Question

How do you solve `4x - y = 30`, `4x + 5y = -6` by graphing and classify the system?

Answer 1

See a solution process below:

Explanation:

To solve this problem, plot two points for each equation and draw a line through the points.

Equation 1:

For `x = 0`

`(4 * 0) - y = 30`

`0 - y = 30`

`-y = 30`

`-1 * -y = -1 * 30`

`y = -30` or `(0, -30)`

For `x = 5`

`(4 * 5) - y = 30`

`20 - y = 30`

`20 - color(red)(20) - y = 30 - color(red)(20)`

`0 - y = 10`

`-y = 10`

`-1 * -y = -1 * 10`

`y = -10` or `(5, -10)`

graph{(x^2 + (y + 30)^2 - 0.75)((x - 5)^2 + (y + 10)^2 - 0.75)(4x - y - 30) = 0 [-100, 100, -50, 50]}

Equation 2:

For `x = 0`

`(4 * 0) + 5y = -6`

`0 + 5y = -6`

`5y = -6`

`(5y)/color(red)(5) = -6/color(red)(5)`

`y = -6/5` or `(0, -6/5)`

For `y = 0`

`4x + (5 * 0) = -6`

`4x + 0 = -6`

`4x = -6`

`(4x)/color(red)(4) = -6/color(red)(4)`

`x = -3/2` or `(-3/2, 0)`

graph{(x^2 + (y + 6/5)^2 - 0.75)((x + 3/2)^2 + y^2 - 0.75)(4x + 5y + 6)(4x - y - 30) = 0 [-100, 100, -50, 50]}

We can see the lines cross at: `(6, -6)`

These two equations are consistent and independent because the have a solution and only one solution.

graph{((x-6)^2 + (y + 6)^2 - 0.05)(4x + 5y + 6)(4x - y - 30) = 0 [-2, 14, -7, 1]}