Is this also simultaneous linear equations? x - y = 5 and xy = 50

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chijioke

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[math]x-y=5 \\ xy=50[/math]When solving
[math]x-y=5~ ....\text{eq.}(1) \\ xy=50~ ....\text{eq.}(2)[/math]from equation (1)
[math]x=5+y[/math]x in equation (2), we have
[math](5+y)y=50 \\ 5y+y^2=50 \\ y^2+5y-50=0[/math]y=5, or y=-10
When y =5, x=5+5=10
Thus y=5, and x=10
When y=-10, x=5+(-10)=5-10=-5
Thus y=-10, x=-5
This has been solved simultaneously by method of substitution.
I know simultaneous linear equations can be solved by elimination method. If the above equation is simultaneous linear equations, can it be solved by elimination method? If yes. How can it be done?
 
Since [imath]xy=50[/imath], is not a linear equation, this clearly is not a system of linear equations!

It is, instead, a system of nonlinear equations. Those are usually solved by substitution, but there are sometimes other things you can do.

In this case, to use something like the addition (elimination) method, try multiplying the first equation by x, then adding!

On the other hand, why bother?
 
Here is a different way to think about this problem. You need to understand what the equations are saying.

x-y means that the two numbers differ by 5 (and that x is the larger of the two).
xy =50 means that the product of the two numbers is 50.

To me the problem is asking to find two numbers that differ by 5 and multiply to 50.
The answer is immediate that x=10 and y=5 OR x=-5 and y=-10.
 
Steven G said:
Here is a different way to think about this problem. You need to understand what the equations are saying.

x-y means that the two numbers differ by 5 (and that x is the larger of the two).
xy =50 means that the product of the two numbers is 50.

To me the problem is asking to find two numbers that differ by 5 and multiply to 50.
The answer is immediate that x=10 and y=5 OR x=-5 and y=-10.
Click to expand...
Yeah but many students would leave it at 5 and 10 and forget about the negatives.
 
I don't agree with the way you wrote your solution.
You wrote that y=-10 or y=5.
Then you said that x =5 or x=10. (x=5 is wrong, it is x = -5).
This seems to me that you feel that you can pick any of the two y values along with any of the two x values(??).

y=-5 and x=10 is not a solution. If y=-5, then must be x must be -5.

The solution is y=-10 and x = -5 OR y=5 and x=10
 
Steven G said:
I don't agree with the way you wrote your solution.
You wrote that y=-10 or y=5.
Then you said that x =5 or x=10. (x=5 is wrong, it is x = -5).
This seems to me that you feel that you can pick any of the two y values along with any of the two x values(??).

y=-5 and x=10 is not a solution. If y=-5, then must be x must be -5.

The solution is y=-10 and x = -5 OR y=5 and x=10
Click to expand...
But at least you agreed that y=-10 or 5. If then you agree, when y= -10 , x=y+5 [math]\rightarrow ~~ x=-10+5=-5[/math]When y=5, [math]\rightarrow~~x=5+5=10[/math]Since, I used two variable, x=5,y=10 or x-5 or y=-10. I think that makes more sense now.
 
chijioke said:
But at least you agreed that y=-10 or 5. If then you agree, when y= -10 , x=y+5 [math]\rightarrow ~~ x=-10+5=-5[/math]When y=5, [math]\rightarrow~~x=5+5=10[/math]Since, I used two variable, x=5,y=10 or x-5 or y=-10. I think that makes more sense now.
Click to expand...
Why not just check for yourself? For each combination does xy=50 and do the numbers differ by 5 (and x is the larger number)?
Did you mean to say what it says in bold?
 
A point that has been referenced all the way back to Harry the Cat’s response is that the solution set consists of exactly TWO elements, each of which is an ORDERED PAIR of numbers. There is a way to be clear about what those elements are.

[math]\text {The solution set is } \{(10, 5), (-5, -10)\}.[/math]
 
Steven G said:
Why not just check for yourself? For each combination does xy=50 and do the numbers differ by 5 (and x is the larger number)?
Did you mean to say what it says in bold?
Click to expand...
The point has already been made in post #12.
 
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