Question

How do you determine whether a linear system has one solution, many solutions, or no solution when given y=1/2x+3 and x+3y=-6?

Answer 1

It depends wether the lines are the same or different, and in this case if they are parallel or not. Your particular example has one solution.

Explanation:

This answer is easily answered with a geometrical point of view. A system is solved when you find a point which satisfies all the equations involved. Visually, your equations are represented by graphs, and such points are the intersections among the graph.

A linear system (that's the keyword!) in two equations and two variables asks for point of intersection between two lines in a plane, and everyone knows that two lines in a plane always cross in a point, unless they are parallel. So, the answer would be: "a linear system in two equations and two variables always have exactly one solutions if the lines are not parallel, and it has no solutions if the lines are parallel".
A tricky case is the one in which the two lines are actually the same line, and so every point is a solution.

But this raises a new question: how do you find out if the lines are parallel? It's very easy.

Write all of your equation in the `y=mx+q` form. Your system will become

`y=m_1x +q_1`
`y=m_2x+q_2`

• If `m_1 \ne m_2`, then the two lines are not parallel, and exactly one solution exists.

• If `m_1=m_2` and `q_1\ne q_2`, then the lines are parallel but different, and so there is no solution.

• If `m_1=m_2` and `q_1=q_2`, then the two lines are actually the same, let's say `r`, and the system has infinite solution: it is in fact asking you: "which points belong to the line `r` and also to the line `r`? All of them of course, since the second request doesn't really add anything.

In your case, the first line is already written in the slope/intercept form, with `m=1/2` and `q=3`.

As for the second equation,
`x+3y=-6 \iff 3y=-x-6 \iff y=1/3x-2`

So, `m_1 \ne m_2` and the system has a solution (namely, `(-6,0)`).

Answer 2

One method: Convert the two equations into slope-intercept form.

Explanation:

If you have two equations in slope intercept form:

`y = a_1x+b_1`
`y=a_2x+b_2`

If `a_1=a_2` and `b_1=b_2`
`color(white)("XXX")`then the equations represent the same line (infinitely many solutions).

If `a_1=a_2` (slopes are equal) but `b_1!=b_2`
`color(white)("XXX")`then the equations represent parallel lines which do not intersect (no solutions).

Otherwise the equations represent lines which will intersect at exactly one point.

For the given example
[1]`color(white)("XXX")y=1/2x+3`
[2]`color(white)("XXX")x+3y=-6`

[1] is already in slope-intercept form.

[2] can be converted into slope-intercept form as
[3]`color(white)("XXX")3y=-x-6`
[4]`color(white)("XXX")y=-1/3x-2`

Since the slopes `(1/2)` and `(-1/3)` are not equal, these equations will have exactly one solution.

Answer 3

This system has one solution: `{(x=-6),(y=0):}`

Explanation:

There are at least 2 methods to find number of solutions of a system of 2 linear equations:

I Using Cramer's Rule

To use this method you have to rearrange the equations first so the corresponding variables are at the same places in all equations, so you have to move `x` in the first equation to the left side to get:

`{(-1/2x+y=3),(x+3y=-6):}`

Now you have to calculate determinant `W`

`W=|(-1/2,1),(1,3)|=(-1/2)*3-1*1=-3/2-1=-2 1/2`

This determinant is not zero, so the system has exactly one solution.
You can contimue and calculate the solution as described later.
Generally the method looks as follows:

Let there be a system: `{(a_1x+b_1y=c_1),(a_2x+b_2y=c_2):}`

Let:
`W=|(a_1,b_1),(a_2,b_2)|=a_1*b_2-a_2b_1`

`W_x=|(c_1,b_1),(c_2,b_2)|=c_1b_2-c_2b_1`

`W_y=|(a_1,c_1),(a_2,c_2)|=a_1c_2-c_1a_2`

Now we can say that:

• If `W!=0` then there is only one solution: `{(x=W_x/W),(y=W_y/W):}`
• If `W=0` and `W_x!=0` or `W_y!=0` then system has no solutions
• If `W=0` and `W_x=0` and `W_y=0` the system has infinitely many solutions.

II Graphical method .

You have to plot both functions in one coordinate system and see if they have 1 common point, are paralell or overlap.

graph{(y-x/2-3)(x+3y+6)=0 [-10, 10, -5, 5]}