# Help with solving linear systems by substitution

• Mar 19th 2008, 06:46 PM
risteek
Help with solving linear systems by substitution
Okay I was doing well until I stumbled across a question that was backwards if that makes sense.

Like:
a+4b=15
5b=2a+3

or

5=2y-x
7=3y-2x

I have been trying at those two for over an hour.

• Mar 19th 2008, 06:58 PM
risteek
Listen I have been this for 5 hours straight I am a poor math student but I try. My teacher is rude and ignorant. Please if it is not to much trouble a little help please. Currently I have no patience. (Headbang)
• Mar 19th 2008, 07:35 PM
Macleef
Quote:

Originally Posted by risteek
Okay I was doing well until I stumbled across a question that was backwards if that makes sense.

Like:
a+4b=15
5b=2a+3

or

5=2y-x
7=3y-2x

I have been trying at those two for over an hour.

I'll do the first one.

(1) a+4b=15
(2) 5b=2a+3

Isolate for either "a" or "b" in (1) or (2).
I'll isolate for "a" because it's easier.

(1) a = 15 - 4b
(2) 5b = 2a + 3

Now, substitute (1) into (2) to find "b"

5b = 2a + 3
5b = 2(15 - 4b) + 3
5b = 30 - 8b + 3
Isolate for "b"
5b + 8b = 33
13b = 33
b = 33/13

Now, substitute "b = 33/13" into (1) or (2) to find "a"

a = 15 - 4b
a = 15 - 4(33/13)
a = 15 - 132/13
Need a common denominator
a = (195 - 132)/13
a = 63/13