Question

Can anyone help with 3 questions?

Answer 1

\[\frac{ 1 }{ 3 }y =\frac{ 7 }{ 3 }x +\frac{ 5 }{ ? }\]

Answer 2

This doesn't make sense... any number can take the place of that question mark... except zero, that is...

Answer 3

\[\frac{ 5 }{ 3 }*\]

Answer 4

okay... so what's the question? :)

Answer 5

Especially lovin' the fact that all the terms have a denominator 3...it's like it's just begging to be cancelled off :D

Answer 6

:)

Answer 7

Solve each system using substitution. Tell whether the system has one solution, infinitely many solutions or No solution.

Answer 8

This is no system if you only have one equation... I'm sure there's a second equation paired with this one?

Answer 9

Real fast hit me with a brick please
x-3y=5

Answer 10

Okay... convenient way to do this?
We have to solve each system by substitution anyway... so... let's get cracking :D
\[\Large \frac13\color{blue}y = \frac73\color{red}x + \frac53\]
\[\Large \color{red}x - 3\color{blue}y = 5\]

Answer 11

Ok so I just stick the fractions in for X and Y

Answer 12

Mind if we simplify the first equation by multiplying 3 to the entire equation? Just to get rid of those denominators...
\[\Large 3\left(\frac13\color{red}y\right)= 3\left(\frac73\color{blue}x\right) +3\left(\frac53\right)\]

Answer 13

\\[\Large \cancel3\left(\frac1{\cancel{3}}\color{red}y\right)= \cancel3\left(\frac7{\cancel3}\color{blue}x\right) +\cancel3\left(\frac5{\cancel3}\right)\]

Answer 14

I understand that
but now what?

Answer 15

Thus giving us a more "acceptable" first equation in the form of...
\[\Large \color{blue}y = 7\color{red}x+5\]

Answer 16

Well, you can go two ways, but the most direct way is to use the fact that the equation above ^
already has y "isolated" or y is on one side of the equation, with everything else on the other.

Answer 17

So, the gist of solving by substitution, is, funnily enough, to SUBSTITUTE that expression for y... in the second equation... where we have
\[\Large \color{red}x - 3\color{blue}y = 5\]
And so we have
\[\Large \color{red}x - 3(7\color{red}x+5)=5\]

Answer 18

So I am really thinking this has infinite solutions

Answer 19

Just look you just told me i could use like 2 different ways so
it's really starting to stick in my mind.

Answer 20

There's a quick and easy way to find out if it has infinitely many solutions, but since this asks us to solve by substitution anyway, might as well do that. Could you solve this equation for x?
\[\Large \color{red}x - 3(7\color{red}x+5)=5\]

Answer 21

If i am correct and my memory is still going
You just take x to both sides of the equal sign

Answer 22

I don't know what that means.. it's probably a method unknown to me... perhaps a demonstration? :)

(It doesn't really matter what method you use, as long as you arrive at the correct value for x, after solving \(\Large \color{red}x - 3(7\color{red}x+5)=5\) )

Answer 23

You move the x on the far left all the way to the 5

Answer 24

Might I remind you that this is just a linear equation in one variable (x) ?
:)
Just distribute, to remove the parentheses, and then move everything WITHOUT x to the right-side.

Answer 25

Ok Now I understand
could you just do it for me
I learn better that way

Answer 26

Sure. You did say you have three questions... if these questions are similar, then I hope the third one, you'll be able to do on your own :)
Let me start from the beginning (after we cancel out the denominators in the original equation, though)

Answer 27

Yes now that you showed me how to do it I can do it on my own
they are the same throughout this whole 7 or 8 question sheets
Just needed some reminders on how to solve

Answer 28

So we start with two equations
\[\Large \color{blue}y =7\color{red}x + 5 \]\[\Large \color{red}x - 3\color{blue}y = 5\]

Answer 29

So, the first equation, we already have y expressed in terms of x... namely that
y = 7x + 5

So, we can replace the y in the second equation with 7x + 5... like so...
\[\Large \color{red}x - 3(7\color{red}x+5)=5\]
We distribute the -3, we get
\[\Large \color{red}x - 21\color{red}x-15=5\]
We add 15 to both sides of the equation to bring it to the right side...
\[\Large \color{red}x - 21\color{red}x-15\color{orange}{+15}=5\color{orange}{+15}\]
Simplifying...
\[\Large \color{red}x - 21\color{red}x=\color{orange}{20}\]

Now we simplify x - 21x, which is simply -20x
\[\Large - 20\color{red}x=20\]


Divide both sides by -20
\[\Large \frac{-20\color{red}x}{-20}= \frac{20}{-20}\]

Leaving us with...
\[\LARGE \color{red}x = \ -1\]

Answer 30

I see now :)
just very long steps

Answer 31

That said, we now have a value for x, so substitute this into ANY of the two equations originally given, let's pick the second...
\[\Large \color{red}x - 3\color{blue}y = 5\]
\[\Large \color{red}{-1} - 3\color{blue}y = 5\]
And now we solve for y

Answer 32

So... we add 1 to both sides....
\[\Large-1 - 3\color{blue}y\color{orange}{+1} = 5\color{orange}{+1}\]
Simplifying...
\[\Large - 3\color{blue}y = 6\]
Dividing both sides by -3
\[\Large \frac{-3\color{blue}y}{-3}= \frac6{-3}\]
\[\LARGE \color{blue}y = -2\]

Answer 33

y=-2
x=-1

Answer 34

Yup :)
and the mere fact that we got a solution as definite as this one means that there is only one solution :D

Answer 35

And the steps are only long because I went into extreme detail. You can find shortcuts for yourself that work for you... but that's up to you :)

Answer 36

:) thank you so much @terenzreignz

Answer 37

No problem :)