Linear equations-help solving for 2 unkown variables

In summary, Redsummers suggested substituting one equation for another in order to solve the homework problem. He provided an example of how to do this using the y=35/x equation and y=12-x equation.
  • #1
Strafespar
47
0

Homework Statement


It seemed simple at first but, how would you solve x+y=12 also given xy=35. I can only seem to solve it by trial and error. Any solutions, I think I may be missing something that's all.

Homework Equations





The Attempt at a Solution

 
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  • #2
Have you tried solving it by substitution?
 
  • #3
What Redsummers means by substitution is to solve one of the equations for one of the variables, then substitute the result into the other equation. You then end up with an equation of a single variable that you can solve. Once you know the value of one variable, you can determine the value of the other variable from either equation.
 
  • #4
Yeah but if i solve for one variable:x+y=12
xy=35 =y-y,x-x? since I am subtracting by the same number, as opposed to a simple problem like 2x+6y=54
2x+3y=45 I can use simple substitution to figure out 3y=9, and then substitute the answer into one of the equations. I'm sorry I am posting this on a physics forum, it was just the only question I had in mind.
 
  • #5
You know x+y=12, so you can rearrange that so you define y in terms of x (leaving only y on the left).

You then substitute your new y into xy=35.

Hope that helps.
 
  • #6
hmm, I don't know how to solve this. Anyone who can, with a formula, solve this please try. Though, I don't think its possible, since x and y are interchangeable in this problem, good luck though. using y=35/x and y=12-x only gets you to--35=12y-y^2, which turns out to be: 35=xy
 
  • #7
Strafespar said:
hmm, I don't know how to solve this. Anyone who can, with a formula, solve this please try. Though, I don't think its possible, since x and y are interchangeable in this problem, good luck though. using y=35/x and y=12-x only gets you to--35=12y-y^2, which turns out to be: 35=xy

You have two equations:

x+y = 12
xy = 35

Now, to do a substitution you will need e.g. to set the x in terms of y, i.e.:

x = 12 - y

Now, putting this equation into the other:

(12 - y)y = -y^2 +12y = 35

Solving for y:

[tex]y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

where a=-1, b=12, and c=-35

And that's definitely not leading you to an imaginary number solution, so it's perfectly doable in [tex]\Re[/tex]. It's not that complex, you could even came up with the two numbers mentally.
 
  • #8
Lol, I just thought about the quadratic formula soon after this. I can't believe I had forgotten it. I was thinking it was complex, whoops. Thanks Redsummers.
 
  • #9
Strafespar said:
Lol, I just thought about the quadratic formula soon after this. I can't believe I had forgotten it. I was thinking it was complex, whoops. Thanks Redsummers.

Yeah, you got to use that formula, is like one of the commandments in linear equation religion. You're welcome ^^
 

1. What is a linear equation?

A linear equation is an algebraic equation that can be written in the form of y = mx + b, where m and b are constants and x is the independent variable. It represents a straight line when graphed.

2. How do you solve a linear equation for two unknown variables?

To solve a linear equation for two unknown variables, you need to have two equations with different variables. You can then use substitution or elimination methods to determine the values of the variables.

3. What is substitution method?

The substitution method is a way to solve a system of linear equations by solving one equation for one variable and then substituting that value into the other equation. This method is useful when one of the equations already has one variable isolated.

4. What is elimination method?

The elimination method is a way to solve a system of linear equations by eliminating one of the variables. This is done by adding or subtracting the two equations to eliminate one of the variables, then solving for the remaining variable.

5. Can you solve a system of linear equations with no solution?

Yes, a system of linear equations can have no solution if the lines represented by the equations are parallel. In this case, the equations have the same slope but different y-intercepts, so they will never intersect and there is no solution.

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