Problem with the system of equations.

K

kittybone

New member
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Apr 21, 2022
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Hi there,
I have a problem with some system of equations. So here you can see it.
Where exactly am I stuck? I tried to solve this system by changing variables, subtracting, adding, and so on. I don't know why, but it doesn't work for me.
The main problem is that I can't bring the right side to zero in any way. math1.JPG
I tried to cut the part with Y^2:
math2.JPG



I tried to cut the X^2 part:

math3.JPG
But it didn`t help and I then keep coming back to the same expression, it is only gradually becoming more complicated by new variables, not freed from them. Looks like a vicious circle.

I need some guidance: what i can start solving the problem from and what is the MAIN IDEA of this system, understanding which i can solve it without any problems. What am I doing wrong? I also tried looking at the solution in the math problem solving apps Photomath and MicrosoftSolver:
math4.jpg

Photomath cannot solve the problem (sorry, i cant bring you an image of the Photomath, because i can add only 4 files, but you can check it on your phone), and MicrosoftMathSolver shows the correct answer, but without a solution, it seems to have solved this system graphically. I need your help please. I will answer any questions. The task is not on any particular topic. Task for the 9th grade. I'm new to this site, I read the rules, if I'm doing something wrong, tell me.
Be polite to me, please. I described the problem to you, how I tried to solve it (not all my attempts, of course, because the rest are written extremely clumsily), what grade it is, what I want from you, etc. Thank you for your understanding. Is there a mistake in the task initially? Please answer me. Or do I need a creative approach to solving a problem? I tried this, I did the replacement of variables, but I can’t competently make it so that everything works like clockwork for me. I will be VERY grateful for any help. Also I am not American or British so my methods of solving the problem can be different from yours, because we have some differences in school program, but I think it won`t disturb you. Thanks.
 
kittybone said:
Is there a mistake in the task initially?
Click to expand...
Hi kittybone. Welcome to the boards. I wouldn't say that your initial approach was a mistake; it just didn't work out. (When we're not sure what to do, initially, there's nothing wrong with experimenting or trying stuff we already know.)

My first thought for solving the system by hand is a substitution method. Are you familiar with 'Completing the Square'? You could complete the square in y – using the first equation. Then, solve the resulting equation for y, and you'll get a pair of expressions containing x. Substitute each for y in the second equation (one at a time), followed by solving for x. Once you have values for x, you may back-substitute to get the corresponding values for y.

Check your answers, using the original equations.

:)

[imath]\;[/imath]
 
YEAH I realized this earlier. Thanks. I have solved this. You can close this theme. Thanks :). I dont know why I missed that I can solve the first equation, It was obvious enough. Sorry for taking your time.
 
kittybone said:
View attachment 32297
Click to expand...
I've found that canceling out the xy terms can be useful. It'll work but it's going to be a bit ugly.

Some steps along the way:
1. Multiply the top equation by 5, multiply the bottom by 7 and add them.

2. Solve for y in terms of x.

3. Put that y value back into the top equation.
Checkpoint: [imath]4x^2\pm 7 x \sqrt{ \dfrac{41 x^2 + 7}{3} } - 2 \left ( \dfrac{41x^2 + 7}{3} \right ) = 0[/imath]

4. Solve for the square root and square both sides.

5. This will simplify into a biquadratic.
Checkpoint: [imath]1127 x^4 - 931 x^2 - 196 = 0[/imath]

6. Solve for x.

See what you can do with this.

-Dan
 
topsquark said:
I've found that canceling out the xy terms can be useful. It'll work but it's going to be a bit ugly.

Some steps along the way:
1. Multiply the top equation by 5, multiply the bottom by 7 and add them.

2. Solve for y in terms of x.

3. Put that y value back into the top equation.
Checkpoint: [imath]4x^2\pm 7 x \sqrt{ \dfrac{41 x^2 + 7}{3} } - 2 \left ( \dfrac{41x^2 + 7}{3} \right ) = 0[/imath]

4. Solve for the square root and square both sides.

5. This will simplify into a biquadratic.
Checkpoint: [imath]1127 x^4 - 931 x^2 - 196 = 0[/imath]

6. Solve for x.

See what you can do with this.

-Dan
Click to expand...
Hi, Dan. Thank you very much. Your method also worked for me, but it is a way longer than what I’ve used recently i think. Anyway it worked, i checked it and understood everything. Thanks.
 
kittybone said:
Hi, Dan. Thank you very much. Your method also worked for me, but it is a way longer than what I’ve used recently i think. Anyway it worked, i checked it and understood everything. Thanks.
Click to expand...
That's great! Be sure to check out Steve G's idea. Having more than one way to attack these is always a plus.

-Dan
 
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