1. ## simultaneous equations

Not sure if I am quite getting this correct?

y = 2x - 3
y = -x^2 + 4x +5

I am trying to solve the above and compare the results to a graph I plotted with a straight line and a parabola. I am looking at where the graph intersects and at the moment there seems to be slight differneces between my values from the graph and the above.

2x - 3
-x^2 + 4x +5

2x - 2 = x - 1
2 2

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

4(-1) + 5 = -4 + 5 = 1

(x + 1) (x - 5)

x = - 1 or x = + 5

These x values conside with previous work I have done and don't seem wrong, please advise.

By comparison to the graph(s), the straight line graph has points (0, 1.5) and the parabola has points (0, 5).

so somewhere I sem to be not quite right?

2. ## Re: simultaneous equations

If the equations are simultaneous and if $\displaystyle y=2x-3$ and $\displaystyle y=-x^2+4x+5$ then:
$\displaystyle 2x-3=-x^2+4x+5$

Solve this equation.

3. ## Re: simultaneous equations

Originally Posted by Siron
If the equations are simultaneous and if $\displaystyle y=2x-3$ and $\displaystyle y=-x^2+4x+5$ then:
$\displaystyle 2x-3=-x^2+4x+5$

Solve this equation.
Is this a follow the BIDMAS rule to solve it?

4. ## Re: simultaneous equations

If the two equations are equal to y then they're equal to each other. Do you understand that?
Can you solve it? It's just a quadratic equation you've to solve. Put all the terms to one side and use the quadratic formula or factore.

5. ## Re: simultaneous equations

Originally Posted by Siron
If the two equations are equal to y then they're equal to each other. Do you understand that?
Can you solve it? It's just a quadratic equation you've to solve. Put all the terms to one side and use the quadratic formula or factore.
I don't seem to get it, this is what I have tried?

2x - 3 = -x^2 + 4x + 5
2x = -x^2 + 4x + 5 + 3
2x = -x^2 + 4x + 8

then

2x + x^2 + 4x + 8
6x + x^2 + 8 = y

Using Ax^2 + BX + c = 0

a = 6, b = 1, c = 8

- b + or - square root b^2 - 4ac / 2a

- 1 + or - square root 1^2 - 4 x 6 x 8 / 2 x 6

= square root 1 - 192

square root of 191 = 13.82 which is divided by 12

= 1.15 + 1 = 2.15 or

1.15 - 1 = 0.15

I believe neither of the above roots are correct, the x- values I think should be -1 and 5 as previously?

6. ## Re: simultaneous equations

hi David Green

you got the coefficients a,b and c wrong:

6x + x^2 + 8 = 0

we can write this as:

x^2 + 6x + 8 = 0

the coefficient with x^2 is 1,so a=1

the coefficient with x is 6,so b=6

the last coefficient is 8,so c=8.

can you solve now?

7. ## Re: simultaneous equations

Originally Posted by anonimnystefy
hi David Green

you got the coefficients a,b and c wrong:

6x + x^2 + 8 = 0

we can write this as:

x^2 + 6x + 8 = 0

the coefficient with x^2 is 1,so a=1

the coefficient with x is 6,so b=6

the last coefficient is 8,so c=8.

can you solve now?
Thanks I will have another go?

8. ## Re: simultaneous equations

hi

your welcome.let me know what you got.

9. ## Re: simultaneous equations

No the results are still incorrect?

Using the formula

-6 + or - square root 6^2 - 4 x 1 x 8
2 x 1

= square root of 4 = 2

2 / 2 = 1

1 + 6 = 7

1 - 6 = - 5

7 is not correct, somewhere I still am not getting the picture?

10. ## Re: simultaneous equations

If $\displaystyle 2x-3=-x^2+4x+5$
$\displaystyle \Leftrightarrow 2x-3+x^2-4x-5=0$
$\displaystyle \Leftrightarrow x^2-2x-8=0$

$\displaystyle D=(-2)^2-4(1)(-8)=4+32=36=6^2$ so we have two different solutions:
$\displaystyle x_1=\frac{2-6}{2}=-2$
$\displaystyle x_2=\frac{2+6}{2}=4$

Check:
$\displaystyle y=2(-2)-3 \Leftrightarrow y=-7$ and $\displaystyle y=2(4)-3=5$
So two solutions (-2,-7) and (4,5)

11. ## Re: simultaneous equations

Thanks for all your helps much appreciated.

I am just going to study what you have done and will let you know when I understand it now. My Graph coordinates now look wrong?

12. ## Re: simultaneous equations

You're welcome!
Take a look here for the graph:
2x-3&#61;-x&#94;2&#43;4x&#43;5 - Wolfram|Alpha