In Problems 17-54, solve each system of equations. If the system has no solution, say that it is

step 1 All right, we're going to solve the following system. So I can see that two of the equations alr

In Problems 17-54, solve each system of equations. If the...

Key Concepts

Systems of Linear Equations

These are collections of one or more linear equations involving the same set of variables. The aim is to find values for the variables that satisfy all the equations simultaneously. This concept is fundamental in algebra and is widely applicable in fields such as engineering, economics, and the physical sciences.

Elimination Method

This method involves adding or subtracting equations to cancel out one or more variables, transforming a multi-variable system into a simpler system with fewer variables. It is a systematic approach for solving systems of linear equations and is especially useful when the coefficients of one of the variables are opposites or can be made opposites.

Substitution Method

The substitution method consists of solving one equation for one variable and then substituting that expression into the other equations. This reduces the number of variables in the remaining equations, making them easier to solve. It is particularly effective when one of the equations is readily solvable for one of the variables.

Consistency and Inconsistency

A system of equations is called consistent if there is at least one set of variable values that satisfies every equation, and inconsistent if no such set exists. Recognizing whether a system is consistent or inconsistent involves checking for contradictions or dependencies among the equations, which is a key part of analyzing linear systems.

Transcript

00:01 All right, we're going to solve the following system.

00:02 So i can see that two of the equations already only have two variables, and the third one has three.

00:08 So all i'm going to do is try to cancel the one of the variables out, the y's, in order to create two equations with x's and zs.

00:21 So i am going to take equation two, and i'm going to add it to equation three.

00:28 So i'm adding equation two to equation three.

00:30 To create a new equation.

00:35 Number four.

00:37 So 2x plus 5x is 7x, y minus y is 0, and then there is no z to add to that, so i end up with just 3z.

00:50 If you want to think about it, you can think about this as i have a 0 z there that i'm adding.

00:56 So i have 0 z plus 3z, which is 3z.

01:00 And then i have negative 41 plus 1 .000, negative 1.

01:06 So that leads me with a negative 42.

01:10 So i'm going to use that one and i'm going to use equation 1 here and i'm just going to rewrite it so we can have that right underneath each other.

01:19 So 6x minus 5 z equals 17.

01:25 So now i either have to cancel out my x's or my z's.

01:29 Your choice how you want to do that.

01:31 Looks like i'm going to have to do operations on both of them so it doesn't really matter which way i go.

01:38 So why don't we cancel out our zs? so i'm going to go 5 times equation 4, and i'm going to add that to 3 times equation 1.

01:51 So i'll write this out here just so we can kind of see what we're doing.

01:54 So 5 times 7 is 35x, 3 times 5 is 15z.

02:02 And we're going to add that to 6 times 3, which is 18x, and negative 5 times 3, which is negative.

02:10 15 z so there is my equation four and my equation one and then i have to do my outside so negative 42 times 5 is negative 210 and i'm adding that to 3 times 17 so three times 17 is 51 so combining this 35x plus 18 x leaves us with 53 xes 15 z minus 15 they are gone and negative 210 plus 51 is negative 159.

02:53 So now i just have to solve for x by dividing by 53 on both sides.

03:00 So that leaves us with an answer of x is negative 3.

03:06 So we're down.

03:08 We've got one of three solved.

03:10 I can now take that and throw it back into equation 1 or equation 4.

03:15 I always like like going back to my original equations, not my created equations.

03:22 So i'm going to substitute it into equation one...