Algebra 1 Question?

Question

Solve the system using the method of your choice and tell how many solutions the system has.

-x+1/3y=-6
and
3x-y=-16

I really don't get this, if you use linear combinations, it reduces down to 0=-26

help please?

Answer 1

x=-289/53

y=16-3(289/53)

Answer 2

I would use the substitution method myself, take the second equation add y to both sides, and add 16 to both sides to get y=3x+16, then substitute 3x+16 for y into the first equation. Solve for x, and incase it isn't obvious, once you have x you can plug that value into either of the original equations and solve for y.

Hmm..., well ok, the answer is there are no solutions, didn't notice that 3x * 1/3 = x and the -x + x = 0 leaving you with, a constant = a different constant. It is either a trick question (no solutions), or you typed it in wrong, but seeing as you probably attempted to solve it from your text book and/or work sheet, not what you typed, my money is on the answer "no solutions".

In algegra and beyond you will find many examples of this where the answer is simply "there is no answer".

Answer 3

There are no values of x and y that will satisfy both equations. To easily recognize this multiply each term of the first equation by -3.The equation will then be 3x-y =18. But since the second equation states 3x-y = -16, they must be two independent equations and not part of a set of equations defining unique values of x and y.

Answer 4

Again, nothing is wrong with you, you are simply having a problem interpreting the result.

the "result" 0 = -34, is absurd and is a contradiction, because we know it is false. The result tells us that these two lines NEVER intersect. Two lines that never intersect are PARALLEL.

Rewrite in slope intercept form ( y = mx +b):

-x +y/3 = -6 --> y = 3x - 18

3x - y = -16 --> y = 3x + 16

So we see that the lines are indeed parallel, so they never intersect, so there are no solutions, that is there exists no ordered pair (x,y) that simultaneously satisfy both equations.

Answer 5

-x + 1/3y = - 6---- -Equation 1
3x - y = - 16- - - - --Equation
- - - - - - - - -
Clear the fraction equation 1

-x + 1/3y = - 6

- 3(x) + 3(1/3y) = - 3(6)

- 3x + y = - 18 new equation 1

- - - - - - - - - - -

- 3x + y = - 18
3x - y = - 16
- - - - - - - - -
0 + 0 = - 34

The x and y cofficients and variables cancell. This problem has no solution.

- - - - - - - - - -s-

Answer 6

using the second equation i came up with
y= -3x - 16

and using that value for y, I came up with

-x + {1/(3 [-3x - 16])} = -6

then

-x + (1 / [-9x -48] = -6

then, but making them a similar fraction
by using the denominator (-9x-48)

(-9x^2 - 48x + 1) / (-9x - 48) = -6

transpose, and you'll have

-9x^2 - 48x + 1 = -6 * (-9x - 48)

then

- 9x^2 - 48x + 1 = 54x + 288

combine all terms with variable x on the left side

- 9x^2 - 48x -54x = 288 - 1

then

- 9x^2 - 102x = 287

there you have your equation..

0 = 9x^2 + 102x - 287

you may now start finding the value for x...

I'll leave the rest of the work to you. gud luck.

Answer 7

Nothing is wrong with you.
But there is something wrong with the equation given above.
Currently,it can't be done.