Not Homework

That is probably the simplest way to get a good answer. You end up having to use the quadratic equation and I seem to get two valid answers for both x and y:

x = 1.5 - y

(1.5 - y)^2 + y^2 = 1.25 ======>2y^2 - 3y + 1 = 0

y = 1 or 1/2 ========> x = 1.5 - (1) = .5 ****or**** x = 1.5 - (.5) = 1

******************* y = 1 ; x = .5 ************* y = .5 ; x = 1

******************* b = 1 ; x = 2 ************* y = 2 ; x = 1

This seems wrong that you can get two equally valid answers that both check. Both are valid when plugged into the 1.25 = x^2 + y^2.

I think when there are fractions like that, it is best to make them into a variable for simplicity-sake.

This seems wrong that you can get two equally valid answers .

Quadratic equations do have two roots. (nth degree equations have n roots).

A line cuts a circle at two distinct points (in a special case the two are same and in anothe infinite number of special cases they are in imaginary area).

Yes. I know that quadratic equations yield two roots. I guess I am so used to just eliminating a root, usually a negative one.

x = 1/a; y = 1/b

1. 1.5 = x + y; x = 1.5 - y

2. 1.25 = x² + y² ; 1.25 = 2.25 - 3y + 2y² OR 2y² - 3y = -1 OR y² - 1.5y = -0.5

So...(y - 0.75)² = -0.5 + 0.5625 = 0.0625

y = +/-(0.0625)^1/2 + 0.75 = 1.0 or 0.5

and x = 0.5 or 1.0

a = 2 ; b = 1 OR a = 1; b = 2

Both solutions satisfy the original equations.

A circle centered on the x and y axis can be represented by x² + y² = R²

Any point on the circle has coordinates x, y. This problem finds the particular solution when the sum of the two coordinates is constant, in this case 1.5.

Why the problem was phrased in terms of 1/x and 1/y is not clear.

Hi ba/ael

"Why the problem was phrased in terms of 1/x and 1/y is not clear."

Maybe this (homework) problem was for the (trigonometry) learner to be forced to identify one or more of the fundamental identities and solve the problem that way.

I cant remember but is 1/a² + 1/b² not related to tan?

Hendrik,

I believe sb had it right in Post #2. One equation is a straight line. The other is a circle of radius 1.25. The simultaneous solution of the two equations determines two intersection points 0.5, 1.0 and 1.0, 0.5 on the x,y axes. This corresponds to 2, 1 and 1, 2 in terms of a, b. But I'm not sure why the equations were written as reciprocals in the first place. Maybe it was done to test the student's cognitive powers.

Good Morning,

Below is the graph for the two equations with intersections at (1,2) and (2,1) representing the solutions to the system of equations.

My graphing utility does not handle discontinuities, so please mentally add a hole at the origin of the graph (0,0) since dividing by zero is a very bad thing!

Have a nice day all. John

If you want to do it without the reciprocal substitution:-

1.5 = (a+b)/ab

so 2.25 = (a+b)²/a²b²

so 2.25 = (a²+2ab+b²)/a²b²..........A

AND 1.25 = (a²+b²)/a²b²).............B

A-B gives 1= 2ab/a²b²

so ab = 2

then substituting in the first equation gives 3 = a + 2/a

so a²-3a+2 = 0

(a-1)(a-2)=0

1/a^2+1/b^2+2/ab=1.5^2=2.25

1/a^2+1/b^2=1.5 by subtracting the bottom from top we get 2/ab=1 or a=2/b

so 1/2/b+1/b=1.5 or b/2+1/b=1.5 or (b^2+2)/2b=1.5 or b^2+2=3b or b^2-3b=-2

so b=2

We know 1/a+1/b=1.5 so 1/a+1/2=1.5 so 1/a=1.5-0.5=1 Therefore 1/a=1 so a=1

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