System of equations with 2 unknowns?

Question

Hi.
I came upon a problem that I am stuck with, I am not even sure how to begin.
I have these two equations:
1) x2 -3y2 + 10y = 19 ; (x squared minus 3y squared . . )
2) x2 -3y2 + 5x = 9; (x squared minus 3y squared . . )
How do I do this? Can someone please walk me through this step by step.
Thank you.

Answer 1

You have two equations and two unknowns. That's good!

A general strategy is 3 steps:
1. solve one of the equations for one of the variables as a function of the other,

2. plug that value into the other equation to solve for a value for the other variable

3. Plug the numeric value for the second variable into one of the equations to get a numeric value for the first variable.

Because of the x^2 and y^2 this would be a bit messy, but if you notice there is a little trick. Both equations have x^2 - 3y^2. So subtracting the bottom equation from the top gives:

10y-5x=19-9 or 10y -5x =10

Solve for x; -5x = 10-10y ; x= -2 + 2y

Plug this expression for x into one of the equations:
(-2+2y)^2 -3y^2 +10 y = 19

Note this just has y's in it - no x's! Multiply out the (-2 +2y)^2
as (-2+2y)(-2+2y) = 4 - 8y + 4y^2 and plug it in:

4 - 8y +4y^2 -3y^2 +10y = 19

Which simplifies as:

y^2 +2y -15 = 0

You may know how to use the quadratic equation, or just factor it: (y-3)(y+5)=0 so y=3 or y= -5.

Now you are going to have two cases for the rest of the problem. One is if y=3, the other is if y=-5. Let's do y=3 first.

Case 1: y=3
Recall x= -2 +2y so for the case y=3 we have
x= -2 +2(3) = -2 +6 = 4.

Case 2: y=-5
x= -2 + 2(-5) = -2 -10 = -12.

So your final answers are (4,3) and (-12,-5). - written in the form (x,y). Now check my math as you work through this - I may have made an error somewhere!!

Answer 2

The equations aren't linear, so you may or may not be able to get a solution. You can try substitution but it could get messy with a lot of square roots. But these look like equations for conic sections, so you might be able to help narrow down a range of values to check. Try to use some completing-the-square on these to get them into another form:

x^2 - 3y^2 + 10y = 19
x^2 - 3(y^2 - 10y) = 19
x^2 - 3(y^2 - 10y + 25) = 19 - 75
x^2 - 3(y-5)^2 = -56
[(y-5)^2 / (56/3)] - [x^2 / 56] = 1

x^2 - 3y^2 + 5x = 9
(x^2 + 5x + 25/4) - 3y^2 = 9
(x + 5/2)^2 - 3y^2 = 9
[(x + 5/2)^2 / 9] - [y^2 / 3] = 1

They're both hyperbolas, one orientated up-down and the other left-right. Find the asymptotes by letting the right hand sides equal "0" and solving for y. Graph the hyperbolas roughly by hand and see if you can find any possible regions where the two graphs could overlap. Treat these regions separately to simplify the original equations and find any solutions.

Answer 3

Although I am not an expert in Mathematics, I will try to help you.

First of all, I will assume that x and y are Real numbers (although
you have not specified this, in your question). As you know, the
set of Real numbers and the operations of "+" and "*" form a
structure of commutative field. This will allow us to make the
following operations:

1. Multiply both members, of the second equation, with the same
number (- 1) [minus one]. You will obtain:
-x^2 + 3y^2 - 5x = -9.
2. Now add the two equations. You will obtain:
10y - 5x = 10; or (which is equivalent with) 2y - x = 2, after both
members have been simplified with 5; or x = 2y - 2; or x = 2(y - 1).
3. Now, I invite you replace this value, for x, in the first, or in the
second equation. For example, if you will replace it in the second
one, you will obtain:
4(y - 1)^2 - 3y^2 + 10(y - 1) = 9. After making the necessary
calculations, you will obtain a second degree equation--which
looks like this: y^2 + 2y - 15 = 0.
4. I suppose you know how to solve it! (tip: the discriminants'
value is 64). It has to Real roots: 3 and - 5.
5. Further is trivial: from the above relation x = 2(y - 1), you will
obtain two values for x: 2 and - 12.
6. You have obtained four "potential" solutions; four pairs of
values; four possible combinations of Xs and Ys. Now you
must check each of them and see if they satisfy BOTH the
initial equations.
7. The couples of Xs and Ys, verified above, form the set of solutions
for the given equation.

I presented you one possible way of solving the system. There
are also other methods. But I will not bore you with new details.-

Answer 4

x^2 - 3y^2 + 10y = 19
x^2 = 3y^2 - 10y + 19
x = (3y^2 - 10y + 19)^(1/2)
x^2 - 3y^2 + 5x = 9
3y^2 - 10y + 19 - 3y^2 + 5(3y^2 - 10y + 19)^(1/2) = 9
5(3y^2 - 10y + 19)^(1/2) = 10y + 9 - 19
5(3y^2 - 10y + 19)^(1/2) = 10(y - 1)
(3y^2 - 10y + 19)^(1/2) = 2(y - 1)
3y^2 - 10y + 19 = 4(y^2 - 2y + 1)
3y^2 - 10y + 19 = 4y^2 - 8y + 4
y^2 + 2y - 15 = 0
(y + 5)(y - 3) = 0
y = - 5, 3
x^2 - 3(-5)^2 + 10(-5) = 19
x^2 - 75 - 50 = 19
x^2 = 144
x = - 12, 12 for y = - 5
x^2 - 3(3)^2 + 10(3) = 19
x^2 - 27 + 30 = 19
x^2 = 16
x = ± 4
(x,y) = (- 12, - 5); ( 12, - 5); ( - 4, 3); (4, 3)

Answer 5

You have two degree equation with two variables. So at least you need another equation system to get solution.

You can use linear optimization to get the solution.