Subsituions and eliminatons

G

Guest

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I am having trouble on subsitutions and elminations. For example 2x+-3y=5
3x+-3y=6 :?:
 
1st: Multipy each equation by numbers to make the 1st variable(x) of each equation have the same coefficient with opposite signs.
3(2x+-3y=5)
-2(3x+-3y=6)
----------------
6x+-9y=15
-6x+6y=-12

2nd: Add each equation and solve to find the 2nd variable(y).
-3y=3
y=-1

Finally: Repeat to eliminate 2nd variable(y) and find solution for 1st variable(x).

3(2x+-3y=5)
-3(3x+-3y=6
----------------
6x-9y=15
-9x+9y=-18
----------------
-3x=-3
x=1

Hope this helps. Good Luck ;-)
 
"System Substitution" (4steps)

1st: Solve 1st equation for 1st variable(x).
2x+-3y=5
2x=3y+5
x=3/2y+5/2

2nd/3rd: substitute it (x) into 2nd equation; then solve to find solution for 2nd variable(y).
3(3/2y+5/2)+-3y=6
9/2y+15/2+-3y=6
3/2y+15/2=6 <get rid of fractions by mult both sides by 2>
3y+15=12
3y=-3
y=-1

4th: Substitute it (y) into 1st solved equation to find (x).
2x+-3(-1)=5
2x+3=5
2x=2
x=1

Hope this helps. Good Luck;-)
 
A Blue Pt said:
I am having trouble on subsitutions and elminations. For example
2x+-3y=5
3x+-3y=6
Kwick start:
2x+-3y=5 : +-3y = 5 - 2x
3x+-3y=6 : +-3y = 6 - 3x

5 - 2x = 6 - 3x
3x - 2x = 6 - 5
x = 1
 
I'm stuck on the same thing, elimination. Here's the problem:

5x+2y=-3
3x+3y=9

:?. My brain feels like it might implode if I try working this thing out any other way then I already have. I'd appreciate some help very much!
 
morraco said:
I'm stuck on the same thing, elimination.
Please post new questions as new threads, not as "hijacks" of other students' threads.

Thank you for your consideration.

Eliz.
 
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