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

[chijioke]

$$x-y=5 \\ xy=50$$When solving
$$x-y=5~ ....\text{eq.}(1) \\ xy=50~ ....\text{eq.}(2)$$from equation (1)
$$x=5+y$$x in equation (2), we have
$$(5+y)y=50 \\ 5y+y^2=50 \\ y^2+5y-50=0$$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?

 

[Dr.Peterson]

Since $xy=50$, 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?

 

[chijioke]

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

That is an excellent idea. I will try it.

On the other hand, why bother?

Nothing really. I just want to learn different ways of doing things.

 

[Steven G]

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.

 

[Harry_the_cat]

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.

Yeah but many students would leave it at 5 and 10 and forget about the negatives.

 

[Steven G]

Yeah but many students would leave it at 5 and 10 and forget about the negatives.

At least they would be thinking!

 

[chijioke]

It is, instead, a system of nonlinear equations.

In other words it a simultaneous but non linear equations.

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

 

[Dr.Peterson]

In other words it a simultaneous but non linear equations.

View attachment 35462

You chose to multiply by y rather than x, making the work a little harder; but it amounts to the same thing. Good.

 

[Steven G]

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

 

[chijioke]

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

But at least you agreed that y=-10 or 5. If then you agree, when y= -10 , x=y+5 $$\rightarrow ~~ x=-10+5=-5$$When y=5, $$\rightarrow~~x=5+5=10$$Since, I used two variable, x=5,y=10 or x-5 or y=-10. I think that makes more sense now.

 

[Steven G]

But at least you agreed that y=-10 or 5. If then you agree, when y= -10 , x=y+5 $$\rightarrow ~~ x=-10+5=-5$$When y=5, $$\rightarrow~~x=5+5=10$$Since, I used two variable, x=5,y=10 or x-5 or y=-10. I think that makes more sense now.

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?

 

[NotJeffM]

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.

$$\text {The solution set is } \{(10, 5), (-5, -10)\}.$$

 

[chijioke]

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?

The point has already been made in post #12.