[SOLVED] Need help solving a system of equation

**Rheanna** · October 19th 2009, 04:17 PM

Trying to figure out how they are coming up with 2,3 and not 5,3
Thanks

4x+3y=17
2x+3y=13

**Rheanna** · October 19th 2009, 05:22 PM

Nevermind, rather than adding the bottom equation they subtracted it which makes it come out to 2,3

**jeremyparker10** · October 20th 2009, 05:09 AM

subtract the 2nd equation from the 1st

you'll get 2x = 4

after solving you'll get x= 2

and put the value of x in any of the above eqn you'll get y

**pacman** · October 20th 2009, 08:15 AM

4x+3y=17
2x+3y=13
______________________________________________

4x+3y=17
(2x+3y=13) - multi ply by -2 to the x terms. Now add . . . .

4x + 3y = 17
-4x - 6y = -26
0 - 3y = -9

y =-9/-3 = 3

Next,

4x+3y=17
2x+3y=13, multiply by -1 to eliminate y terms,

4x +3y = 17
-2x - 3y = -13 then add,

?

**Defunkt** · October 20th 2009, 09:04 AM

Originally Posted by Rheanna

Nevermind, rather than adding the bottom equation they subtracted it which makes it come out to 2,3

Just a note: It should make no difference whether they add or substract the equations -- either way there is one unique solution.

If you choose to add them, you get:

$4x+3y+2x+3y=17+13 \Rightarrow 6x+6y = 30 \Rightarrow 6(x+y) = 6*5$ $\Rightarrow x+y = 5 \Rightarrow x = 5-y$

Now substitute that into either equation, let's say the first:

$4(5-y)+3y = 17 \Rightarrow 20-4y+3y=17\Rightarrow -y = -3 \Rightarrow y = 3 \Rightarrow x + 3 = 5$ $\Rightarrow \boxed{x=2, y=3}$

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