How to solve these 3 equations for three unknowns x,y,z

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up vote
3
down vote

favorite

Question:

solve :

$xy+x+y=23tag1$

$yz+y+z=31tag2$

$zx+z+x=47tag3$

My attempt:

by adding all we get

$sum xy +2sum x =101$

multiplying $(1)$ by $z \$ $(2)$ by $x$ and $(3)$ by $y$ and adding altogeather gives

$3xyz+ 2sum xy =31x+47y+23z$

then, from above two equations after eliminating $sum xy$ term we get

$35x+51y+27z=202+3xyz$

after that subtracting $(1)times 3z$ from equation just above (to eliminate $3xyz$ term) gives

$35x +51y-3z(14+x+y)=202implies (x+y)[35-3z]+16y-42z=202$

i tried pairwise subtraction of $(1),(2)$ and $(3)$ but it also seems to be not working

please give me some hint so that i can proceed or provide with the answer

thanks for your help

regards

asked 24 mins ago

veeresh pandey

782314

add a commentÂ |Â

up vote
3
down vote

favorite

Question:

solve :

$xy+x+y=23tag1$

$yz+y+z=31tag2$

$zx+z+x=47tag3$

My attempt:

by adding all we get

$sum xy +2sum x =101$

multiplying $(1)$ by $z \$ $(2)$ by $x$ and $(3)$ by $y$ and adding altogeather gives

$3xyz+ 2sum xy =31x+47y+23z$

then, from above two equations after eliminating $sum xy$ term we get

$35x+51y+27z=202+3xyz$

after that subtracting $(1)times 3z$ from equation just above (to eliminate $3xyz$ term) gives

$35x +51y-3z(14+x+y)=202implies (x+y)[35-3z]+16y-42z=202$

i tried pairwise subtraction of $(1),(2)$ and $(3)$ but it also seems to be not working

please give me some hint so that i can proceed or provide with the answer

thanks for your help

regards

asked 24 mins ago

veeresh pandey

782314

add a commentÂ |Â

up vote
3
down vote

favorite

Question:

solve :

$xy+x+y=23tag1$

$yz+y+z=31tag2$

$zx+z+x=47tag3$

My attempt:

by adding all we get

$sum xy +2sum x =101$

multiplying $(1)$ by $z \$ $(2)$ by $x$ and $(3)$ by $y$ and adding altogeather gives

$3xyz+ 2sum xy =31x+47y+23z$

then, from above two equations after eliminating $sum xy$ term we get

$35x+51y+27z=202+3xyz$

after that subtracting $(1)times 3z$ from equation just above (to eliminate $3xyz$ term) gives

$35x +51y-3z(14+x+y)=202implies (x+y)[35-3z]+16y-42z=202$

i tried pairwise subtraction of $(1),(2)$ and $(3)$ but it also seems to be not working

please give me some hint so that i can proceed or provide with the answer

thanks for your help

regards

asked 24 mins ago

veeresh pandey

782314

Question:

solve :

$xy+x+y=23tag1$

$yz+y+z=31tag2$

$zx+z+x=47tag3$

My attempt:

by adding all we get

$sum xy +2sum x =101$

multiplying $(1)$ by $z \$ $(2)$ by $x$ and $(3)$ by $y$ and adding altogeather gives

$3xyz+ 2sum xy =31x+47y+23z$

then, from above two equations after eliminating $sum xy$ term we get

$35x+51y+27z=202+3xyz$

after that subtracting $(1)times 3z$ from equation just above (to eliminate $3xyz$ term) gives

$35x +51y-3z(14+x+y)=202implies (x+y)[35-3z]+16y-42z=202$

i tried pairwise subtraction of $(1),(2)$ and $(3)$ but it also seems to be not working

please give me some hint so that i can proceed or provide with the answer

thanks for your help

regards

algebra-precalculus contest-math systems-of-equations

asked 24 mins ago

veeresh pandey

782314

asked 24 mins ago

veeresh pandey

782314

asked 24 mins ago

veeresh pandey

782314

asked 24 mins ago

veeresh pandey

782314

asked 24 mins ago

veeresh pandey

782314

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
5
down vote

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

answered 18 mins ago

TonyK

38.9k348127

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

add a commentÂ |Â

up vote
3
down vote

We can use Simon's favorite factoring trick.

$$xy+x+y+1=(x+1)(y+1)$$

This tells us

$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = fracsqrt24cdot32cdot4832to x=5$. Likewise, you can find the other variables.

answered 17 mins ago

Rushabh Mehta

2,755222

1

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

add a commentÂ |Â

up vote
0
down vote

hint

$(2)-(1)$ gives

$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$

$(3)-(2)$ becomes

$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$

$(3)-(1)$ yields to

$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$

From here, we put
$$X=x+1,;Y=y+1,; Z=z+1$$
thus

$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
5
down vote

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

answered 18 mins ago

TonyK

38.9k348127

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

add a commentÂ |Â

up vote
5
down vote

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

answered 18 mins ago

TonyK

38.9k348127

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

add a commentÂ |Â

up vote
5
down vote

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

answered 18 mins ago

TonyK

38.9k348127

Hint: Put
$$X=x+1$$
$$Y=y+1$$
$$Z=z+1$$

Then we have

$$XY=24$$
$$YZ=32$$
$$ZX=48$$

Can you take it from there?

answered 18 mins ago

TonyK

38.9k348127

answered 18 mins ago

TonyK

38.9k348127

answered 18 mins ago

TonyK

38.9k348127

answered 18 mins ago

TonyK

38.9k348127

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

add a commentÂ |Â

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

K : that was nice trick ......now i can take it
â€“Â veeresh pandey
15 mins ago

add a commentÂ |Â

up vote
3
down vote

We can use Simon's favorite factoring trick.

$$xy+x+y+1=(x+1)(y+1)$$

This tells us

$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = fracsqrt24cdot32cdot4832to x=5$. Likewise, you can find the other variables.

answered 17 mins ago

Rushabh Mehta

2,755222

1

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

add a commentÂ |Â

up vote
3
down vote

We can use Simon's favorite factoring trick.

$$xy+x+y+1=(x+1)(y+1)$$

This tells us

$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = fracsqrt24cdot32cdot4832to x=5$. Likewise, you can find the other variables.

answered 17 mins ago

Rushabh Mehta

2,755222

1

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

add a commentÂ |Â

up vote
3
down vote

We can use Simon's favorite factoring trick.

$$xy+x+y+1=(x+1)(y+1)$$

This tells us

$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = fracsqrt24cdot32cdot4832to x=5$. Likewise, you can find the other variables.

answered 17 mins ago

Rushabh Mehta

2,755222

We can use Simon's favorite factoring trick.

$$xy+x+y+1=(x+1)(y+1)$$

This tells us

$$(x+1)(y+1)=24$$$$(y+1)(z+1)=32$$$$(x+1)(z+1)=48$$So, we know that $x+1 = fracsqrt24cdot32cdot4832to x=5$. Likewise, you can find the other variables.

answered 17 mins ago

Rushabh Mehta

2,755222

answered 17 mins ago

Rushabh Mehta

2,755222

answered 17 mins ago

Rushabh Mehta

2,755222

answered 17 mins ago

Rushabh Mehta

2,755222

1

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

add a commentÂ |Â

1

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

amazing thank you ...i will take it from here .....
â€“Â veeresh pandey
14 mins ago

@veereshpandey No problem! If this was helpful, make sure to hit that green check mark!
â€“Â Rushabh Mehta
14 mins ago

add a commentÂ |Â

up vote
0
down vote

hint

$(2)-(1)$ gives

$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$

$(3)-(2)$ becomes

$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$

$(3)-(1)$ yields to

$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$

From here, we put
$$X=x+1,;Y=y+1,; Z=z+1$$
thus

$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

add a commentÂ |Â

up vote
0
down vote

hint

$(2)-(1)$ gives

$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$

$(3)-(2)$ becomes

$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$

$(3)-(1)$ yields to

$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$

From here, we put
$$X=x+1,;Y=y+1,; Z=z+1$$
thus

$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

add a commentÂ |Â

up vote
0
down vote

hint

$(2)-(1)$ gives

$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$

$(3)-(2)$ becomes

$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$

$(3)-(1)$ yields to

$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$

From here, we put
$$X=x+1,;Y=y+1,; Z=z+1$$
thus

$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

hint

$(2)-(1)$ gives

$$(z-x)(y+1)=8=(z+1-(x+1))(y+1)$$

$(3)-(2)$ becomes

$$(x-y)(z+1)=16=(x+1-(y+1))(z+1)$$

$(3)-(1)$ yields to

$$(z-y)(x+1)=24=(z+1-(y+1))(x+1)$$

From here, we put
$$X=x+1,;Y=y+1,; Z=z+1$$
thus

$$YZ-YX=8$$
$$ZX-ZY=16$$
$$XZ-XY=24$$

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

edited 28 secs ago

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

answered 15 mins ago

Salahamam_ Fatima

34.7k21431

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

add a commentÂ |Â

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

OP mentioned he took pairwise differences and couldn't make progress.
â€“Â Rushabh Mehta
15 mins ago

add a commentÂ |Â

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