Algebra help again!

[Nocturnal]

Right now I'm trying to find a solution to each system of equations by substitution.

the problem I'm having is finding which variable to solve for and then putting it back into the equation itself.

2x + 3y = 7
6x - 2y = 10

Do I need to turn these equations into the slope-intercept form like y = mx + b or can I just solve for whatever variable I feel like it in order for it to get a regular answer or real number answer?

I would first try to turn it into y = mx + b.

So, I would subtract 2x from both sides of the equation which would bring the equation to 3y = -2x + 7

Which in turn I would divide the 3y out by 3 to get "y" by itself and then I would have (I'm going to spell it out) negative 2x over 3 + 7 over 3.

Now I have what y is equal to. I plug that into the second equation or, can I plug that into the first equation and then solve for x?

HELP!

 

[Nocturnal]

What happens when 2x and -2x cancel each other out? I don't have a X anymore, or is that still "X?"

 

[dighn]

dont plug into 1st again, you get no result

use2nd

 

[Supermercado]

If I read all that correctly, yes, once you solve one equation in terms of the other variable (in this case, y in terms of x), you substitute the right side of that equation into the other variable in the other equation and solve. Figure out what that variable is and then go back to one of the original equations with it and use it to find the other. Use both equations to check your answer.

 

[Nocturnal]

I'm getting confused because it ends up cancelling each other out. In one of the problems, that is.

 

[Nocturnal]

Do I need to turn it into y = mx + b form though to solve the problem?

 

[kehi]

2x + 3y = 7
6x - 2y = 10



(-3)2x + 3y=7
+ 6x - 2y=10
------------------------
y=17 (because -6x cancels 6x)



(original equation) 2x + 3y=7

2x + 3(17)=7
2x +51 = 7
2x= -44
x= -22


Does this help?

 

[Nocturnal]

NEVERMIND I FIGURED IT OUT, I OWN MYSELF HAHAHA

 

[NewSc2]

Originally posted by: Nocturnal
NEVERMIND I FIGURED IT OUT, I OWN MYSELF HAHAHA

You call yourself an ATOTer, and you PWN yourself? *sigh* gotta have more pride

 

[Nocturnal]

Originally posted by: kehi
2x + 3y = 7
6x - 2y = 10



(-3)2x + 3y=7
+ 6x - 2y=10
------------------------
y=17 (because -6x cancels 6x)



(original equation) 2x + 3y=7

2x + 3(17)=7
2x +51 = 7
2x= -44
x= -22


Does this help?

You got the equation wrong. You need to solve it in y = mx + b.

 

[Syringer]

Originally posted by: Nocturnal

Originally posted by: kehi
2x + 3y = 7
6x - 2y = 10



(-3)2x + 3y=7
+ 6x - 2y=10
------------------------
y=17 (because -6x cancels 6x)



(original equation) 2x + 3y=7

2x + 3(17)=7
2x +51 = 7
2x= -44
x= -22


Does this help?

You got the equation wrong. You need to solve it in y = mx + b.

For substitution yes, but otherwise that's a fine method. You can also put it in x = ay + b

 

[speg]

What was the answer? I got x=1, y=5/3. But graphing it i got x=2 y =1

NM, i mixed up x and y, i did get x=2, y=1

 

[dighn]

Originally posted by: Syringer

Originally posted by: Nocturnal

Originally posted by: kehi
2x + 3y = 7
6x - 2y = 10



(-3)2x + 3y=7
+ 6x - 2y=10
------------------------
y=17 (because -6x cancels 6x)



(original equation) 2x + 3y=7

2x + 3(17)=7
2x +51 = 7
2x= -44
x= -22


Does this help?

You got the equation wrong. You need to solve it in y = mx + b.

For substitution yes, but otherwise that's a fine method. You can also put it in x = ay + b

the method is right but the original poster did it incorrectly

 

[AthlonXP]

omg algebra 1!

 

[Nocturnal]

Ok so I have this now 3x = 8/5x - 8/5 - 3

I combine like terms which is - 8/5 and - 3.

And then How do I get one X on one side ONLY?

Do I a) multiply both sides by the reciprical of 8/5x???

 

[tikwanleap]

Originally posted by: Nocturnal
Ok so I have this now 3x = 8/5x - 8/5 - 3

I combine like terms which is - 8/5 and - 3.

And then How do I get one X on one side ONLY?

Do I a) multiply both sides by the reciprical of 8/5x???

subtract 8/5 x from both sides: --> 3x - 8/5 x = -3 - 8/5

or to get rid of the fractions, multiply both sides by 5 first: --> 15x = 8x - 8 - 15

 

[speg]

Is that 8/(5x) or (8/5)x?

Muliply everythign by 5 to get rid of fraction and rearrange for x. I got x= -23/7

 

[almega]

nocturnal, get the first equation in either x= or y= then plug into second equation for that vairable then solve for one of the answers. Plug that answer into the first equation then you will have answers for both x and y, then check your answer. For example.

2x+3y=7
6x-2y=10
solve first equation for x:
2x=7-3y
x=7/2-3/2y
plug this into 2nd equation
6(7/2)-6(3/2y)-2y=10
21-9y-2y=10
-11y=-11
y=1

plug 1 into the first equation as the y variable
2x+3(1)=7
2x+3=7
2x=4
x=2
check your answers
2(2)+3(1)=7 4+3=7 7=7
6(2)-2(1)=10 12-2=10 10=10

 

[hdeck]

solve for one variable. substitute that into the other equation. solve.

 

[Muzzan]

Originally posted by: Nocturnal
Ok so I have this now 3x = 8/5x - 8/5 - 3

I combine like terms which is - 8/5 and - 3.

And then How do I get one X on one side ONLY?

Do I a) multiply both sides by the reciprical of 8/5x???

Sort of hard to know exactly which equation you mean, since you didn't place any parantheses around 8/5x.

3x = (8/5)x - 8/5 - 3
5( 3x ) = 5( (8/5)x - 8/5 - 3 )
15x = 8x - 8 - 15
7x = -23
x = -23 / 7

or:

3x = 8/(5x) - 8/5 - 3
3x = 8/(5x) - 4,6
3x * 5x = 8 - (4,6 * 5x)
15x^2 = 8 - 23x
15x^2 + 23x = 8
x^2 + (23/15)x = 8/15

*insert quadratic formula here*

*edit* worked it out with fractions.

 

[Pilsnerpete]

I think x=2 and y=1

 

[SociallyChallenged]

2x + 3y = 7
6x - 2y = 10

Solve for either variable and just plug them in.

 

[amishhonda]

Originally posted by: Nocturnal
NEVERMIND I FIGURED IT OUT, I OWN MYSELF HAHAHA

sometimes it takes a while to figure that out

 

[Ameesh]

using subsitution i get x= 2 and y = 1

 

[Sid59]

Originally posted by: Ameesh
using subsitution i get x= 2 and y = 1

me too ..

you can take one equation and isolate a variable and then plug that variable into the other equation

1. 6x-2y=10
y=3x-5

2. 2x + 3(3x-5) = 7
11x - 15 = 7
x=2 .. plug that into any original equation

or you can add the equations

(-3)(2x+3y=7) = -6x -9y = -21

-6x - 9y = -21
6x - 2y = 10
-11y = -11
y=1 .. plug it back