Question

How do you solve the system of equations by graphing and then classify the system as consistent or inconsistent `y= -1/3x-3` and `y= -7/6x-8`?

Answer 1

If the system consists of two linear equations such as the ones you have given then there are three cases, which correspond to three situations in the graph: intersecting, parallel and co-incident.

Explanation:

Each linear equations is represented by a straight line. If the two lines have different gradients, they are not parallel and will cross at a unique point, which represents the unique solution. If the lines have the same gradient, then either the lines are parallel yet distinct, in which case they have no point in common, and the system has no solution at all.

If the lines have the same gradient then it is possible that they represent the same line, in which case there are an infinite number of solutions, because any pair of values `(x,y)` which satisfy one equation will necessarily satisfy the other on. This will happen if the ratios of the coefficients of `x`, `y` and the constant are all the same, such as:
`x+y=1`
`2x+2y=2`.
These equations represent the same line.

On the other hand,
`x+y=1`
`x+y=2`
represent parallel lines, and is the situation where there are no solutions, no points in common, otherwise the same pair of numbers would be adding up to different numbers.

If the ratio of the coefficients of the `x` and `y` differ between the two equations, then there will be a unique solution and the graphs will cross. Your example is one such because `1::-1/3` is different to `1:-7/6`.

The whole idea extends to more than two variables.

Answer 2

`(-6,-1)" and consistent"`

Explanation:

Since both equations are given as y in terms of x, we can equate the right hand sides of the equations.

`rArr-1/3x-3=-7/6x-8`

To eliminate the fractions, multiply ALL terms on both sides by the lowest common multiple ( LCM) of 3 and 6 which is 6

`(cancel(6)^2xx(-x)/cancel(3)^1)-(6xx3)=(cancel(6)^1xx(-7x)/cancel(6)^1)-(6xx8)`

`rArr-2x-18=-7x-48`

add 7x to both sides.

`-2x+7x-18=cancel(-7x)cancel(+7x)-48`

`rArr5x-18=-48`

add 18 to both sides.

`5xcancel(-18)cancel(+18)=-48+18`

`rArr5x=-30`

To solve for x, divide both sides by 5

`(cancel(5) x)/cancel(5)=(-30)/5`

`rArrx=-6`

To find the corresponding value of y, substitute x = - 6 into either of the 2 equations.

`x=-6toy=-7/6xx(-6)-8=7-8=-1`

`rArr" solution is " (-6,-1)`

Since there is 1 `color(blue)"unique solution"` then the system is `color(blue)"consistent and independent"`