math
posted by Holly .
Can someone help me solve the following....using elimination methods
1)
5x+7y=43
4x+y=25
2)9x9y=36
7y3x=14
I think the answers are
would the answer to number 1 be
(4.26,1.52)?
would the answer to number 2 be
(0.5,6.5)?
Can someone please verify or tell me what I did wrong if they are not right!
Thanks

Substitute the calculated values of x and y into the lefthandside of the equation and see if the answer match the righthandside.
For example,
5(4.26)+7(1.52)=10.66 ≠ 43
4(4.26)+1.52=18.56 ≠ 25
Therefore the answers are not correct.
You could, and you should do this check whenever you solve simultaneous equations, especially during exams.
The answers for the second problem are not correct either.
I will tackle the first one, and you can solve the second one similarly. Post your answers for a check if you can.
Solve by elimination:
5x+7y=43.....(1)
4x+y=25.....(2)
Find the LCM (lowest common multiple) of the coefficient of x, which in this case, is 20. Multiply by the appropriate factors to make the coefficients of x equal to 20.
(1)*4:
20x+28y=172 ....(1A)
20x+5y=125 ....(2A)
Add (1A) and (1B) to eliminate the xterms:
20x20x+28y+5y=172+125 ...(3)
33y = 297
y=9 ....(4)
Substitute the value of y from (4) into equation (1) to get the value of x:
5x+7(9)=43
5x=4363
x=4
Therefore x=4, y=9
Substitute into (1):
5(4)+7(9)=6320=43 OK
Substitute into (2):
4(4)+9 = 16+9 = 25 OK. 
There is no LCM for 9 and 7 though so how do you work the second one if there is no lcm?

There are no common factors between 9 and 7, so the LCM is simply the product of the two numbers, namely 63.
However, noting that the coefficients of y are 9 and 3, one being already the multiple of the other. So the job is simpler if you eliminate y instead of x.
Whichever variable you eliminate, you should get the same answers. 
You have me so confused!

Instead of looking for the LCM of 9 and 7, which are the coefficient of x, you could use the coefficients of y, which are 9 and 3, of which the LCM is simply 9. This way, you will be eliminating y from the equations to find x. Then you will substitute x into the initial equations to find y.

Now I see why you were confused!
In the second question, the variables in the second equation are inverted (y before x).
So the coefficients of x are 9 and 3. The LCM is simply 9.
Proceed as in the example above and you should get your answers of 7/2 and 1/2 for x and y respectively.
I apologize for the error and the resulting confusion. 
Lol