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# AS simultaneous equations help watch

1. Hiya.

I'm totally stuck on this question and so I'd be really grateful if someone could please give me a few hints on how to work it out.

Solve this pair of simultaneous equations:

x = 3 - 2y

2xy - y^2 + 8 = 0

My working out is as follows (but is probably wrong somewhere):

2y (3 - 2y) - y^2 + 8 = 0

6y - 4y^2 - y + 8 = 0

- 5y^2 + 6y + 8 = 0

y^2 - 6/5y - 8/5 = 0

(y - 3/5) (y - 3/5) = 0

y^2 - 6/5y - 9/25 = 0

(y - 3/5)^2 - 31/25 = 0

(y - 3/5)^2 = 31/25

y - 3/5 = +/- the root of 31/25

Thank you.

Cathy
2. That looks sound apart from you expanding the brackets. Should be + 9/25

The best way to check is find out the two corresponding x values (don't mix them up) and plug them into the equations. If they work out, then your answers are correct
3. (Original post by CathyLou)
Hiya.

I'm totally stuck on this question and so I'd be really grateful if someone could please give me a few hints on how to work it out.

Solve this pair of simultaneous equations:

x = 3 - 2y

2xy - y^2 + 8 = 0

My working out is as follows (but is probably wrong somewhere):

2y (3 - 2y) - y^2 + 8 = 0

6y - 4y^2 - y + 8 = 0 <-- error in this line, y needs to be y^2, but corrected below.

- 5y^2 + 6y + 8 = 0

y^2 - 6/5y - 8/5 = 0

(y - 3/5) (y - 3/5) = 0 <-- incorrectly factorised

y^2 - 6/5y - 9/25 = 0

(y - 3/5)^2 - 31/25 = 0

(y - 3/5)^2 = 31/25

y - 3/5 = +/- the root of 31/25

Thank you.

Cathy

Always factorise using whole numbers, its easier.
4. Thanks so much guys! For some reason I find it impossible to check my own work.

Cathy
5. You're working out is certainly correct up to the step before you factorise.

It is at that point my method differs...I choose to find y using the quadratic formula rather than try factorising something horrid with factions or with multiples the y^2 term.

Working through the quadratic formula I get

y = 2 or y = -8/10
6. CathyLou, about the following lines:

(y - 3/5) (y - 3/5) = 0

y^2 - 6/5y - 9/25 = 0

Are they your attempted factorisations, or side working in working out what should go outside the brackets? I thought the latter, which would mean it was just the sign you muddled up and your method and understanding is sound, but if the former I'd suggest a little more practice on this
7. (Original post by Trangulor)

(y - 3/5) (y - 3/5) = 0

y^2 - 6/5y - 9/25 = 0

Are they your attempted factorisations, or side working in working out what should go outside the brackets? I thought the latter, which would mean it was just the sign you muddled up and your method and understanding is sound, but if the former I'd suggest a little more practice on this
It was just the sign that I muddled up. Sorry for not making myself clear enough but the top line was me trying to complete the square and should have read (y - 3/5)^2.

The next line was the result when I expanded the brackets and therefore should have read: y^2 - 6/5y + 9/25 = 0

Cathy
8. (Original post by Roger Kirk)
You're working out is certainly correct up to the step before you factorise.

It is at that point my method differs...I choose to find y using the quadratic formula rather than try factorising something horrid with factions or with multiples the y^2 term.

Working through the quadratic formula I get

y = 2 or y = -8/10
Yeah, they are the answers that I got eventually.

Thanks.

Cathy
9. I was also quite stuck on the following two questions as well and would really appreciate it if someone could possibly help me out a little.

3.

(a) Show that y - x = 5 is a tangent to the curve y = 2x^2 + 13x + 23.

I figured this part out as follows:

Line is a tangent if there is only one real solution to the equation:

2x^2 + 13x + 23 = x + 5

2x^2 + 12x + 18 = 0

x^2 + 6x + 9 = 0

= (x + 3)^2 = 0.

x = -3 so the equation has one root and the line is a tangent.

However, I got totally stuck on part b:

(b) Find the coordinates of the point where the straight line meets the curve.

I'd be really grateful for any help.

Thanks.

Cathy
10. Also, what method would I use to figure out:

7. The line y = 3x - 5 is a tangent to the curve y = 2x^2 + 9x + k. Find the value of k.

Thank you.

Cathy
11. a) hasn't been finished. you need to show that the gradients are equal at that point (or it could be just any old line, such as a normal)

b) you have found the x coordinate. substitute that in either of the original equations to get y. Check in the other equation if you want to be sure

7. Find when the gradients are equal (should only be one point), then substitute
12. Thank you very much!

Cathy
13. You're welcome
14. (Original post by Trangulor)

7. Find when the gradients are equal (should only be one point), then substitute
Sorry to be a pain (this is my last question!) but how would I go about finding when the gradients are equal?

Cathy
15. Differentiate both. Remember that constants 'disappear'. The value of x will be the same for the same gradient
16. Okay, I get it now.

Thanks again.

Cathy

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