simultaneous equation tips pls

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nanase

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halo Guys, I need help in solving the following simultaneous equations manually in an easy way as I am in grade 10 and can't use GDC.



what I have tried :
1. tried subtracting them, but I ended up with big equations to the power of 6
2. tried making c or K as a subject for substitution but ended up like no 1
3. I tried looking for cancellation since I noticed some parts of the equations are exactly the same but still end up with difficult equations.
thanks for your help! :)
 
I don't even know what GDC means. (Well, now I do, having looked it up.)

Sixth-degree equations are rather hard (in general, impossible); but that's what this is! I'm wondering if something in the context that you omitted would help; maybe you don't even really have to solve these equations. Please copy the entire problem you are working on, as given to you, so we can know for sure.

Also, tell us about your context -- if this is for a class, what topics have been covered that it might be intended to exercise?

What you tried is what I tried. But you might just consider trying some numbers to see if you can find a solution "by inspection", and then continue from there. (Or maybe use the rational root theorem.) Also, see if you can sketch a graph of your equation from 1.
 
sure Dr. Peterson, I am actually working on integral question. it looks like this, I have been trying several manual workings on this.
calculus.jpeg

maybe you are right, there should be a way around it than solving something to the power of 6
 
nanaseailie said:
sure Dr. Peterson, I am actually working on integral question. it looks like this, I have been trying several manual workings on this.
View attachment 21829

maybe you are right, there should be a way around it than solving something to the power of 6
Click to expand...
Look carefully at my last paragraph. As I wrote it, I was carrying out my suggestions, and was able to solve it. It actually isn't hard. I presume you got to the equation [MATH](k-1)^6=21k+1[/MATH]; sketch the graph of each side to convince yourself that there are only two solutions, and try a few possible values of k to find them. Trial and error, within the right framework, is entirely mathematical!
 
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Dr.Peterson said:
Look carefully at my last paragraph. As I wrote it, I was carrying out my suggestions, and was able to solve it. It actually isn't hard. I presume you got to the equation [MATH](k-1)^6=21x+1[/MATH]; sketch the graph of each side to convince yourself that there are only two solutions, and try a few possible values of k to find them. Trial and error, within the right framework, is entirely mathematical!
Click to expand...
do you mean 21k +1 on the right side?
is there a possible way to solve this algebraically without sketching the graph, assuming we don't want to know the shape of power of 6.
I used trial and error for k and was able to obtain k=3 and solve c, but just wondering if there is an actual steps, since i will get mark for the steps.
 
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nanaseailie said:
do you mean 21k +1 on the right side?
Click to expand...
Yes, fixed.

nanaseailie said:
is there a possible way to solve this algebraically without sketching the graph, assuming we don't want to know the shape of power of 6
Click to expand...
Try it and see. You should know the general shape of the function (it's just a very blunt parabola), but you can do this without graphing anything. The main point was to convince yourself that you have found all the solutions, once you have done so. But you can get by without it.

If you want to do more algebraically, you can expand the left-hand side (partially will be good enough) to discover one (trivial) solution algebraically.

But then just try a few simple numbers!
 
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