x+y=4
2x-y=5
What is the solution to this system?

- anonymous

x+y=4
2x-y=5
What is the solution to this system?

- katieb

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- anonymous

use elimination method

- anonymous

add down using elimination method

- anonymous

add
solve for x
then solve for y
I think it will be easier

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## More answers

- anonymous

multiply both side of x+y=4 by 5
and
multiply both side of 2x-y=5 by 4

- anonymous

I think adding two equation will give you in x terms only... no ned to go in a roundabout way

- anonymous

- anonymous

welcome :)

- anonymous

welcome..so u got it... O glad..

- anonymous

Is the answer suppose to be a decimal tho??

- anonymous

because i got -0.3333333333.. etc.

- anonymous

what did u get it??

- anonymous

show in -1/3

- anonymous

by using the elimination method ...
Oh okay @gerryliyana
then plug that in to x in x+y=4 ?

- anonymous

yes...
first you nd to eliminate one variable...
sinc there are two equations and two unknowns.. it would is possible...
but you won't get -0.3333.. in this case

- anonymous

yes

- anonymous

NO!!! you have two use the equations which are available to you

- anonymous

What would it be @Rosh007 because thats what i got, am i doing it wrong then??

- anonymous

ur right @JessNicole

- anonymous

Try using substitution and see if you got the same as me. (:

- anonymous

@jazy you made a mistake here
Solve for x in the first equation.
x + y = 4
Subtract y.
x = 4 - y
Substitute the value of x into the second equation. Solve for y.
2(4 - y) = 5 ----- mistake

- anonymous

be careful @jazy
2(4 - y) = 5 ????
the right is
2(4 - y) - y= 5

- anonymous

Oh okay now I see.

- anonymous

be careful @jazy
the mistake at
2(4 - y) = 5 ????
the right is
2(4 - y) - y= 5

- anonymous

@JessNicole please look gerryliyana's reply

- anonymous

jazzy missed one y there

- anonymous

@gerryliyana Thank youuuu!!:) for catching that mistake, i fixed it on my paper!

- anonymous

no

- anonymous

I prefer substitution.
x + y = 4
2x - y = 5
Solve for x in the first equation.
x + y = 4
Subtract y.
x = 4 - y
Substitute the value of x into the second equation. Solve for y.
2(4 - y) - y = 5
Distribute.
8 - 2y - y= 5
Subtract 8.
-2y - y = -3
-3y = -3
Divide both sides by -3.
y = 1
Now plug in the value of y(1) into any equation to solve for x.
x + y = 4
x + 1 = 4
Subtract 1
x = 3
----------------------------------
So
x = 3
and
y = 1

- anonymous

@gerryliyana What would it be then?

- anonymous

2(4 - y) - y= 5
8 - 3y = 5

- anonymous

2(4 - y) - y= 5
8 - 3y = 5

- anonymous

Thank you catching that mistake! (:

- anonymous

It should be right now!

- anonymous

yes you got it jazzy.. but a very simpl method is there

- anonymous

@JessNicole
2(4 - y) - y= 5
8 - 3y = 5

- anonymous

@jazy @Rosh007 @gerryliyana THANK YOUUUUUUUUU!! SO MUCH EVERYONE!!:D .. I'd give all of you guys each a medal if i could!!

- anonymous

sice answer is already wrote here( which I don't like to do write directly) it is for you to know that there is a simple method than this

- anonymous

@jazy you are welcome

- anonymous

@JessNicole ur welcome..., happy to help u :)

- anonymous

@Rosh007 There's an easier way ? cause you can teach it to me ..this stuff is hard for me, i'm always confused.

- anonymous

actually it is derivd from the method mentioned here.. It is actually a special case of this method..
it is applicabl here since
in one equation +y and in other equation -y is there
actually the problem is simple
x+y=4
2x-y=5
|dw:1353003234368:dw|
so when we add both equation (LHS with LHS and RHS with RHS)
you get
3x =9
which can be simplified quickly and reach the answer fast

- anonymous

You understood?

- anonymous

actually the problem is simple
x+y=4
2x-y=5
add
x+y+2x-y = 5+4
3x+0 = 9
3x =9
if you substitute the value of x in equn 1 you get th value of y....

- anonymous

@Rosh007, yess! i understood perfectly. It is much simpler!..Thank you very much. I really appreciate it.!:)

- anonymous

Welcome...But remember it is a special case ;)
all the best

- anonymous

Ha ha okay, thank youagain !:p

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