The equations 3 x minus 4 y = negative 2, 4 x minus y = 4, 3 x + 4 y = 2, and 4 x + y = negative 4 are shown on the graph below. On a coordinate plane, there are 4 lines. Green line goes through (0, negative 4) and (2, 4). Blue line goes through (negative 2, 2) and (2, negative 1). The pink line goes through (negative 2, negative 1), and (2, 2). The purple line goes through (negative 2, 4) and (0, negative 4). Which is the approximate solution for the system of equations 3 x + 4 y = 2 and 4 x + y = negative 4? (–1.4, 1.5) (1.4, 1.5) (0.9, –0.2) (–0.9, –0.2)

  (–1.4, 1.5)

Step-by-step explanation:

The blue line and the purple line are the lines corresponding to the equations of interest. Their point of intersection is in the 2nd quadrant, so is nearest to ...

  (–1.4, 1.5)

__

It can be useful to understand that for equations in standard form:

  ax +by = c

the x- and y-intercepts are ...

__

For the equations of interest, the first has intercepts of ...

  x=2/3, y=1/2 . . . . graphed line makes a 1st-quadrant triangle with the axes (blue line)

And the second has intercepts of ...

  x=-1, y=-4 . . . . graphed line makes a 3rd-quadrant triangle with the axes (purple line)

Since the purple line has a steeper slope, the point of intersection of the lines will be in the 2nd quadrant. There is only one 2nd-quadrant answer choice: (-1.4, 1.5).

b

step-by-step explanation:

if a was wrong then it has to be b it cant be c either because not all the y intercepts are the same!

ok now , i would like brainlest plz

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl