Question

Guys, can you give me spare time helping me answer this

Thanks, medal will be GIVEN.


Solve for the unknown


By Substitution

1. 5x-3y-16=0
4x-5y-5=0

By Add a Subt

1a. x-3y=2
2x+3y=13

2b. 2/x+3/y=7
6/x-6/y=1

Answer 1

To solve for substitution, first, do you know how to isolate a variable?

Answer 2

To solve *by* substitution... sorry.

Answer 3

is this correct 5x+3y+16?

Answer 4

I thought it was
5x - 3y - 16 = 0
?

Answer 5

yes and i change the sign, isnt that the 1st step? to find x?

Answer 6

Sure... find x.

Manipulate the equation so that x stands alone on one side, with the rest on the other.

Answer 7

this lesson is linear equation with two variable btw

Answer 8

Yup... I got that :)

Answer 9

so 5x=3y+16

Answer 10

Here, I'll give you an example...

2x - 12y + 10 = 0

So, bring everything without "x" to the right side, like so...
2x - 12y = -10
2x = 12y - 10

then divide everything by 2...
x = 6y - 5

And that's how you isolate a variable :)

Answer 11

Oh, it seems you have it down... great :D

Answer 12

Just one more step to isolate that x.

Answer 13

divide?

Answer 14

Divide, yes, by what? :)

Answer 15

can you help me on that :D

Answer 16

??
You have
\[\Large 5x = 3y + 16\]
so dividing both sides by 5 should give you
\[\Large x = \frac{3y+16}{5}\]

Answer 17

ok i get that then is this right ?\[2\left(\begin{matrix}3y+16 \\ 5\end{matrix}\right)\]

Answer 18

Why 2?
(also, if you wanna put fractions, type \frac{numerator}{denominator} in your equation editor)

Answer 19

i mean 4. sorry

Answer 20

That's right :)
\[\Large 4\left(\frac{3\color{red}y+16}{5}\right)-5\color{red}y - 5 = 0\]
As you can see, now, it is just a linear equation in one variable. Solve for y, and you're good-to-go :)

Answer 21

i dont know how to solve \[4\frac{3y+16}{5}\]

Answer 22

I see... you get rid of the denominator by multiplying everything by 5, first...

Or you could rely on this fact first...
\[\Large p\cdot \frac{a}b = \frac{pa}b\]

IE, just mulitply 4 to the numerator first.

Answer 23

so it will be 4(15y+80)?

Answer 24

Yes... techincally
BUT
we didn't multiply 5 just so you could distribute it to the numerator, we multiply 5 to CANCEL THE DENOMINATOR

Don't confuse those two methods I mentioned... they're not the same... either you distribute the 4 first OR you multiply everything by 5.

Your call.

Answer 25

so my answer is valid? 4(15y+80)?

Answer 26

that's not the answer yet, that's just a step.
It is a valid step, but also unnecessary, and actually rather unhelpful, because you have to reverse what you did anyway...

What I meant was this
\[\Large 4(5)\left(\frac{3\color{red}y+16}{5}\right)-5(5)\color{red}y - 5(5) = 0\]
So that we can do this
\[\Large 4(\cancel5)\left(\frac{3\color{red}y+16}{\cancel5}\right)-25\color{red}y -25 = 0\]
And so we're left with this
\[\Large 4(3\color{red}y+16)-25\color{red}y - 25 = 0\]

Answer 27

ok then after that we multiply now? 12y+64-25y-25?

Answer 28

is this correct also?
x+2y-5=0
4x+y-6=0

x=2y+5

4x+y-6=0

4(-2y+5)+y-6=0

-7 =-20+6
-7y=-14
7 7

(x=2)

x=-2(2)+5
=-4+5
(y=1)

Answer 29

Hang on a sec...

Answer 30

It's wrong.

Answer 31

why

Answer 32

From
x + 2y - 5 = 0
you got
x = 2y + 5


faulty... redo that step.

Answer 33

On second thought, that may have been a typo, as I see you used the correct substitution when it mattered... recalibrating :D

Answer 34

It's correct.
Great job :D

Answer 35

thanks

Answer 36

but if i apply that method in what were working on will i be correct?

Answer 37

Of course... you just need a refresher when it comes to dealing with fractions :)

Answer 38

12y+64-25y-25? so this becomes = 37y=-89?

Answer 39

37y?
Note the minus sign next to 25y...

Answer 40

Great... now solve for x and head-on home :D

Answer 41

13y 39
13 13 = 3y

Answer 42

5x-3(3)-16
-9-16? is this the right equation i think its wrong

Answer 43

can you help me out with the 2 number please. specially the fraction one

Answer 44

lol that was a pretty nice trick question, number 2... let's write it out...

Answer 45

\[\Large \frac2x + \frac3y=7\]\[\Large \frac6x - \frac6y = 1\]

Answer 46

This does look confusing, doesn't it? :D

Answer 47

yes, but im not done finding the x in number one :D

Answer 48

Oh, by all means... do so :D

Answer 49

i dont know the equation if im right.

5x-3(3)-16
-9-16?

Answer 50

Yeah, that's right... and all that's equal to zero...
5x - 9 - 16 = 0

Answer 51

can it be x=0?

Answer 52

Of course not.
Solve for x...
You solved for y earlier, and you solved it well... utilise those same skills to solve for x now :D

Answer 53

but this is my equation so whats next for this?

Answer 54

5x-3(3)-16

Answer 55

5x - 9 - 16 = 0
?
Much like earlier, of course, bring everything without x to the other side, and solve.

Answer 56

5x=-9-16
5x=24?

can you help me out on this and before you get going can you help me solve the fraction in no 2 1st so that i can review it later.

Answer 57

Still wrong...
5x - 9 - 16 = 0

Redo it... :P

Answer 58

5x=9+16?

Answer 59

Okay, that's better.
Now solve for x.

Answer 60

Quickly now... time's not on my side.

Answer 61

i dont know whats next can you help me out

Answer 62

Add 9 and 16?

Answer 63

25

Answer 64

x=5

Answer 65

so y=3 x=5

Answer 66

so y=3 x=5 is the final answer in 1?

Answer 67

Good. And you're done.

As for the one involving fractions,
you can substitute first, if it makes it simpler.
Let
\[\Large \bar x = \frac1x\]\[\Large \bar y = \frac1y\]
So it then becomes...

\[\LARGE 2\bar x + 3\bar y = 7\]\[\LARGE 6\bar x - 6\bar y = 1\]

and now it looks more familiar... solve normally from here, and then substitute back.

Answer 68

I need to go now... I'm sure someone else would be willing to take up this challenge :)

----------------------------------------
Terence out

Answer 69

x 6 - 12x+18y = 49
x 2 - 12x-12y =2?

Answer 70

terenz can you leave out the solution here and answer so i can review it later use. thans