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Solving system 5r - 2s = 27, 2r + 5s = 63 by elimination
- Thread starter cal1911
- Start date
arthur ohlsten
Full Member
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- Feb 20, 2005
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Re: Solving by elimination
5r-2s=27
2r+5s=63
I will do it by eliminating the r terms. i suggest you do it by eliminating the s terms
We will make the r terms in each equation to have the same coefficients, and then either add or subtract to eliminate
multiply the 1st eq. by 2 and the second eq. by 5
10r-4s=54
10r+25s=315 subtract the eq.2 from eq.1
--------------
-29s=-261 divide both sides by -29
s=9 answer
substitute into eq1 or eq2
5r-2[9]=27
5r=45
r=9
I suggest you redo the problem
Arthur
5r-2s=27
2r+5s=63
I will do it by eliminating the r terms. i suggest you do it by eliminating the s terms
We will make the r terms in each equation to have the same coefficients, and then either add or subtract to eliminate
multiply the 1st eq. by 2 and the second eq. by 5
10r-4s=54
10r+25s=315 subtract the eq.2 from eq.1
--------------
-29s=-261 divide both sides by -29
s=9 answer
substitute into eq1 or eq2
5r-2[9]=27
5r=45
r=9
I suggest you redo the problem
Arthur
Re: Solving by elimination
I suggest you are also having trouble with your text. If it calls this the 'elimination' method, you need a better text.
1) If you add or subtract (possibly after a multiplication or two) that is 'elimination by addition'.
2) If you solve one of the equations for r (or s) and substitute that into the other, that is 'elimination by substitution'.
You want to know both methods; some examples SHOULD be done by method 1, some by method 2. It is up to you (and your book and especially your teacher) to find out which is which.
cal1911 said:Hello I am having trouble with this problem. Thanks for the help.
Solve by the elimination method
5r-2s=27
2r+5s=63
I suggest you are also having trouble with your text. If it calls this the 'elimination' method, you need a better text.
1) If you add or subtract (possibly after a multiplication or two) that is 'elimination by addition'.
2) If you solve one of the equations for r (or s) and substitute that into the other, that is 'elimination by substitution'.
You want to know both methods; some examples SHOULD be done by method 1, some by method 2. It is up to you (and your book and especially your teacher) to find out which is which.
arthur ohlsten
Full Member
- Joined
- Feb 20, 2005
- Messages
- 852
Re: Solving by elimination
Paulik is probably correct.
I agree with his definition of " by substitution" .
I thought that "elimination " probably meant eliminating one term by the method I suggested.
If you could ,would you post what your instructor meant by ,"elimination method ", for my clarification. [ Other students might request help and I would give them erronious information. I never taught high school or undergraduates thus I am not familiar with terms used by instructors.]
Arthur
Paulik is probably correct.
I agree with his definition of " by substitution" .
I thought that "elimination " probably meant eliminating one term by the method I suggested.
If you could ,would you post what your instructor meant by ,"elimination method ", for my clarification. [ Other students might request help and I would give them erronious information. I never taught high school or undergraduates thus I am not familiar with terms used by instructors.]
Arthur