Turn on thread page Beta
 You are Here: Home >< A-levels

# AS C1 Maths Problem {REPRETURNS} watch

1. Solve simultaneously
xy=1
3x+2y=5

i've done 5xy=3x+2y so they both equal 5
don't know what to do now
2. (Original post by AsianG99)
Solve simultaneously
xy=1
3x+2y=5
You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
3. (Original post by Plagioclase)
You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
just updated it
4. (Original post by AsianG99)
just updated it
Okay, so what you've done is technically correct but it's not really helping you because you've still got an equation in two unknowns, which you can't solve. You need to get down to an equation in one unknown, which you can solve.

I'll give you a clue: try to find an expression for y in terms of x, and substitute it into the other equation.
5. (Original post by Plagioclase)
You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
just solved it
got x=1 and y=1
i did x=0 y=5/2
y=0 x= 5/3

3/5x+2/5y=xy
3/5*5/3= 1 for x
5/2*2/5= 1 for y
6. (Original post by Plagioclase)
Okay, so what you've done is technically correct but it's not really helping you because you've still got an equation in two unknowns, which you can't solve. You need to get down to an equation in one unknown, which you can solve.

I'll give you a clue: try to find an expression for y in terms of x, and substitute it into the other equation.
Find set of values for k where x^2-kx+k has no real roots
I've done
-k^2 - 4*1*k < 0
so -k^2 -4k < 0

4k < k^2
2k < k

i don't get this question at all
7. (Original post by AsianG99)
just solved it
got x=1 and y=1
i did x=0 y=5/2
y=0 x= 5/3

3/5x+2/5y=xy
3/5*5/3= 1 for x
5/2*2/5= 1 for y
That's one set of answers, but there are actually two sets of (x,y) that fulfil this equation. Again, I recommend that you find an expression for y in terms of x, substitute it into the other equation, and solve it. I'll do the first couple of steps.

8. (Original post by AsianG99)
Find set of values for k where x^2-kx+k has no real roots
I've done
-k^2 - 4*1*k < 0
so -k^2 -4k < 0

4k < k^2
2k < k

i don't get this question at all
Firstly, it should be because . To get further, factorise that quadratic.
9. (Original post by Plagioclase)
That's one set of answers, but there are actually two sets of (x,y) that fulfil this equation. Again, I recommend that you find an expression for y in terms of x, substitute it into the other equation, and solve it. I'll do the first couple of steps.

converted it into a quadratic and got an x value of 2/3
y=1
so they both equal 5
10. (Original post by Plagioclase)
Firstly, it should be because . To get further, factorise that quadratic.
k< +-2
11. (Original post by AsianG99)
converted it into a quadratic and got an x value of 2/3
y=1
so they both equal 5
If x = 2/3, y is not equal to 1!
12. (Original post by AsianG99)
k< +-2
I'm not really sure how you got there... please try factorising the expression. It is the only way you are going to get to an answer.
13. (Original post by Plagioclase)
If x = 2/3, y is not equal to 1!
3x+2y=5
3x=2
2y=3
y=3/2
14. (Original post by Plagioclase)
I'm not really sure how you got there... please try factorising the expression. It is the only way you are going to get to an answer.
crapppp i did d.o.t.s by accident
k(k-4)<0
k<0 or k<4
15. (Original post by AsianG99)
3x+2y=5
3x=2
2y=3
y=3/2
Yep exactly, or even easier just use the fact that
16. (Original post by Plagioclase)
Yep exactly, or even easier just use the fact that
Thank You!!!!!!
17. (Original post by AsianG99)
crapppp i did d.o.t.s by accident
k(k-4)<0
k<0 or k<4
Nearly. If the question were then you'd be right, . However, we've got the inequality. According to your answer, -1 should fulfil the equation but if you plug -1 into your expression for k, you will see that it's greater than 0 so therefore the constraint k<0 is wrong.

Instead, try sketching the graph of k(k-4) (using the roots you've already found) and find where that is less than 0.
18. (Original post by Plagioclase)
Nearly. If the question were then you'd be right, . However, we've got the inequality. According to your answer, -1 should fulfil the equation but if you plug -1 into your expression for k, you will see that it's greater than 0 so therefore the constraint k<0 is wrong.

Instead, try sketching the graph of k(k-4) (using the roots you've already found) and find where that is less than 0.
would it have to be inbetween
0<k<4
19. (Original post by AsianG99)
would it have to be inbetween
0<k<4
Yep! Well done.
20. (Original post by Plagioclase)
Yep! Well done.
i have 1 last question

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: November 20, 2017
Today on TSR

### University open days

1. University of Bradford
Wed, 25 Jul '18
2. University of Buckingham
Psychology Taster Tutorial Undergraduate
Wed, 25 Jul '18
3. Bournemouth University
Clearing Campus Visit Undergraduate
Wed, 1 Aug '18
Poll
Help with your A-levels

## All the essentials

### Student life: what to expect

What it's really like going to uni

### Essay expert

Learn to write like a pro with our ultimate essay guide.

### Uni match

Our tool will help you find the perfect course for you

### Create a study plan

Get your head around what you need to do and when with the study planner tool.

### Resources by subject

Everything from mind maps to class notes.

### Degrees without fees

Discover more about degree-level apprenticeships.

### Study tips from A* students

Students who got top grades in their A-levels share their secrets

## Study help links and info

Can you help? Study help unanswered threadsRules and posting guidelines

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE