Thread: Simultaneous Equations

2xy + y = 10 (1)
x + y = 4 (2)

I got as far as:

(2) y = 4 - x

substitute into (1)

2x(4-x) + (4-x) = 10

2x(4-x) + (4-x) -10 = 0

Question: how to know when/ why to make it equal 0?

Also, I multiply out and get

8x - 2x^2 + (4 - x)- 10 = 0

then, 7x - 2x^2 - 6 = 0

but the answer has 2x^2 - 7x + 6 = 0 at this stage

Confused.

Originally Posted by GAdams

2xy + y = 10 (1)
x + y = 4 (2)

I got as far as:

(2) y = 4 - x

substitute into (1)

2x(4-x) + (4-x) = 10

2x(4-x) + (4-x) -10 = 0

Question: how to know when/ why to make it equal 0?

Also, I multiply out and get

8x - 2x^2 + (4 - x)- 10 = 0

then, 7x - 2x^2 - 6 = 0

but the answer has 2x^2 - 7x + 6 = 0 at this stage

Confused.

If you are asking, in terms of a procedure, when you should rearrange an equation such that one side is equal to 0, it's going to depend on the method.

In your case, you simply haven't gone far enough. If you can't factor what you have, use the quadratic formula (or completing the square). I'll factor it:
2x^2 - 7x + 6 = (2x - 3)(x - 2) = 0

x = 3/2 or x = 2

So we need y values for each of these points:
x = 3/2 ==> y = 4 - 3/2 = 5/2
x = 2 ==> y = 4 - 2 = 2

So your solutions are:
(x, y) = (3/2, 5/2)
(x, y) = (2, 2)

-Dan

OK. I don't understand this:

after substituting you get:

8x - 2x^2 + 4 - x - 10 = 0

How do we end up with

2x^2 - 7x + 6 = 0

shouldn't it be:

7x -2x^2 -6 = 0 ???

Originally Posted by GAdams

OK. I don't understand this:

after substituting you get:

8x - 2x^2 + 4 - x - 10 = 0

How do we end up with

2x^2 - 7x + 6 = 0

shouldn't it be:

7x -2x^2 -6 = 0 ???

Step by step then.

7x - 2x^2 - 6 = 0

Multiply both sides by -1:

(-1)*(7x - 2x^2 - 6) = (-1)*0

-7x + 2x^2 + 6 = 0

2x^2 - 7x + 6 = 0

-Dan