system of equations

Question

1.) 6x-2y=10                   3.)  3x-9y=3

3x-2=y                         2x=16-y

 

 

 

2.) y=3x+1                   4.)   2x+3y=9

x+2y=-5                    x-2y=1

 

 

Answer 1

1.) 6x-2y=10

     3x-2=y

 

We can solve by substitution.  In the 2nd equation

we have y = 3x-2.  Substitute 3x-2 in place of y in

the 1st equation to solve for x

 

6x-2(3x-2) = 10

6x-6x+4 = 10

4=10

Since we come up with a false statement these two

lines do not cross and there is no solution to this

system of equations.

 

This makes sense actually.  Put both lines in

slope intercept form of y = mx + b

6x-2y = 10        y = 3x-2

-2y = -6x+10

y = (-6x+10)/(-2)

y = 3x - 5

Both lines have a slope of 3 so they are parallel

and have different y-intercepts... ((0,-5) and (0,-2).

Thus they never meet.

 

3)  3x-9y=3

     2x=16-y

 

If we solve the 2nd equation for y we get

y = -2x+16

Substitute that into the first equation

 

3x-9(-2x+16) = 3

3x +18x - 144 = 3

21x - 144 = 3

21x = 147

x = 7

 

y = -2x+16 = -2(7) + 16 = 2

The lines cross at the point (7,2)

 

Let's take a look at problem 4 and do it by elimination:

 

4)  2x+3y=9

       x-2y=1

 

Multiply the 2nd equation by -2 and add the equations.

 

  2x + 3y = 9

-2x  + 4y = -2

----------------

         7y = 7

           y = 1

 

Substitute 1 for y in one of the original equations to solve for x

 

2x + 3(1) = 9

2x + 3 = 9

2x = 6

x = 3

 

The lines cross at the point (3,1)

 

Try problem 2 yourself.  It can be done either by substitution or elimination.