Question

math problem

Answer 1

@AZ

Answer 2

\[ \begin{cases} 5x+7y=6 \\ -10x-14y=-12 \end{cases} \]
we can see that we can multiply by negaive 2 the 1st equation;
\[ \begin{cases} (5x+7y=6) \times 2 \\ -10x-14y=-12 \end{cases} \Rightarrow \begin{cases} -10x-14y=-12 \\ -10x-14y=-12 \end{cases} \]
that said, we can understand that there will be 2 lines, but both of them will have the same position; so there will only be 1 line in the grap which means there are \( \infty\) solutions to that

Answer 3

there isnt an infinite option

Answer 4

oh, i thought it was a similar question to the previous one 🤦‍♂️

Answer 5

\[ \begin{cases} 5x+7y=6 \\ -10x-14y=-12 \end{cases} \]
this time ,we're solving for x
\[ \begin{cases} 5x=6-7y \\ -10x-14y=-12 \end{cases} \Rightarrow \begin{cases} \cancel{5}x==\frac{6}{5}+\frac{-7 y}{5} \\ -10x-14y=-12 \end{cases} \]
lets plug that value of x in the 2nd equation:

\[ \begin{cases} -10\left(\frac{6}{5}-\frac{7 y}{5}\right)-14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]
remember \( a(b+c)=a b+a c \) ?? thus,
\[ \begin{cases} -10\left(\frac{6}{5}\right)-10\left(-\frac{7 y}{5}\right)-14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]

Answer 6

so this is the part that we want to start simplifying

\[ \begin{cases} \cancel5 \cdot\left(-2\left(\frac{6}{\cancel5}\right)\right)-10\left(-\frac{7 y}{5}\right)-14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]
by canceling that, we get:

\[ \begin{cases} -2 \cdot 6-10\left(-\frac{7 y}{5}\right)-14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]

Answer 7

ok am lazy do to the latex so ill be explaining it -- so now we do \( -2 \cdot6 = -12 \) -- after that to get rid of the other denominator, my writing \( -10 \) in the form of \( -2 \cdot 5 \) after we cancel the 5ves as well we end up getting this :

\[ \large \begin{cases} -12-2(-7 y)-14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]

Answer 8

multiply \( 2 \) and \( -7y \)
\[ \begin{cases} -12+ 14y -14 y=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]
\[ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \begin{cases} -12=-12 \\ \frac{6}{5}+\frac{-7 y}{5} \end{cases} \]

since we have \( -12=-12 \) , which is equal to \( 0=0 \) , it means that that statement will always be true

that tells us that \( x = x \) -- so you can see even in the options, that we are writing \(x\) just as \(x \) lol

what we should do now, is solve for y

Answer 9

Substitute the x as a variable type it in in the calculator exactly like that then put (,x) at the and and it should give you the answer

Answer 10

so yeah lets solve for \(y\)

\[ \frac{6}{5}+\frac{-7 y}{5} \]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ -\frac{7 y}{5}=x-\frac{6}{5} \]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ -\frac{7 y}{5}=x-\frac{6}{5} \]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \text{multiply both sides by} -\frac{7 y}{5} } \]
\[ \frac{-5}{7} \cdot\left(-\frac{7 y}{5}\right)=-\frac{5}{7} \cdot\left(x-\frac{6}{5}\right) \]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \text{factor 5 out of the negative 5 so we can calncel the 5ves} \]
\[ \frac{\cancel 5 \cdot-1}{7} \cdot \frac{-7 y}{\cancel{5}}=-\frac{5}{7} \cdot\left(x-\frac{6}{5}\right) \]
\[ \ \ \ \ \ \ \ \ \ \Updownarrow \]
\[ \frac{-1}{7} \cdot(-7 y)=-\frac{5}{7} \cdot\left(x-\frac{6}{5}\right) \]lapsus*

Answer 11

ugh nvm you can read it anyway

Answer 12

wait, i just noticed -- were not finished with the left side yet, because we can siplify the \( \frac{number}{7} \) by factoring \(7\) out of \( -7 y \):

\[\frac{-1}{\cancel { 7}} \cdot(\cancel{7}(-y))=-\frac{5}{7} \cdot\left(x-\frac{6}{5}\right) \]
\[\ \ \ \ \ \ \ \ \ \Updownarrow\]
\[ y=-\frac{5}{7} \cdot\left(x-\frac{6}{5}\right) \]

and now were totally finished with the left side. i mean we got to y, where we want it to be

Answer 13

so its B or the second option

Answer 14

wait i havent finished it yet myself as im going with the same pace as my latex😹 -- i will skip this second part -- too lazy

Answer 15

ok, so in the end we are going to have \( y= -\frac{5 x}{7}+\frac{6}{7} \)

we said \( x=x \)

that said, we would get something like:
\[ \left(x, - \frac{5 x}{7} + \frac{6}{7}\right) \]
which option is that?

Answer 16

@klmjnmkj ??

Answer 17

f or the last option

Answer 18

yeah, that looks right