Systems of Linear Equations

[jonb]

I'm currently stuck on the problems involving two equations that don't have an isolated variable. I just don't know what to do and my book is not helping.

An example...

6x-4y=3
3x-2y=9

For 3x-2y=9, if there was no 3 with the x I would add 2y to both sides and be left with x=2y+9 and continue on but I have no idea what to do with the 3. Thanks in advance and I hope I've explained it well enough to receive help.

 

[mmm4444bot]

… if there was no 3 with the x I would add 2y to both sides and be left with x=2y+9 and continue on …

That is the substitution method.

Another method for solving systems is the elimination method.

You multiply one or more equations each by an appropriate constant such that the coefficients on one of the variables wind up as additive opposites.

3x - 2y = 9

Multiply this equation by -2.

Then you have one equation with a 4y and a second equation with a -4y.

Add these two equations, and 4y - 4y = 0. The result is an equation that contains only x.

 

[mmm4444bot]

Whoops, I just realized that you will not wind up with an equation that contains only x with this system.

You'll end up with 0 = 9.

In other words, both variables are eliminated, and you wind up with a false statement.

This is a sign that the system has no solution.

We can confirm this by putting each of the given equations into Slope-Intercept Form.

y = (3/2)x - 3/4

y = (3/2)x - 9/2

Note that the slopes are equal and the y-intercepts are not.

This given system represents two parallel lines; therefore, they never intersect, and thus there are no (x,y) pairs common to both.

I'll give another example of the elimination method.

EG:

2x + 3y = 4

7x + 8y = 9

First, we choose which variable to eliminate. I choose to eliminate y.

Note that (-8)(3y) = -24y and (3)(8y) = 24y.

In other words, if we multiply the first equation by -8 and we multiply the second equation by 3, then we end up with two new equations that contain -24y and 24y.

Adding these two new equations will eliminate the y variable.

(-8)[2x + 3y = 4] gives -16x - 24y = -32.

(3)[7x + 8y = 9] gives 21x + 24y = 27.

-16x - 24y = -32
21x + 24y = 27
---------------------
5x = -5

x = -1, and so it goes.

 

[jonb]

I'm even more confused now. I don't think this other method was ever mentioned in class.

 

[mmm4444bot]

I'm even more confused now …

Our posts crossed in cyberspace.

Let's get you unconfused.

Look at the example I added to my previous post, and tell me what step(s) you do not understand.

 

[mmm4444bot]

… I don't think [that the elimination] method was ever mentioned in class.

If you have not learned other methods, then you can still use the substitution method.

EG1:

Solve the first equation for y.

y = (3/2)x - 9/2

Now, substitute this expression for y in the second equation.

6x - 4[(3/2)x - 9/2] = 3

6x - 6x + 18 = 3

0 = -15

This is a false result. Therefore, there is no solution.

EG2:

Solve the first equation for x.

x = (2/3)y + 2

Now, substitute this expression for x in the second equation.

6[(2/3)y + 2] - 4y = 3

4y + 12 - 4y = 3

12 = 3

This is a false result. Therefore, there is no solution.