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

math 
MathMate
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. 
math 
Holly
There is no LCM for 9 and 7 though so how do you work the second one if there is no lcm?

math 
MathMate
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. 
math 
Kristen
You have me so confused!

math 
MathMate
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.

mathcorrection 
MathMate
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. 
math 
Epic
Lol
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