Solve each system of equations. x-2 y+4 z=3 x+3 y-2 z=6 x-4 y+3 z=-5

Name: Problem 15, Chapter 4: Algebra for College Students
Uploaded: 2020-05-01T20:18:34-08:00
Duration: 3 min 57 s
Channel: Jake Zanazzi
Description: Solve each system of equations. x-2 y+4 z=3 x+3 y-2 z=6 x-4 y+3 z=-5

Solve each system of equations. x-2 y+4 z=3 x+3 y-2...

Key Concepts

Cramer’s Rule

This is a method that uses determinants of matrices to solve for the variables in a system of linear equations. It applies when the system has a unique solution and provides a formulaic way to express each variable, although it is less efficient for larger systems.

Matrix Methods and Gaussian Elimination

Systems of linear equations can also be solved by representing them in matrix form and performing row operations to convert the matrix into row-echelon or reduced row-echelon form. Gaussian elimination is a systematic procedure that simplifies solving the system and is especially useful for systems with many variables.

Substitution Method

In this method, one of the equations is solved for one variable in terms of the others, and that expression is substituted into the remaining equations. This process reduces the number of variables in the subsequent equations, allowing the solution to be determined step by step.

Elimination Method

A common method for solving systems of linear equations involves adding or subtracting entire equations to cancel one of the variables. This reduction simplifies the system into fewer equations with fewer variables, eventually leading to the solution.

System of Linear Equations

These are collections of two or more linear equations involving the same set of variables. The objective is to find values for the variables that satisfy all given equations simultaneously. Systems of linear equations are foundational in algebra and appear in various fields of mathematics and applied sciences.

Transcript

00:01 All right, so here are our system equations x minus 2y plus 4 z equals 3.

00:06 We have x plus 3y minus 2 z equals 6.

00:11 Finally, we have x minus 4y plus 3z equals negative 5.

00:17 Well, you notice we have an x in every single one of these equations.

00:21 So what i'm going to do is i'm going to solve for x in this first equation, 2y minus 4 z plus 3.

00:30 And i'm also going to do it in the second equation.

00:33 It's going to give us 6 minus 3y plus 2z.

00:39 Now, you can see that these two equations are equivalent because they both equal x.

00:43 So what you can do is you can actually set them equal to each other.

00:50 You're going to be able to solve for z in terms of y and the other way around.

00:57 So now we can see that this is going to simplify and it's going to give us 3.

01:01 If we get rid of this 3, minus 3 y plus 2.

01:05 2y is going to be minus y, and if we move this 2 z over, equals 2z.

01:11 Therefore, y is going to equal 3 minus 2 z.

01:16 So now we have y in terms of z.

01:19 Now what we can do is we can go back to that original problem, and we said that x minus 2, and now we have y.

01:28 We can substitute this equation in for y, 3 minus 2 z plus 4 z equals 3.

01:38 X minus 6 plus 4 z plus 4 z equals 3 this is going to give us x equals 9 minus 8 z so now we have x and y in terms of z and we can plug these equations into either any of these top three to actually solve for z once we have z you plug back in to solve for x and y so let's do that here i'm going to use the second equation.

02:07 We have x, which we know is 9 minus 8 z plus three times y, which is 3 minus 2 z, minus 2z equals 6...

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