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Solve each system.

$$\begin{array}{l}2 x-4 y+6 z=12 \\6 x-12 y+18 z=36 \\-x+2 y-3 z=-6\end{array}$$

$\{(x, y, z) \mid-x+2 y-3 z=-6\}$

Algebra

Algebra 2

Chapter 4

Systems of Linear Equations

Section 3

Systems of Linear Equations in Three Variables

Polynomials

Systems of Equations and Inequalities

Introduction to Algebra

Graph Linear Functions

Write Linear Equations

Linear Equations and Functions

Matrices and Determinants

University of Texas at Austin

University of California, Berkeley

University of North Carolina at Chapel Hill

University of Michigan - Ann Arbor

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but the system is two x minus four y close six z equals 12 six x minus 12 y e t z equals 36 and negative acts close to Why, yes. Three. The equals negative. Six. So what you could notice about these equations is they're all multiples of each other. Two times 36 Negative. Four times negative three times three is now 12 6 times Stories 18 12 times three is 36 negative one times native sixes 62 times negative. Six is next 12? Make it three times negative. Six is 18 negative six times negative. Six is 36. This means that they're gonna infinitely many solutions, right? Because any solutions to this first equation is also going to satisfy the second and third equation, since they're just multiples of each other. Um, so actually find one solution if you ask to, um, you might be asked if they're infinitely many solutions Final One because not any X y and Z is going to solve this. But I think you just plug in zero for X and Y or for y and Z and then solve for the third variable. So if you plug in zero fraction Why you get 60 equals 12 z's got equal to. So you know, zero you're too. Is an example of the solution. Um, and then just plug. Listen to double check. Six z is 12 that works. 18 z equals 36 That works negative. Three z equals negative. Six that works. So there you found your infinitely many solutions to this.

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