Question

use the substitution method to solve the system of linear equations.
3a=6-b
3a=5-b

Answer 1

First you pick one of the equations and solve for a variable. Let's pick the first one and solve for a.
3a=6-b , divide both sides by 3 and \[a=\frac{ 6-b }{ 3 }\]
Next you're going to substitute that equation into the second one in the place of a in order to solve for b.
For example, 3a=5b will now be \[3(\frac{ 6-b }{ 3 }) =5b\]
Distribute the 3, \[\frac{ 18-3b }{ 3 } =5b\]
Again multiply both sides by 3 which will give you 18-3b = 15b
Then add 3b to both sides giving you 18 = 18b.
Divide by 18 and b=1

Now that you've done the hard part go back to either of the original equations and input b to find a. Let's use the first equation again, 3a=6-1.
3a=5
\[a=\frac{ 5 }{ 3 }\]

I know it's a long process but this is the substitution method. The elimination method is actually the easiest.

Answer 2

um, since 3a = ______

then just substitute 3a in one equation for the 3a of the other equation

Answer 3

3a=6-b
3a=5-b

6-b = 5-b

Answer 4

That's not how the substitution method works...

Answer 5

yes, it is ....

Answer 6

Not in college algebra.

Answer 7

youre substituting "3a" in one equation, for its equivalent "value" in the other ....

Answer 8

your method of solving for just "a" is fine .... but it tends to be too restrictive on the principle employed; and you forgot the "-" ... 5-b, not 5b

Answer 9

6-b=5-b would cancel out b though.

Answer 10

correct, which means there is no solution

Answer 11

lol my bad, I just noticed I forgot the minus.

Answer 12

your way will provide a "no solution" result as well :)

Answer 13

Yeah you're right lol.