# How to solve these equation?I try but getting wrong answer

• vkash
In summary, the conversation discusses finding the number of ordered pairs (x,y) that are solutions to the equation (x+y)+(x/y)=1/2 and (x+y)(x/y)=-1/2, with options for the answer being 0, 1, 2, or none of these. The individual providing the answer suggests setting u = x+y and v = x/y to simplify the equations, and points out that the solutions will not necessarily satisfy all values where y=x or y=-2x. They also mention finding the solutions by inserting y=x or y=-2x into the original equations, and acknowledge a mistake they made in their initial work. The conversation ends with the use of Wolfram Alpha to
vkash
Question
Number of ordered pair (x,y)(for y not equal to zero) solution of this equation.
(x+y)+(x/y)=1/2
(x+y)(x/y)=-1/2
options:
(a) 0
(b) 1
(c) 2
(d) none of these {this is my answer}

from second equation (x+y)=(-1/2)(y/x)
putting this in first equation.
(-1/2)(y/x)+(x/y)=1/2
let y/x=t;
-t/2+1/t=1/2
=>t2+4t-1=0;
Discriminant of this equation is greater than zero so there will two values for y/x;
hence there can be infinite sets of values y,x that can satisfy this equation, so answer should (d) none of these..
But this is not correct where am i doing it wrong.
thanks!

Just a quick look and it looks like you've multiplied t by 4 rather then 2..? Does this help?

HmBe said:
Just a quick look and it looks like you've multiplied t by 4 rather then 2..? Does this help?

You are pointing out it?
-t/2+1/t=1/2
=>t2+4t-1=0;
Oh! sorry that's typo error:"
it will t2+t-2=0
But even after that same problem is there. It has discriminant >0; so there must infinite solutions...

Try solving that quadratic, y/x=t, so you will have two solutions to y/x. Try rearranging these solutions so you can put them in one of the equations.

vkash said:
You are pointing out it?

Oh! sorry that's typo error:"
it will t2+t-2=0
But even after that same problem is there. It has discriminant >0; so there must infinite solutions...

Now the equation is correct. You will obtain two values for t. You just need to insert them in anyone of the original two equations and substitute x for y/t and then each of the two values for t will give you an ordered pair (x,y). Thus, you will have not have an infinite number of ordered (x,y) pairs but only two.

vkash said:
Question
Number of ordered pair (x,y)(for y not equal to zero) solution of this equation.
(x+y)+(x/y)=1/2
(x+y)(x/y)=-1/2
options:
(a) 0
(b) 1
(c) 2
(d) none of these {this is my answer}

from second equation (x+y)=(-1/2)(y/x)
putting this in first equation.
(-1/2)(y/x)+(x/y)=1/2
let y/x=t;
-t/2+1/t=1/2
=>t2+4t-1=0;
Discriminant of this equation is greater than zero so there will two values for y/x;
hence there can be infinite sets of values y,x that can satisfy this equation, so answer should (d) none of these..
But this is not correct where am i doing it wrong.
thanks!

Why not set u = x+y and v = x/y, to get the system
u + v = 1/2 and u v = -1/2 ? After you know u and v you can get x and y.

RGV

spf said:
Now the equation is correct. You will obtain two values for t. You just need to insert them in anyone of the original two equations and substitute x for y/t and then each of the two values for t will give you an ordered pair (x,y). Thus, you will have not have an infinite number of ordered (x,y) pairs but only two.

solution of the equation is y/x=1 and y/x= -2 so all values where y=x &y= -2x should be satisfied. BUT why these two conditions are not satisfying the main two equations.
got confused?><?
did it required to find out the relation in x+y too?
RGV said:
Why not set u = x+y and v = x/y, to get the system
u + v = 1/2 and u v = -1/2 ? After you know u and v you can get x and y.

RGV
excellent help but what's wrong in my work...

vkash said:
solution of the equation is y/x=1 and y/x= -2 so all values where y=x &y= -2x should be satisfied. BUT why these two conditions are not satisfying the main two equations.
got confused?><?
did it required to find out the relation in x+y too?

excellent help but what's wrong in my work...

I don't know; I did not check. RGV

Last edited:
vkash said:
solution of the equation is y/x=1 and y/x= -2 so all values where y=x &y= -2x should be satisfied. BUT why these two conditions are not satisfying the main two equations.
got confused?><?
did it required to find out the relation in x+y too?

Finding the two conditions for y/x is only the first step in solving the original equations. With this step you did not find the solutions to the original equations yet, you just narrowed them down. By now, you only know that the solutions to the original two equations satisfy either y=x or y=-2x (and NOT that ALL the values which satisfy y=x or y=-2x are all solutions to your original equations, most of them aren't). To find the the solutions to the original equations, you have to first insert y=x in anyone of the two original equations and solve it in order to obtain the first ordered (x,y) pair, and then insert y=-2x in anyone of the two original equations and solve it in order to obtain the second ordered (x,y) pair.

vkash said:
solution of the equation is y/x=1 and y/x= -2 so all values where y=x &y= -2x should be satisfied. BUT why these two conditions are not satisfying the main two equations.

Just return that equation in one of the first equation and you will found all solutions.

You can also see SOLUTION

EDIT: oh, i didn't see that @spf post

spf said:
Finding the two conditions for y/x is only the first step in solving the original equations. With this step you did not find the solutions to the original equations yet, you just narrowed them down. By now, you only know that the solutions to the original two equations satisfy either y=x or y=-2x (and NOT that ALL the values which satisfy y=x or y=-2x are all solutions to your original equations, most of them aren't). To find the the solutions to the original equations, you have to first insert y=x in anyone of the two original equations and solve it in order to obtain the first ordered (x,y) pair, and then insert y=-2x in anyone of the two original equations and solve it in order to obtain the second ordered (x,y) pair.
thanks spf..
that's genuine mistake i did.
actually just assembling equation together aways does not give answer.every step may add something new thing or may remove some answers.

karmata said:
Just return that equation in one of the first equation and you will found all solutions.

You can also see SOLUTION

EDIT: oh, i didn't see that @spf post
Here i am concerned about mistake i did.

after all i understood my mistake from spf's tiny lecture.
thanks to all of you for helping me.

vkash said:
after all i understood my mistake from spf's tiny lecture.
thanks to all of you for helping me.

Glad that I could help. Thanks to Ray Vickson for the elegant alternative substitution and to Karamata for the nice visual presentation of the solutions with wolfram-alpha.

## 1. How do I know which method to use to solve an equation?

There are several methods you can use to solve an equation, including substitution, elimination, and graphing. The best method to use depends on the type of equation and its complexity. It is important to understand all the different methods and when they are most effective in order to choose the best approach for a particular equation.

## 2. What should I do if I keep getting a wrong answer when solving an equation?

If you are consistently getting a wrong answer when solving an equation, it could be due to a simple math error or a misunderstanding of the concept. Double check your calculations and make sure you are following the correct steps for the chosen method. If you are still stuck, seek help from a tutor or teacher who can guide you through the problem.

## 3. How do I handle fractions or decimals when solving an equation?

If your equation contains fractions or decimals, it is important to simplify or convert them to whole numbers before starting the solving process. This can make the equation easier to work with and reduce the chance of making a mistake. If you are using a graphing method, some calculators have the ability to handle fractions and decimals, but it is always best to simplify them beforehand.

## 4. What if the equation has variables on both sides?

If the equation has variables on both sides, the goal is to get all the variables on one side and the constants on the other side. This can be done by using methods such as elimination or combining like terms. Once the variables are on one side, you can proceed with solving the equation as usual.

## 5. How do I check my answer to make sure it is correct?

To check your answer, you can substitute the value you obtained for the variable back into the original equation. If both sides of the equation are equal, then your solution is correct. In some cases, you can also use a graphing method to visually verify your solution. It is always important to double check your work to ensure accuracy.

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