this problem??

Question

could someone please help!?

x-4y= -4
3x+3y= -4

Answer 1

x - 4y = - 4- - - - - -Equation 1
3x + 3y = - 4- - - - -Equation 2
-- - - - - - - -

Substitute Method

x - 4y = - 4

x - 4y + 4y = - 4 + 4y

x = - 4 + 4y

Insert the x value into equation 2

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

3x + 3y = - 4

3(-4 + 4y) + 3y = - 4

- 12 + 12y + 3y = - 4

- 12 + 15y = - 4

- 12 + 15y + 12 = - 4 + 12

15y = 8

15y/15 = 8/15

y = 8/15

The answer is y = 8/15

Insert the y value into equation 1

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

x - 4y = - 4

x - 4(8/15) = - 4

x - 32/15 = - 4

x - 32/15 + 32/15 = - 4 + 32/15

x = - 4 + 32/15

x = - 60/15 + 32/15

x = - 28/15

The answer is x = - 28/15

Insert the x value into equation 1

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

Check for equation 1

- 28/15 - 4(8/15) = - 4

-28/15 - 32/15 = - 4

- 60/15 = - 4

- 4 = - 4

- - - - - - - -

Check for equation 2

3x + 3y = - 4

3(-28/15) + 3(8/15) = - 4

- 84/15 + 24/15 = - 4

- 60/15 = - 4

- 4 = - 4

- - - - - - - -

The solution set { - 28/15, 8/15 }

- - - - - - - s-

Answer 2

Solve for x:

x=4y-4

Substitute in the second equation:

3(4y-4)+3y=-4
12y-12+3y=-4

Combine like terms:

15y=8
y= 8/15

Insert y value into first equation to determine x:

x=4(8/15)-4
x= 32/15 - 4
x= 32/15 - 60/15
x= -28/15

So...
x= -28/15
y = 8/15

Strange problem...not easily solved algebraically. Is this a geometric problem?

Answer 3

Take the first eq'n and solve for 'x'.

x = 4y - 4

Sub this into the 2nd eq'n.

3(4y - 4) + 3y = -4

Now, solve for 'y'.

12y - 12 + 3y = -4
==> 15y = 8
==> y = 8/15

Sub this back into the eq'n for 'x'.

x = 4(8/15) - 4 = 32/15 - 4 = (32 - 60)/15 = -28/15 = -28/15

x = -28/15
y = 8/15


Odd numbers...check my arithmetic, but that's one way of doing this kind of problem.

Answer 4

systems of equations are easily solved using a matrix form of the equation
you can represent this equation with the matrix

[1 -4] [x] = [-4]
[3 3] [y] [-4]
and use Cramer's rule, but that's too much work for a simple problem like this

or what I do for this type of problem is put it into an augmented matrix

[1 -4 -4]
[3 3 -4]


then put it into its reduced row echelon form, which will give you the values for x and y
you can do this quickly in a calculater usually using the command
rref(

where you type your matrix after the rref( command
on ti 83's and 89's you type
rref([1,-4,-4;3,3,-4])

then it gives you the solution

[1 0 -28/15]
[0 1 8/15]

where x= -28/15 and y = 8/15

for more complicated equations where the variables aren't as neat as this, put them in the form ax + by = c
then make the matrix the form
[a b c]

each row of the matrix gets its own a b and c... if there is no y or x in one of the equations use a 0 in its place, but make sure you keep them in order x in the first column, y in the second column and then whatever is left over in the last column for each equation,

Answer 5

x - 4y = - 4------------X by 3
3x + 3y = -4----------X by 4

3x - 12y = -12
12x + 12y = - 16

15x = -28
x = -28/15

Now 3x + 3y = -4, therefore:-
3 x -28/15 - 12y = - 12

- 28/5 - 12y = -12

28/5 + 12y = 12

12y = 12 - 28/5

12y = 60/5 - 28/5

12y = 32/5

y = 32/5 x 1/12

y = 8/5 x 1/3 = 8 /15

Answer is x = - 28 /15 , y = 8 /15

Answer 6

You could also solve this by elimination
x-4y= -4
3x+3y= -4

Multiply the first eqn. by -3

-3x + 12y = 12
3x + 3y = -4
--------------------
15y = 8
y = 8/15

Then sub 8/5 in for y
x - 4(8/5) = -4
x -6.4 = -4
x = 2.4