Question

math help

Answer 1

@florisalreadytaken

Answer 2

What is it

Answer 3

were given this system:
\[ \begin{cases} -2x+7y=-6\\-4x+14y=-12 \end{cases} \]
first step is to make the 1st equation be ralatable to the 2nd one -- in our case, we will have to multiply by -2:
\[ \begin{cases} ( -2x+7y=-6 ) \times (-2) \\ -4x+14y=-12 \end{cases} \]

by adding it together we get:
\[ (4x-14y=12) + (-4x+14y=-12) = 0 \]
that tells us that this equation, intersects at an infinite number of points.

lets solve for y (im going to use the 1st equation):
\[ 4x-14y=12 \overset{\text{take 4x to the other side}}{======== \Rightarrow} -14y = 12 - 4x\]
\[ \overset{\text{divide by -14 on both sides}}{========== \Rightarrow} \frac{-14 y}{-14}=\frac{12}{-14}+\frac{-4 x}{-14} \]
\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \updownarrow \]
\[ y=\frac{12}{-14}+\frac{-4 x}{-14} \]
thus,
\[ y=-\frac{6}{7}+\frac{2 x}{7} \]
which can also be written as:
\[ y= \frac{2 x}{7}-\frac{6}{7} \]
that said, we get:
\[ \left( x,-\frac{6}{7}+\frac{2 x}{7} \right) \]
which option is that?

Answer 4

i told you it would take long to explain : D

Answer 5

c?

Answer 6

no... the last one

Answer 7

what?😹😹😹