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.

Reply 4

AsianG99

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

(edited 6 years ago)

Reply 5

AsianG99

Original post by Plagioclase

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

Reply 6

Plagioclase

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.

y = \frac{1}{x}

3x + 2 \left( \frac{1}{x} \right) = 5

3x + \frac{2}{x} = 5

Reply 7

Plagioclase

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

k^2 -4k < 0

because

(-k)^2 = k^2

. To get further, factorise that quadratic.

Reply 8

AsianG99

Original post by Plagioclase

y = \frac{1}{x}

3x + 2 \left( \frac{1}{x} \right) = 5

3x + \frac{2}{x} = 5

converted it into a quadratic and got an x value of 2/3
y=1
so they both equal 5

Reply 9

AsianG99

Original post by Plagioclase

Firstly, it should be

k^2 -4k < 0

because

(-k)^2 = k^2

. To get further, factorise that quadratic.

k< +-2

Reply 10

Plagioclase

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!

Reply 11

Plagioclase

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.

Reply 12

AsianG99

Original post by Plagioclase

If x = 2/3, y is not equal to 1!

3x+2y=5
3x=2
2y=3
y=3/2

Reply 13

Plagioclase

Original post by AsianG99

3x+2y=5
3x=2
2y=3
y=3/2

Yep exactly, or even easier just use the fact that

y = \frac{1}{x} \rightarrow \frac{1}{\frac{2}{3}} = \frac{3}{2}

Reply 14

AsianG99

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

Reply 15

AsianG99

Original post by Plagioclase

Yep exactly, or even easier just use the fact that

y = \frac{1}{x} \rightarrow \frac{1}{\frac{2}{3}} = \frac{3}{2}

Thank You!!!!!!

Reply 16

Plagioclase

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

k(k-4) = 0

then you'd be right,

k = 0, 4

. 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.