Find all solutions of the system of equations. { (2)/(x)-(3)/(y)=1 -(4)/(x)+(7)/(y)=1 .

Name: Problem 35, Chapter 10: Algebra and Trigonometry
Uploaded: 2020-08-19T00:22:01-08:00
Duration: 5 min 25 s
Channel: Erin Kearney
Description: Find all solutions of the system of equations. { (2)/(x)-(3)/(y)=1 -(4)/(x)+(7)/(y)=1 .

Find all solutions of the system of equations. { ...

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. The goal is to find values for the unknowns that satisfy all of the equations simultaneously. This concept is fundamental in algebra and serves as a basis for understanding more complex systems encountered in mathematics and applied sciences.

Rational Equations

Rational equations are equations that involve rational expressions, which are fractions containing polynomials in the numerator and/or denominator. Solving these equations requires strategies such as finding a common denominator and being careful to account for values that might cause division by zero, as these create restrictions on possible solutions.

Elimination Method

The elimination method is a systematic approach used to solve systems of equations. It involves combining the equations in such a way as to cancel out one of the variables, reducing the system to a simpler equation that can be solved more easily. This technique is particularly useful when dealing with linear systems.

Domain Restrictions

When dealing with equations that involve variables in the denominator, it is crucial to identify and apply domain restrictions. These restrictions ensure that the variables do not take on values that would make any denominator zero, as division by zero is undefined. Recognizing these limitations is essential for arriving at valid solutions.

Transcript

00:01 So my goal here is to find all the solutions to the system of equations, 2 over x minus 3 over y equals 1, and negative 4 over x plus 7 over y equals 1.

00:12 Now, when i'm finding a solution to a system, i can either use the graphic method, i can use elimination, or i can substitute.

00:20 But no matter which of the methods that i'm using, my goal is to find any and all of the points of intersection in x, comma, y format.

00:28 So i need both an x coordinate and a y coordinate.

00:33 Now given the way that this system is written, i'm going to use the elimination method, which means that i'm going to set up my system so that way i can either add the two equations or subtract them in the hopes of eliminating a variable.

00:47 Now what i notice is that 2 over x and negative 4 over x are first opposites and that this term is positive and this term is negative.

00:55 And i can probably pretty easily change this term, to match this one by multiplying by two.

01:02 And so that's what i'm going to try.

01:04 I'm going to see what happens here.

01:05 Sometimes we try things and it doesn't always work out.

01:09 But i'm going to multiply by two.

01:11 Now the key here is that when i multiply by two, i have to make sure that i multiply every single term by two.

01:17 So two times two over x is going to give me 4 over x minus two times three over y is going to give me six over y because remember multiplying by two is really multiplying by like two over one and then two times one would give me two.

01:38 Now my second equation of my system, i'm not changing at all.

01:42 I'm leaving it the way it is but i am going to rewrite it.

01:44 I like when things are stacked.

01:46 I think that makes it a little bit easier to kind of see everything that's going on.

01:51 I'm going to draw my line and i'm going to put circle.

01:55 Now when i eliminate in this circle, i can either put a plus or a minus, meaning that i'm either going to be adding these equations together or subtracting them from each other depending on which operation would get me to eliminate a variable.

02:07 Now since four over x and negative four over x are opposites, that means that i should add because addition of opposites becomes zero.

02:17 So i just want to make sure that this is actually going to work.

02:21 So off to the side, i'm going to add 4 over x plus negative 4 over x.

02:26 Let's see what happens.

02:27 So 4 over x plus negative 4 over x.

02:32 I have a common denominator, so that would stay x and 4 plus negative 4 is 0 and 0 divided by anything is 0...

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