Solve each system. { (1)/(x)+(1)/(y)+(1)/(z)=3 (2)/(x)+(1)/(y)-(1)/(z)=0 (1)/(x)-(

step 1 All right. Let's look at question number 72. So interesting issue with 72 is that the variables

Solve each system. { (1)/(x)+(1)/(y)+(1)/(z)=3 ...

Key Concepts

System of Equations

A system of equations involves finding values for multiple variables that satisfy all the given equations simultaneously. This concept is foundational in algebra and is applicable across various fields such as physics, engineering, and economics to model multiple interconnected relationships.

Reciprocal Substitution

When variables appear in the denominators of equations, a common strategy is to introduce new variables that represent their reciprocals. This substitution transforms a system with fractional relationships into a more manageable form, often converting it into a system of linear equations.

Linearization

Linearization is the process of converting a non-linear system into a linear one. By applying substitutions such as letting a new variable equal the reciprocal of an original variable, complex systems become simplified, allowing the use of linear algebra techniques to find solutions.

Elimination Method

The elimination method is a systematic procedure used in solving systems of linear equations by adding or subtracting equations to cancel out one or more variables. This method is essential for reducing the complexity of a system and finding the values of the unknowns sequentially.

Transcript

00:01 All right.

00:02 Let's look at question number 72.

00:06 So interesting issue with 72 is that the variables are in the denominator.

00:13 Remember, when we solved systems before, i simplified the equations that had fractions by multiplying by a common denominator.

00:24 Well, the only common denominator that x, y, and z will all go into is the product x, y, z.

00:37 So we're going to multiply every term from every equation by x, y, and z.

00:45 So i want to show you what that looks like very quickly.

00:48 If i multiply 1 over x by x, y, z, the x is cancel out, and i'm left with y over z.

00:56 Sorry, y, z.

00:58 If i multiply x, y, z over 1 by 2 over z, the z's cancel out and i'm left with 2xy.

01:10 So i'm going to multiply every term from every equation by x, y.

01:21 So this is what that looks like.

01:22 Equation number one becomes y z plus x z plus x y equals x y equals 3 x y z equation number two becomes 2 y z plus x z minus x y y equal 0 so i'm going to stop right here and i'm going to add these two together really fast because that will let me eliminate the x y there's an imaginary one here so that leaves me with 3 y z plus 2 x z equals 3x y z all right now i'm going to look at the third equation the third equation when i multiply it by x y z equals y z minus 2 x z plus 4 x y equals 21 x z plus 4 x y equals 21 x x x, y, z.

02:47 Now, i am going to take equation number two and multiply it by four.

02:54 That will let me have additive inverses.

02:59 So four times two is eight.

03:02 That gives me 8 y, z, four times one plus four x, z minus four x, y, y, equals zero.

03:15 And i'm going to add these two equations together.

03:17 When i add these two together, 8 plus 1 is 9y z plus a negative 2 and a positive 4 is 2 xz equals these cancel out 21 x, y, z.

03:37 So 3, 2 .3.

03:40 Let's go to a new screen.

03:44 Y, z, x, z.

03:59 And then 9 to 21...