can I use to graph a system of equations? I need one.

Question

y = –2x + 3 and 2x + 4y = –8?

Answer 1

Given equations are y = -2x + 3 and 2x + 4y = -8

Substitute y = -2x + 3 in 2x + 4y = -8 we get,

2x + 4(-2x + 3) = -8

2x - 8x + 12 = -8

-6x + 12 = -8

-6x = -8 -12

-6x = -20

x = 20/6 = 3.33

Substitute x = 20/6 in y = -2x + 3 we get,

y = -2(20/6) + 3

  = -20/3 + 3

  = (-20 + 9)/3

  = -11/3 = -3.67

Therefore (x, y) = (3.33, 3.67)

The graph for the given system of equations is as follows:  

Answer 2

The system of equations are y = - 2x + 3 and 2x + 4y = - 8.

Slope - intercept form of the line equation is y = mx + b, where, m is slope and b is y - intercept.

First write the equations in slope - intercept form.

The first equation is y = - 2x + 3.

Compare the equation with slope - intercept form : y = mx + b.

slope = - 2 and y - intercept is 3.

y - intercept is 3, so the line crosses the y - axis at (0, 3).

Using slope find the next point.

Slope = rise/run = - 2/1

Start at point (0, 3), move 2 units down and 1 unit right, then plot the point (1, 1).

The second equation is 2x + 4y = - 8.

Divide each side by 2.

x + 2y = - 4.

2y = - x - 4

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

Compare the equation with slope - intercept form : y = mx + b.

slope = - 1/2 and y - intercept is - 2.

y - intercept is - 2, so the line crosses the y - axis at (0, - 2)

Using slope find the next point.

Slope = rise/run = - 1/2

Start at point (0, - 2), move 1 unit down and 2 units right, then plot the point (2, - 3).

Draw the lines through these points.

Graph :

Observe the graph,

The solution of the system (x, y) = (3.33, - 3.67).