Solve this System of Equations: .5= (x-40)/(x-y) and .5=(40-y)/(x-y)

In summary, the student is attempting to solve a system of equations but has not been successful. They have tried various methods and gotten different results, which suggests they might be missing a step.
  • #1
bmed90
99
0

Homework Statement



Solve for x and y

.5= (x-40)/(x-y) and .5=(40-y)/(x-y)

Homework Equations





The Attempt at a Solution



I have been trying to solve this system of equations for hours and I can't seem to get it. Can anyone please give it a shot and see what they get? I'm sure I am overlooking a simple mistake. Thanks
 
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  • #2
.5= (x-40)/(x-y) and .5=(40-y)/(x-y)

You need to first isolate a single variable for the substitution method, or put the equations in the standard form for elimination. I chose to use the substitution method (and what better variable to chose then Y to double check with a calc). I will show you my work for the first equation.

.5 = (x-40)/(x-y)

//Multiply both sides of the left eq. by the denominator of the right to cancel the denominators/fractions.
.5(x-y) = (x-40)

//Distribute out the .5(x-y)
.5x-.5y = x-40

//Subtract a .5x from both sides
-.5y = .5x - 40

//Finally divide both sides by a -.5 to isolate the variable y, resulting in:

y = (.5x-40)/(-.5)

Here is what I got for the two in terms of Y.

Y1= (.5x-40)/(-.5)

Y2= (40+.5x)/(.5)

Surely you know where to go from here? (Substitution, Elimination, Matrice, or a graphing calculator etc)

How do your steps look? Perhaps if you show your steps up until, and after this point we can figure out where you are going wrong.

Cheers
 
Last edited:
  • #3
Personally, I would get rid of all fractions by multiplying each equation by 2(x- y).

That gives x- y= 2(x- 40) and x- y= 2(40- y). Multiplying those out, x- y= 2x- 80 or x+ y= 80, and x- y= 80- 2y or x+ y= 80. Those two equations are the same so the original equations were not "independent". There are an infinite number of solutions. Given any x, (x, y)= (x, 80- x) is a solution (except (40, 40) for which x- y= 0 and the original equation makes no sense).
 
  • #4
HallsofIvy said:
Personally, I would get rid of all fractions by multiplying each equation by 2(x- y).

That gives x- y= 2(x- 40) and x- y= 2(40- y). Multiplying those out, x- y= 2x- 80 or x+ y= 80, and x- y= 80- 2y or x+ y= 80. Those two equations are the same so the original equations were not "independent". There are an infinite number of solutions. Given any x, (x, y)= (x, 80- x) is a solution (except (40, 40) for which x- y= 0 and the original equation makes no sense).

Again with giving out the answer! I thought that wasn't allowed in the homework forum!
 

What is a system of equations?

A system of equations is a set of two or more equations that contain multiple variables. The goal of a system of equations is to find the values of the variables that satisfy all of the equations simultaneously.

What is the solution to a system of equations?

The solution to a system of equations is a set of values for the variables that satisfy all of the equations in the system. This can be a single solution, multiple solutions, or no solution at all.

How do you solve a system of equations?

There are several methods for solving a system of equations, including substitution, elimination, and graphing. The most common method is substitution, where you solve for one variable in one equation and plug that value into the other equation to solve for the other variable.

What is the difference between consistent and inconsistent systems of equations?

A consistent system of equations has at least one solution, while an inconsistent system has no solutions. Inconsistent systems can occur when two or more equations contradict each other, making it impossible to find a solution.

Why are systems of equations important in science?

Systems of equations are used in science to model and solve real-world problems that involve multiple variables. They are particularly useful in fields such as physics, chemistry, and engineering, where complex systems and relationships between variables need to be analyzed and understood.

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