Question

* PLEASE HURRY* Here is the graph for one equation in a system of equations:

(0,4) (5.5,0)

Write an equation for the system so it has infinitely many solutions.

Write a second equation whose graph goes through (0,1) so the system has no solutions.

Write a third equation whose graph goes through (0,2) so the system has one solution at (4,1).

Answer 1

1) The equation that has infinitely many solution is y = -8/11·x + 4

2) The equation of the graph that has no solution is y = -8/11·x + 1

3) The equation of the line that has only one solution is y = -1/4·x + 2

Explanation:

1) The points on the graph are;

(0, 4), and (5.5, 0)

The slope of the line defined by the given two points, we have;

(0 - 4)/(5.5 - 0) = -40/55 = -8/11

The equation of the straight line graph in point and slope form, we have;

y - 4 = -8/11 × (x - 0)

y - 4 = -8/11·x

For the equation for the system that has infinitely many solution, we have;

y = -8/11·x + 4

2) For the equation whose graph foes through the point (0, 1) and the system has no solution, we have;

An equation that has no solution with the equation y = -8/11·x + 4, is parallel to y = -8/11·x + 4, and therefore, the slope are equal

Therefore, the equation whose graph foes through the point (0, 1) and the system has no solution, in point and slope form is given as follows;

y - 1 = -8/11 × (x - 0)

y - 1 = -8/11·x

The equation of the graph that has no solution is therefore;

y = -8/11·x + 1

3) For the equation of the graph that goes through the point (0, 2), and has a common solution with the system at (4, 1), we have;

The slope of the graph is (1 - 2)/(4 - 0) = -1/4

The equation of the line that has only one solution with the system in point and slope form is given as follows;

y - 2 = -1/4×(x - 0)

y - 2 = -1/4·x

The equation of the line that has only one solution in standard form is therefore;

y = -1/4·x + 2