only one more algebra question!?

Question

is this solveable?....and if so, how do u do these kind of equations?: y-2x=-6
2y-x=5
( y-2x=-6 is over 2y-x=5)

Answer 1

These equations are called Simultaneous Equations iwith two Variables, x & y.

We solve them by elimimating either x or y & getting the value of the other by sustitution of that value in any one of the given equations..

Here is the sure solution:

y - 2x = - 6 .... (1)

2y - x = 5 .... (2)

4y - 2x = 10.... (3) .........(1) * 2
y - 2 x = - 6 ... (1)

Subtracting, (1) from (3),
3 y = 16

y = (16/3) or 5 & 1/3 .... (A)

Substituting the value of 'y' in equation # 2 we get,

2 (16/3) - x = 5
=> 32/3 - x = 5
-x = 5 -32/3
-x = (15-32) / 3
-x = -17/3
x = 17/3..or 5 & 2/3. (B)

Answer: x = 17/3..or 5 & 2/3

y = (16 / 3) or 5 & 1/3

(x, y) = ( 17/3, 16/3)

Give me my 10.

Answer 2

This is a set of equations. You have two equations with two unknowns. This is solvable.

y-2x = -6 and 2y-x = 5.

You could solve like this.

From equation (1), we have that y = 2x-6. Putting this into equation (2), we have

2(2x-6)-x = 5. or 4x-12-x=5 or 3x=17. Then x = 5 2/3.

put this result into either equation, for example, equation (1) and

y-2(5 2/3) = -6
y-11 1/3 = -6
y = 5 1/3.

Answer 3

this is a linear equation already.
you have to plot these equations on a graph.
in order to do this, you have to assign random values to x while y is valued at 0, or vice versa.

but i think there's something wrong with your given equation. one of the variables should have exponents.

coz if not, there'd be no equation to solve because the variables would be cancelled out.

Answer 4

I beleive this is soveable. If this is a division problem, you don't have an equation though- you have two equations to solve and then divide them. So you simplify the them separately...

Answer 5

this is called systems of equations

here is a link to help
http://www.unlv.edu/faculty/bellomo/Math124/Notes/Ch05-Sect01.pdf#search='dividing%20equations%20with%20two%20variables'