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

Name: Problem 17, Chapter 7: Beginning and Intermediate Algebra
Uploaded: 2022-02-22T20:08:49-08:00
Duration: 4 min 4 s
Channel: Alison Rodriguez
Description: Solve each system. x+y-z=-2 2 x-y+z=-5 -x+2 y-3 z=-4

step 1 For this problem, we're going to be solving the system provided. In order to do this, we're goin

Solve each system. x+y-z=-2 2 x-y+z=-5 -x+2...

Key Concepts

Gaussian Elimination

Gaussian elimination is a method that uses row operations on the augmented matrix of a system to simplify it into row echelon form or reduced row echelon form. This matrix-based approach provides a structured way to determine the solution of the system, identify inconsistencies, or discover if there are infinitely many solutions.

Substitution Method

The substitution method consists of solving one of the equations for one variable and then substituting the resulting expression into the other equations. This reduces the number of variables and equations gradually and is particularly helpful when one of the equations is easily solved for one variable.

System of Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. The solution to the system is a set of values for the variables that satisfies every equation simultaneously. Understanding these systems is fundamental in algebra and linear algebra as they model numerous practical problems and serve as a basis for more advanced topics.

Elimination Method

The elimination method involves adding or subtracting equations in order to cancel out one or more variables. This step-by-step reduction transforms a complex system into a simpler one with fewer variables, ultimately leading to a solution. It's a widely-used, systematic approach that can be applied to systems of various sizes.

Transcript

00:01 For this problem, we're going to be solving the system provided.

00:04 In order to do this, we're going to use the process of elimination.

00:07 So the first thing we want to do is choose a couple of these to eliminate a variable.

00:12 Then we'll choose another two and eliminate the same variable, and then we'll go from there.

00:18 So i'm first going to start off with equations 1 and 2, because i see if the x values have the exact same coefficients, just one's positive and one's negative.

00:26 So let me rewrite this over here.

00:27 So x plus y minus z is equal to negative two and then negative x plus 2 y minus 3 z is equal to negative 4 and so what i'm going to do is decide do i have to add or subtract well if i add then what will happen is that these xs will cancel now combine these together it gets me 3y minus 4 z is equal to negative 4 is equal to negative 6 okay so then i'm going to take two other equations so what i did was i chose the 2x minus y plus 2 z plus z equals negative 5 so it's going to be one of them and then i'm going to use the second one again but i need to get it so the coefficients of x are the same so what i'm going to do is multiply this through by 2 so my result is going to be 2x minus y plus z equals negative 5 so that's from the second equation there and then when i multiply i get negative 2x plus 4y minus 6 z is equal to negative 8 now when i combine again i see through addition that these will cancel so negative plus 4 y is going to get me uh three three y one minus six is minus 5 z and then negative 5 minus or plus negative 8 is negative 13 so these are the two equations we're going to be focusing on now uh both of the ys have the exact same variable so i want to put those together now and try to eliminate the next one.

01:54 So i'm going to take the 3y minus 4z equals negative 6.

01:59 3y minus 5z is equal to negative 13.

02:05 And then from there, i'm going to solve through once again either adding or subtracting.

02:10 So in this case, i'm going to subtract everything because in order for the why is to cancel, we have to subtract to get zero.

02:16 So that means i'm going to subtract everything else.

02:18 So negative 4z minus negative 5z gets me just 1 z or z.

02:24 And then and negative 6 minus negative 13 is 7.

02:27 So i have my first of my answers, which is z equals 7.

02:30 So now i can play it back into either equation.

02:33 It doesn't really matter.

02:33 I'm going to kind of put it back into this one.

02:35 And i'm going to make sure i put an in for z.