Simultaneous equations

[maali5]

1. The problem statement, all variables and given/known data

How do yo solve simultaneos equations?

2. Relevant equations

Am I right ?


3. The attempt at a solution

y = x2 – 5x + 5 question a



5x + 3y = 30. question b



Find location of points collide?


Add a+b

x2 + 2y -25=0


quadratic formula use

x= -2+ square root(2^2-(4x1x(-25))/2 x = 4.09901

or

x= -2 - square root(2^2-(4x1x(-25))/2 x=-6.09


subsitue y etc

[berkeman]

1. The problem statement, all variables and given/known data

How do yo solve simultaneos equations?

2. Relevant equations

Am I right ?


3. The attempt at a solution

y = x2 – 5x + 5 question a



5x + 3y = 30. question b



Find location of points collide?


Add a+b

x2 + 2y -25=0


quadratic formula use

x= -2+ square root(2^2-(4x1x(-25))/2 x = 4.09901

or

x= -2 - square root(2^2-(4x1x(-25))/2 x=-6.09


subsitue y etc

Looks like you are on the right track...

[maali5]

Looks like you are on the right track...

Cheerio

[eumyang]

Add a+b

x2 + 2y -25=0
I would use substitution. You already have equation a solved for y. Plug it into equation b:
5x + 3y = 30
5x + 3(x2 – 5x + 5) = 30
... and so on.

[NascentOxygen]

Looks like you are on the right track...

Errrm, if eumyang is on the right track, then I'll eat my hat!

... and I'll post the video of it on youtube.

[SammyS]

...

x2 + 2y -25=0

quadratic formula use

x= -2+ square root(2^2-(4x1x(-25))/2, x = 4.09901

or

x= -2 - square root(2^2-(4x1x(-25))/2, x=-6.09

substitute y etc.
You can't use the quadratic formula on x2 + 2y -25=0, because it has two different variables in it.

B.T.W: Yesterday, Mark44 pointed out to you that you should not use the lower case letter, x, to indicate multiplication.

[maali5]

Hey ladies ;) + gentleman :)

Now you have "REALLY" lost me.

I went with eumyang- subsituations and this is what I have done.

NB: Allright I WILL USE . FOR MULTIPLY NOW

Ok, lets start

1)

y = x^2 – 5x + 5 question a



5x + 3y = 30. question b


Find location of points collide?



Answer;


5x + 3(x^2 – 5x + 5) = 30

5x+ 15x^2 -15x+15=30

15x^2-10x+15=30

15x^2-10x -15 = 30

5(3x^2 -2x -3) = 30

use formula

x= 2 + sqrt{40} / 6

x= 1.3874

or

x=2 - sqrt{40} / 6

x= -0.72075


Is this the right directions?

[SammyS]

...

Answer;
5x + 3(x^2 – 5x + 5) = 30

5x+ 15x^2 -15x+15=30
The coefficient of the x2 term is 3, not 15.
15x^2-10x+15=30

15x^2-10x -15 = 30

5(3x^2 -2x -3) = 30

use formula

x= 2 + sqrt{40} / 6

x= 1.3874

or

x=2 - sqrt{40} / 6

x= -0.72075

Is this the right directions?Once you get a solution for x, you should go back to an original equation to find y. To check your answer for the pair x & y, you should then plug those values into the other original equation.

[NascentOxygen]

Errrm, if eumyang is on the right track, then I'll eat my hat!

... and I'll post the video of it on youtube.

CORRECTION: I intended to write
Errrm, if maali5 is on the right track, then I'll eat my hat!

... and I'll post the video of it on youtube.

[NascentOxygen]

You must make the right hand side = 0 before using the quadratic formula.

Taking as an example: 15x^2 - 10x - 15 = 30

Make RHS=0:

15x^2 - 10x - 15 - 30 = 30 - 30

x = (10 ± sqrt(100 + 4•15•45)) / 30