Question

What did I do wrong: Solve the following simultaneous equation: y = x^2 and y = 5x - 6

Answer 1

x^2 - 5x + 6 = 0
(x + 6) (x - 1) = 0
x + 6 = 0, x - 1 = 0
x = -6, x = 1, y = 36, y = 1

The answers say that the answers are x = 3, y = 9 and x = 2 and y = 4

Answer 2

You factored wrong. The second line of your working is wrong. It should factor to (x-3)(x-2)=0

Answer 3

How?

Answer 4

You need two numbers that multiply to give +6 and add to give -5. So the factors of 6 are 6 and 1, or 2 and 3. Either both factors are positive, or both are negative (to get positive 6 as the 'c' in your quadratic equation. We choose 2 and 3, and make both of them negative because -2 + -3 = -5, which is the 'b' in your quadratic equation, i.e.the coefficient of the x term.

Try expanding the second line of your working and see that you do not get back the original quadratic.

Answer 5

Oh I remember that

Answer 6

Now I am stuck on y = 2x^2 and y = 8x

Answer 7

I thought the last number could be 4 but it doesn't add up to 8

Answer 8

Make them equal to each other.

Answer 9

2x^2 - 8x - _ = 0

Answer 10

That is what I'm up to now

Answer 11

Your last number, your c, is 0, right?

Answer 12

So x is common to both 2x^2 and -8x, which means we can factor out just the x, which leaves us with x(2x-8)=0

Answer 13

ok

Answer 14

u can also factor out 2 along vth x vch will giv u 2x(x-4)=0