Question

What is the solution to the system of equations represented by the two equations?

y=2/3x
y=-2/3x+4


( , )
@robtobey

Answer 1

@ParthKohli

Answer 2

\[\large y=\frac{2}{3}x\ ..................(1)\]
\[\large y=-\frac{2}{3}x+4\ ........(2)\]
To solve for x, you need to plug the value for y given in equation (1) into equation (2).

Answer 3

idk can you tell me the answer and then explain how you got it?

Answer 4

Plugging in as described gives:
\[\large \frac{2}{3}x=-\frac{2}{3}x+4\ .........(3)\]
To solve equation (3) for x, the first step is to add (2/3)x to both sides of equation (3). Can you do that?

Answer 5

To help a little more, I will set out the step as follows:
\[\large \frac{2}{3}x+\frac{2}{3}x=(\frac{2}{3}x-\frac{2}{3}x)+4\ ........(4)\]

Answer 6

:( i have to finish this soon. ughhh

Answer 7

Solving (4) can be done in your head.

Answer 8

@kittiwitti1 can you tell me the answer i have to finish this soon

Answer 9

i will fan and medal

Answer 10

\[\text{Don't bribe people into helping you.}\]

Answer 11

No, I don't know the answer. I'm still looking through the posts.

Answer 12

welp i honestly don't kno what to do, i don't understand this and i have to finish this soon soo

Answer 13

@Bossimbacon I have explained exactly what to do. Are you prepared to do any work on this at all?

Answer 14

@kropot72 \[\text{It seems your explanation is too complex for him.}\]

Answer 15

@kittiwitti1 if you're hear to "insult" me, then you can leave.

Answer 16

here*

Answer 17

I'm not here* to "insult" you, I'm telling you to quit testing my patience when I'm trying to freaking HELP YOU.

Unless your entire point was to annoy me.
It's plausible.

Answer 18

...I'm still writing out the whole thing. You can go do something else while you wait.

Answer 19

thank you for helping i appreciate it.

Answer 20

Well, whatever he tried to explain to you, I can't understand it either...
and I tried to solve by making the equations equal each other, but that just deleted the x off the map entirely...

Answer 21

\[\text{WAIT I JUST HAD AN IDEA}!\]\[y = \frac{ 2 }{ 3 }x\]\[y= - \frac{2}{3} + 4\]Isolate x from the first equation\[x= \frac{3}{2}y\]and insert that x-value into the second equation.
Does this help at all?

Answer 22

WHOOPS did it wrong!
I solved and got -4 = 0 which is obviously wrong xD
Use the idea that @kropot72 gave you:\[\frac{2}{3}x=- \frac{2}{3}x + 4\]\[\text{Now move all x-coefficients to the left side:}\]\[\frac{2}{3}x+\frac{2}{3}x=4\]\[\text{this becomes}\]\[\frac{4}{3}x=4\]\[\text{now isolate the x}\]\[x=4 \div \frac{4}{3}~becomes~x=4 \times \frac{4}{3}\]\[x= \frac{16}{3}\]\[\text{Did this help?}\]