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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
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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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