help with steps in maths

[Anvity]

hello, i need help to complete these maths problems with steps. I couldn't understand these, lol. If someone could maybe help...

1.

2.
Example of a equation with 3 systems with three unknowns that are incompatible. explaining what did you think.

Thankkk.

 

[payaso]

Oh lord this gave me an huge throwback, i remember you could use an app called photomath or sum like that and it will solve it for you showing every step

 

[Ally]

Question 1. https://sevr.org/i/82284ca1-ddb9-4d59-8067-767bd62fec33.pdf

Revised: https://sevr.org/i/8b321234-9471-4094-8ea1-cfeb293ec2b0.pdf

Edit: I think 1a) +-sqrt(2) isn't a valid solution.

Didn't do Gauss or matrices. Not familiar with it.

 

[Function]

mathway.com or download mathway :>

 

[Existing Potato]

Just trying my best here, please dont yeet me, Ill add more when I solve them starting from the easier ones(If I solve them):

Starting off 1C the easiest question
1 .C) 2x^2 * (x-3)^2 * (3x+4)=0
If any one of the three (2x^2), (x-3)^2, (3x+4) is 0, the entire equation is 0 cus 0*anything is 0.
Case 1: 2x^2=0
x=0, kinda obvious
Case 2: (x-3)^2=0
(x-3)=0
x=3, also kinda obvious
Case 3: 3x+4=0
3x=-4
x=-4/3

So for C) X=0,3,-4/3

 

[ItzKamilo]

This quarantine shit made me forget all this. I think they need to send me to summer school...

 

[Existing Potato]

For 1 E

3 logx - log(2x^2+x-2)=0

Since Y*log(X)= log(X^Y) you get

log(x^3) - log(2x^2+x-2)=0

Log(a)-Log(b)= Log(a/b) so you get

log( x^3 )
----------(divide) =0
(2x^2+x-2)

Since X^0=1 Log(1)=0 Sooooo

( x^3)/(2x^2+x-2) =1

So they equal and we get
( x^3)=(2x^2+x-2)
and thennnnn
(x^3)-(2x^2)-x+2=0
This is a level 3 polynomial so you can't use your quadratic formula, but if you look at the coefficients, in order they are 1,-2,-1,2.
Since there is a Pattern in the coefficients (1,2 positive signs and negative signs don't matter), we now know (x^3)-(2x^2)-x is divisible by (+/-x+/-2, eg x-2) because they are the coefficients we see in the patterns.
Its hard for me to draw out, but ((x^3)-(2x^2)-x)/x-2 = (x^2-1)(x-2)
Soooooooooooo (x^3)-(2x^2)-x)=(x^2-1)(x-2)(x-2)

Now all we need either (x^2-1) or (x-2) to be 0(we don't need to consider the last x-2 cus its the same lmao),
(x^2-1)=0
x^2=1
x=+/-1
-------------
(x-2)=0
x=2



So x=+/-1,2

 

[Ally]

Just trying my best here, please dont yeet me, Ill add more when I solve them starting from the easier ones(If I solve them):

Starting off 1C the easiest question
1 .C) 2x^2 * (x-3)^2 * (3x+4)=0
If any one of the three (2x^2), (x-3)^2, (3x+4) is 0, the entire equation is 0 cus 0*anything is 0.
Case 1: 2x^2=0
x=0, kinda obvious
Case 2: (x-3)^2=0
(x-3)=0
x=3, also kinda obvious
Case 3: 3x+4=0
3x=-4
x=-4/3

So for C) X=0,3,-4/3

For 1 E

3 logx - log(2x^2+x-2)=0

Since Y*log(X)= log(X^Y) you get

log(x^3) - log(2x^2+x-2)=0

Log(a)-Log(b)= Log(a/b) so you get

log( x^3 )
----------(divide) =0
(2x^2+x-2)

Since X^0=1 Log(1)=0 Sooooo

( x^3)/(2x^2+x-2) =1

So they equal and we get
( x^3)=(2x^2+x-2)
and thennnnn
(x^3)-(2x^2)-x+2=0
This is a level 3 polynomial so you can't use your quadratic formula, but if you look at the coefficients, in order they are 1,-2,-1,2.
Since there is a Pattern in the coefficients (1,2 positive signs and negative signs don't matter), we now know (x^3)-(2x^2)-x is divisible by (+/-x+/-2, eg x-2) because they are the coefficients we see in the patterns.
Its hard for me to draw out, but ((x^3)-(2x^2)-x)/x-2 = (x^2-1)(x-2)
Soooooooooooo (x^3)-(2x^2)-x)=(x^2-1)(x-2)(x-2)

Now all we need either (x^2-1) or (x-2) to be 0(we don't need to consider the last x-2 cus its the same lmao),
(x^2-1)=0
x^2=1
x=1
-------------
(x-2)=0
x=2



So x=1,2

Plaintext is not the best way to convey this .

 

[Existing Potato]

Plaintext is not the best way to convey this .

yes but Im too lazy[DOUBLEPOST=1590112490][/DOUBLEPOST]For number 2 if you want to be lazy, you could just do this:
x+y+z=0
x=y+z
2y+2z-2x=1

First 2 are just random formulas, but from those you know y+z-x=0, so you just make it y+z-x=1 so it doesn't work, but since that looks tooooo lazy, you just multiple both sides by 2(Original formula) to get 2y+2z-2x=0 (from 2*0) and to make it incompatible, just make the 0 a 1.

 

[Anvity]

Thankkkkkk you alll!!!!!