Quadratic formula

[homeschool girl]

I have a question asking "
Find one pair

of real numbers such that

and

"

I did the problem, but I have two pairs, not just one and I don't know where I went wrong.

 

[homeschool girl]

this is how I got two pairs:
[MATH][/MATH][MATH]x+y=4[/MATH][MATH]x=4-y[/MATH]plug in value for x
[MATH](4-y)^3+y^3=100[/MATH][MATH](4-y)(4-y)(4-y)+y^3=100[/MATH][MATH]-y^3+12y^2-48y+64+y^3=100[/MATH][MATH]-y^3+y^3+12y^2-48y+64=100[/MATH][MATH]12y^2-48y+64=100[/MATH]and solve the quadratic

[MATH]y=2-sqrt7 ,2+sqrt7[/MATH]
and plug in the values of y

[MATH]x=4-y[/MATH]
[MATH]x=4-(2+sqrt7)[/MATH]
[MATH]x=4-(2+sqrt7)[/MATH]

 

[Deleted member 4993]

this is how I got two pairs:
[MATH][/MATH][MATH]x+y=4[/MATH][MATH]x=4-y[/MATH]plug in value for x
[MATH](4-y)^3+y^3=100[/MATH][MATH](4-y)(4-y)(4-y)+y^3=100[/MATH][MATH]-y^3+12y^2-48y+64+y^3=100[/MATH][MATH]-y^3+y^3+12y^2-48y+64=100[/MATH][MATH]12y^2-48y+64=100[/MATH]and solve the quadratic

[MATH]y=2-sqrt7 ,2+sqrt7[/MATH]
and plug in the values of y

[MATH]x=4-y[/MATH]
[MATH]x=4-(2+sqrt7)[/MATH]
[MATH]x=4-(2+sqrt7)[/MATH]

The problem asked you to find one pair of solutions.

However, it did not specify how many pairs could be potential solution.

You report one pair - out of those.

Do not forget to check your solution (by putting those back in the equation) for accuracy.

 

[homeschool girl]

I got it, thank you very much

 

[MarkFL]

What you did was fine. I would approach this problem as follows:

[MATH](x+y)^3=4^3[/MATH]
[MATH]x^3+3x^2y+3xy^2+y^3=64[/MATH]
[MATH]xy(x+y)=-12[/MATH]
[MATH]xy=-3[/MATH]
[MATH]x-\frac{3}{x}=4[/MATH]
[MATH]x^2-4x-3=0[/MATH]
[MATH]x=2\pm\sqrt{7}[/MATH]
Hence:

[MATH](x,y)=(2\pm\sqrt{7},2\mp\sqrt{7})[/MATH]
It can be shown that the graph of \(x^3+y^3=100\) is asymptotic to the line \(y=-x\), which is parallel to the given line.