Solve the system x-3 y+2 z =14 2 x+5 y-z =-9 -3 x-y+2 z =2 (2) (a) by substitu

Solve the system x-3 y+2 z =14 2 x+5 y-z =-9 -3...

Key Concepts

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 such a system is the set of variable values that simultaneously satisfies all the equations. Systems of linear equations form the basis for many areas in mathematics and its applications, including algebra, calculus, and linear programming.

Substitution Method

The substitution method is an approach to solving a system of equations where one solves one of the equations for one variable in terms of the others, and then substitutes that expression into the remaining equations. This method reduces the number of equations and variables, simplifying the problem into one that can be solved sequentially. It is particularly useful when one equation is easily solvable for one of the variables.

Elimination Method

The elimination method involves adding or subtracting equations, or multiplying them by appropriate factors, to eliminate one or more variables from the system. By strategically combining the equations, the method reduces the system into a simpler one with fewer variables, eventually leading to a solution that satisfies the original set of equations. This technique is essential in solving larger systems where substitution might be cumbersome.

Transcript

00:01 All right, so in this problem, we're given another system of equations, three variables.

00:05 2x minus y plus 3z equals 14, x plus y minus 2z equals negative 5, 3x plus y minus z equals 2.

00:19 So here we have to solve for one of the variables first, plug this in, solve for the other variable, plug it in and solve for the third variable.

00:28 So what i'm going to do first is i'm going to take the second equation and i'm going to solve for x.

00:34 So x equals 2z minus y minus 5.

00:43 So now we have x in terms of z and y.

00:46 Okay? so let's plug that in to this first equation.

00:51 And then we can solve for y in terms of z, if that makes sense.

00:55 So what we're going to do is we're going to have two times 2 z minus y.

00:58 Minus 5 minus y plus 3 z equals 14 so now we have 4 z minus 2y minus 10 minus y plus 3 z equals 14 this is going to simplify and this is going to give us 7 z minus 3 y equals 24 delete this is going to give us 3y equals 7 z minus 24 y equals 7 over 3 z minus 8 okay so now we have y in terms of z and we have x in terms of y and z so what we can do now is plug in this equation for y into this equation right here and then we have x in terms of z and y in terms of z so this is going to equal x equals 2z minus 7 thirds z minus 7 thirds z minus 8 minus 5 and this equals 2 z minus 7 thirds z plus 8 minus 5.

02:21 This is going to simplify to 3.

02:24 And this is the same as 2 and 1 thirds.

02:27 So this is going to equal negative 1 third z.

02:36 In terms of z and y in terms of z.

02:39 I'm just going to leave this here.

02:43 And so now we can plug this into any of the equations and actually solve for z.

02:48 I'm going to use this equation right here because we have a 3x and we're multiplying by 1 3.

02:53 So that's going to be an easy one to plug in.

02:56 So we have 3 times negative 1 3, z plus 3 plus y, which is 7 thirds z minus 8.

03:09 Minus z equals two.

03:12 This is going to give us negative z plus nine plus seven -thirds z minus eight equals two...

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