r-3=8
-11r=11
r= -1
m=1

Hope that helps.