I'll do a couple for you, but I ain't doing your whole assignment...

#1.

−x+2y = 4

−4x+y = −5

The 1st equation says that x = 2y-4. So, use that in the 2nd equation, and you have

-4(2y-4) + y = -5

-8y+16+y = -5

-7y = -21

y = 3

So, x = 2y-4 = 2

The solution is this (2,3)

Check to be sure it works in both equations.

#2.

−2x+y = 3

4y−4 =x

Rearranging things a bit, we have

y = 2x+3

y = 1/4 x + 1

The slopes are different, so the two lines must intersect somewhere. Two lines can only intersect in single point, so one solution.

So, what can you do on the others? When you get an answer, do a check to be sure it's right.

Some suggestions:

I you have a system of equations and one of the equations has a variable with 1 as a coefficient

such as −4x+y = −5 in #1, arrange the equation so the variable sits by itself and use substitution.

e.g.

y = 4x - 5

now plug that into −x+2y = 4

-x + 2(4x-5) = 4

-x + 8x - 8 = 4

7x = 14

x = 2

now back into y = 4x-5

y = 4(2) - 5 = 3

I the two equations have the same slope, they are parallel, so they couldn't possibly intersect

but....

it could be that they gave you two equations that are really the same, one having been disguised.

e.g. look at #3

3x+y= 5

−2y=6x − 10

working on the 2nd

-6x - 2y = -10

divide each term by -2

3x + y = 5, which is the same as the first.

In that case there will be an infinite number of solutions, that is, any point that lies on the

line is a solution:

(1,2), (3,-1), (0,5) ....

Look at #6

3x+3y= 27

x−3y= −11

You could use substitution, and follow my steps that I outlined, BUT, notice that you have

-3y and +3y

That is perfectly set up for elimination, since adding the two equations will "eliminate" the y's

4x + 0 = 16

x = 4

now sub that back into either of the given equations, I will go for the firs

3x+3y= 27

12 + 3y = 27

3y = 15

y = 5 , so you have (4,5) as a solution.

did you notice that the first equation could have been simplified ?

3x+3y= 27

x−3y= −11

x +y = 9 , by dividing each term by 3

x - 3y = -11

now by subtracting them, I can eliminate the x's

0 +y - (-3y) = 9 - (-11)

y + 4y = 9+11

4y = 20

y = 5, now go back into one of the first equations, sub in y = 5 to find the x

I could also have used substitution for #6

using x−3y= −11 ------> x = 3y-11

into the other equation:

3x + 3y = 27

3(3y-11) + 3y = 27

9y - 33 + 3y = 27

12y = 60

y = 5

etc

Showing that there is not ONE way to do these, but you will always get the same

answer no matter what method you use.

Hope this helps