You are not logged in.
Post a reply
Topic review (newest first)
- mathsyperson
- 2005-12-02 08:46:30
For that one, there's no way to get an equivalent term by changing just one of them (without involving fractions), so you need to change both.
2x - 3y = 2 --> 6x - 9y = 6
3x + 2y = 16 --> 6x + 4y = 32
Take the first from the second:
13y = 26 ∴ y = 2
Substitute in y = 2: 2x - (3*2) = 2, 2x = 8, x = 4
Check: 3*4 + 2*2 = 16.
x = 4, y = 2
You always use the method of making variables match when the two equations are linear. If they are quadratic, you need to use substitution.
The way to use substitution for the example above would be to rearrange the first equation to give x as the subject:
2x - 3y = 2
2x = 2 + 3y
x = (2 + 3y)/2
Then substitute that into the second equation.
3[(2 + 3y)/2] +2y = 16
That can be used to solve for y and then you would go back and use the y value to find x, as before.
- RickyOswaldIOW
- 2005-12-02 06:08:35
3x + 2y = 16
3x + y = 8
y = 8 - 3x
My mistake was, I was always trying to take 3x over first.
- RickyOswaldIOW
- 2005-12-02 06:06:05
What is the first step for the following simultaneous equation:
2x - 3y = 2
3x + 2y = 16
- RickyOswaldIOW
- 2005-12-02 05:24:35
Do we always use the method of multiplying one of the equations so that two of the variables match in value? Unless the coeffiecient of one of the variables is 1?
- RickyOswaldIOW
- 2005-12-02 05:18:48
*slaps himself*
- mathsyperson
- 2005-12-02 05:04:43
You're taking away the -10, but as it's already negative then it's equivalent to you adding 10.
3y - (-10y) = 3y + 10y = 13y
- RickyOswaldIOW
- 2005-12-02 05:01:37
Ah thankyou, I did not know of this method of making a pair of variables match.
4x + 3y = 19
4x - 10y = 6
surely
0x + (3y - 10y) = 13
-7y = 13 and not 13y = 13 as you suggest. Your answer is indeed correct but I'm not quite sure how you got this part.
- mathsyperson
- 2005-12-02 04:58:07
The simplest way to get one of the equations to have an equal term to the other is to double the second one.
4x - 10y = 6
Pair this up with the other:
4x + 3y = 19
4x - 10y = 6
Subtract the second equation from the first: 13y = 13 ∴ y = 1
Use this to work out x: 4x - 10 = 6, 4x = 16, x = 4
Check with the other equation: (4*4) + 3*1 = 19. So, it does have whole numbers.
- RickyOswaldIOW
- 2005-12-02 04:48:14
I know the basics of solving these, but I can't seem to get the following one right:
4x + 3y = 19
2x - 5y = 3
I've tried getting each of the letters on their own and putting it back into the equations but for each, I end up with a fractional answer. The answers in the back of the book say they should be whole numbers!