# sim equations

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• Sep 11th 2009, 09:42 PM
realistic
sim equations
Solve the following equations algebraically
2x + y = 0
−2x −3y = 8 im stuck i can do the easy ones when in the format y=mx+c
• Sep 11th 2009, 09:47 PM
VonNemo19
Quote:

Originally Posted by realistic
Solve the following equations algebraically
2x + y = 0
−2x −3y = 8 im stuck i can do the easy ones when in the format y=mx+c

Thee's generally two ways to do these, but I'll hint towards the addition method because it is alreaddy set up

Add the two equations like term by like term

$2x+(-2x)=0$

$y+(-3y)=-2y$

$0+8=8$

Now set up what's left

$-2y=8$

Can you finish?
• Sep 11th 2009, 09:49 PM
realistic
Quote:

Originally Posted by VonNemo19
Thee's generally two ways to do these, but I'll hint towards the addition method because it is alreaddy set up

Add the two equations like term by like term

$2x+(-2x)=0$

$y+(-3y)=-2y$

$0+8=8$

Now set up what's left

$-2y=8$

Can you finish?

yeah that would be -4
what is the other way cause i think thats what we are doing at school

3x-5y=1
x +5y=7

=
3x+x =4x
-5y+5y=0
1+7=8

0=4x+8?
x=-2?

however the answer is 2 what did i do wrong?
• Sep 11th 2009, 09:57 PM
VonNemo19
Quote:

Originally Posted by realistic
yeah that would be -4
what is the other way cause i think thats what we are doing at school

It's called the substitution method.

EG

We have the sim equations

$x+y=1$ and $3x-2y=7$

Then we solve for a variable in one of the equations and substitute for the variable in the other

$x+y=1$
$y=1-x$

Now we sub for y in the other

$3x-2\overbrace{(1-x)}^y=7$

solve for x and we can determine the solution
$3x-2(1-x)=7$
$3x-2+2x=7$
$5x=9$
$x=\frac{9}{5}$

Now we put $\frac{9}{5}$ back in the original for x.

Do you see?
• Sep 11th 2009, 10:00 PM
realistic
Quote:

Originally Posted by VonNemo19
It's called the substitution method.

EG

We have the sim equations

$x+y=1$ and $3x-2y=7$

Then we solve for a variable in one of the equations and substitute for the variable in the other

$x+y=1$
$y=1-x$

Now we sub for y in the other

$3x-2\overbrace{(1-x)}^y=7$

solve for x and we can determine the solution
$3x-2(1-x)=7$
$3x-2+2x=7$
$5x=9$
$x=\frac{9}{5}$

Now we put $\frac{9}{5}$ back in the original for y.

Do you see?

yeah thats what we are doing but your first way is alot easier

can you view the one i just tried? plz

3x-5y=1
x +5y=7

=
3x+x =4x
-5y+5y=0
1+7=8

0=4x+8?
x=-2?[/B]

also how would you tackle this one

2x=11 -3y
0=6x-2y +11
• Sep 11th 2009, 10:06 PM
VonNemo19
Quote:

Originally Posted by realistic
yeah that would be -4
what is the other way cause i think thats what we are doing at school

3x-5y=1
x +5y=7
Adding these two I'm left with

4x+0=8

Then x=2

If we substitute back into the original we get

3(2)-5y=1
6-5y=1
-5y=-5
y=1

So the equations have a common solution of

(2,1)

...
• Sep 11th 2009, 10:13 PM
realistic
Quote:

Originally Posted by VonNemo19
...

ah i c my mistake with that one

2x = 11-3y
0= 6x-2y+11

=

2x=-3xy+11
-6x= -2y+11

would you times the top by 3 then just add them ?
6x= -9y+33

+
-6x=2y +11

0= -11y +44
11y= 44
y= 4 ?

then just substitute y =4 into one of above equations to get x?
• Sep 11th 2009, 10:24 PM
VonNemo19
Quote:

Originally Posted by realistic
ah i c my mistake with that one

2x = 11-3y
0= 6x-2y+11

=

2x=-3xy+11
-6x= -2y+11

im stuck

Think of it like a game, with the object being to line thx's and y's up underneath of each other and then add or subtract them to get rid of one of them.
$2x = 11-3y$
$0= 6x-2y+11$

You can see here that the x's and y's are not right underneath of each other, but wait...

$2x =-3y+11$
$-6x=-2y+11$

Now the x's and y's are lined up. But remember that the object is to add (or subtract) so that I eliminate one of the variables. To do this I have to do a little trick with multiplication...

$(-3)(2x) =(-3y+11)(-3)$
$-6x=-2y+11$

$-6x =9y-33$
$-6x=-2y+11$

Now look! The x's and y's are still lined up, but now if I subtract the two equations, the x's will dissapear and I can solve for y!

Do you see?
• Sep 11th 2009, 10:29 PM
realistic
Quote:

Originally Posted by VonNemo19
Think of it like a game, with the object being to line thx's and y's up underneath of each other and then add or subtract them to get rid of one of them.
$2x = 11-3y$
$0= 6x-2y+11$

You can see here that the x's and y's are not right underneath of each other, but wait...

$2x =-3y+11$
$-6x=-2y+11$

Now the x's and y's are lined up. But remember that the object is to add (or subtract) so that I eliminate one of the variables. To do this I have to do a little trick with multiplication...

$(-3)(2x) =(-3y+11)(-3)$
$-6x=-2y+11$

$-6x =9y-33$
$-6x=-2y+11$

Now look! The x's and y's are still lined up, but now if I subtract the two equations, the x's will dissapear and I can solve for y!

Do you see?

thanks a bunch i now understand that i can either subtract or plus i always just thought i could just minus.

this has helped me alot and i thank you for you time (Bow)
• Sep 11th 2009, 10:31 PM
VonNemo19
Quote:

Originally Posted by realistic
thanks a bunch i now understand that i can either subtract or plus i always just thought i could just minus.

this has helped me alot and i thank you for you time (Bow)

No problem!(Wink)
• Sep 12th 2009, 12:28 AM
realistic
been doing alot of them and just got stuck at this one

3y +2x +10 = 0
2y = x −2

=
3y +2x +10 = 0
2y - x +2 = 0

• Sep 12th 2009, 06:27 AM
VonNemo19
Quote:

Originally Posted by realistic
been doing alot of them and just got stuck at this one

3y +2x +10 = 0
2y = x −2

=
3y +2x +10 = 0
2(2y - x +2) = (0)2

3y +2x +10 = 0
4y -2x +4 = 0

Now add the two equations to eliminate the x-terms

.
• Sep 12th 2009, 06:56 AM
realistic
Quote:

Originally Posted by VonNemo19
.

actually i did that i got 6 but it was wrong?

what did you get?
• Sep 12th 2009, 07:00 AM
Matt Westwood