Solve each system using elimination and back-substitution....

Key Concepts

Elimination Method

The elimination method involves combining equations in a linear system in order to cancel out one of the variables. By adding or subtracting equations, you simplify the system step by step, reducing the number of unknowns until you arrive at an equation with a single variable. This method is a core technique for solving systems of equations in algebra.

Back-Substitution

Back-substitution is the process of solving a simplified system of equations beginning with an equation that contains only one variable. After determining the value of this variable, you substitute it back into preceding equations to solve for the other unknowns. This step-by-step approach ensures that all variables are solved sequentially once the system has been reduced through elimination.

Linear Systems

A linear system consists of two or more linear equations involving the same set of variables. The goal is to find values for the variables that satisfy all the equations simultaneously. These systems are fundamental in algebra and are used in various fields to model problems where multiple relationships must be met concurrently.

Transcript

00:01 In this problem, we are given a system of equations and asked to provide the solution that we can plug in for x, y, and z in order to generate the same number on the left side as on the right side.

00:11 Now, normally we look for some equation with x, y, and z that has a 1 as a coefficient for x.

00:20 However, as you can see, there is no such equation in this system of equations.

00:26 Now, you are more than welcome to divide one of the equations by a number in order to give x a coefficient to.

00:32 Of 1.

00:33 However, i believe that an easier way to go about it is to avoid fractions altogether by multiplying the equations by certain numbers and adding them, or that product, two multiples of other equations.

00:45 What do i mean by this? i mean, in order to eliminate the x's in the second and third equation, we might multiply the first equation times negative 2, and add that to 3 times the second equation.

01:03 That will give us 0 as a coefficient for x.

01:06 In the second equation.

01:07 We might then do the same thing for the third equation.

01:19 And now when we write our new equations, we get 3x minus y plus c equals 6.

01:31 Negative 2 times 3x is negative 6x.

01:35 Add that to 3 times 2x, negative 6x plus 6x is 0x, which is what we wanted.

01:41 Negative 2 times negative y is 2y.

01:43 Add that to 3 times 2y in the second equation.

01:47 We got 2y plus 6y, that's 8y, and then we get negative 2 times z, negative 2z, add that to 3 times negative z.

02:01 So we get negative 2 z minus 3 z, that's negative 5 z.

02:10 Then we get negative 2 times 6, negative 12, plus 3 times 5 in the second equation, negative 12 plus 15, that's equal to 3.

02:22 And then we take negative 2 times 3x again, negative 6x, add that to 3 times the 2x in the 3x in the third equation.

02:32 We get 6x minus 6x, that's 0x, which is what we wanted.

02:37 Then we take negative 2 times negative y.

02:42 That's 2y.

02:45 Add that to 3 times negative y.

02:49 That's 2y minus 3y.

02:51 That's negative y.

02:54 Then we take negative 2 times the z in the first equation, negative 2z, plus 3 times the z.

03:00 In the third equation, we get negative 2 z plus 3 z, that's plus z.

03:09 All that is equal to negative 2 times 6, negative 12 plus 3 times 15, negative 12 plus 15, that's 3.

03:21 Now we want to eliminate all the ys below the first and second equation.

03:30 And in order to do that, we want to find an equation with y where it has a coefficient of 1...

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