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Algebra - Solving equations with 3 variables.
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a,b,c $in Z^+$ satisfying the equations: $$ 5a + 5b + 2ab = 92$$ $$ 5b+5c+2bc=136$$ $$5c+5a+2ac=244$$ Find the value of 7a + 8b + 9c.
I took cases like a,b are odd numbers, even numbers and using the properties of their last digits i found the answer. But i want to know if there is some other method to solve this question.
linear-algebra functions
asked yesterday
Identicon
808
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up vote
2
down vote
favorite
a,b,c $in Z^+$ satisfying the equations: $$ 5a + 5b + 2ab = 92$$ $$ 5b+5c+2bc=136$$ $$5c+5a+2ac=244$$ Find the value of 7a + 8b + 9c.
I took cases like a,b are odd numbers, even numbers and using the properties of their last digits i found the answer. But i want to know if there is some other method to solve this question.
linear-algebra functions
asked yesterday
Identicon
808
Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
a,b,c $in Z^+$ satisfying the equations: $$ 5a + 5b + 2ab = 92$$ $$ 5b+5c+2bc=136$$ $$5c+5a+2ac=244$$ Find the value of 7a + 8b + 9c.
I took cases like a,b are odd numbers, even numbers and using the properties of their last digits i found the answer. But i want to know if there is some other method to solve this question.
linear-algebra functions
asked yesterday
Identicon
808
a,b,c $in Z^+$ satisfying the equations: $$ 5a + 5b + 2ab = 92$$ $$ 5b+5c+2bc=136$$ $$5c+5a+2ac=244$$ Find the value of 7a + 8b + 9c.
I took cases like a,b are odd numbers, even numbers and using the properties of their last digits i found the answer. But i want to know if there is some other method to solve this question.
linear-algebra functions
asked yesterday
Identicon
808
edited yesterday
asked yesterday
Identicon
808
asked yesterday
Identicon
808
asked yesterday
Identicon
808
808
Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday
add a comment |Â
Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday
Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday
Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
4
down vote
accepted
If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11times 19=-11times -19= 1times 209dots$$
Treat the others the same, and see what drops out.
$(2b+5)(2c+5)=297=11times 27$ (and various other possible factorisations) and $(2a+5)(2c+5)=513=27times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
answered yesterday
Mark Bennet
76.3k772169
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
add a comment |Â
up vote
3
down vote
In the same sprit as David G. Stork, consider
$$ 5a + 5b + 2ab = 92tag 1$$ $$ 5b+5c+2bc=136tag 2$$ $$5c+5a+2ac=244tag3$$
From $(1)$ elimintate $b$ to get
$$b=frac92-5 a2 a+5$$ Plug in $(2)$ and eliminate $c$ to get
$$c=frac27 a+20119 $$ Plug in $(3)$ and simplify to get
$$frac5419 (a-7) (a+12)=0$$
answered yesterday
Claude Leibovici
111k1054126
add a comment |Â
up vote
2
down vote
We can write the system of equations as following $$(2a+5)(2b+5)=209\(2b+5)(2c+5)=297\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
answered yesterday
Mostafa Ayaz
8,5203530
add a comment |Â
up vote
0
down vote
Two solutions (found by elimination of variables), only one with all positive values:
a -> 7, b -> 3, c -> 11
so $7 a + 8 b + 9 c = 172$
answered yesterday
David G. Stork
7,3102728
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
add a comment |Â
up vote
0
down vote
From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1le a,ble 12 textand\
a+b=frac92-2ab5=18+frac2(1-ab)5 Rightarrow abequiv 1pmod5.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation.
From the second and third equations one can easily find the complete set.
answered yesterday
farruhota
13.4k2632
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11times 19=-11times -19= 1times 209dots$$
Treat the others the same, and see what drops out.
$(2b+5)(2c+5)=297=11times 27$ (and various other possible factorisations) and $(2a+5)(2c+5)=513=27times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
answered yesterday
Mark Bennet
76.3k772169
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
add a comment |Â
up vote
4
down vote
accepted
If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11times 19=-11times -19= 1times 209dots$$
Treat the others the same, and see what drops out.
$(2b+5)(2c+5)=297=11times 27$ (and various other possible factorisations) and $(2a+5)(2c+5)=513=27times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
answered yesterday
Mark Bennet
76.3k772169
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11times 19=-11times -19= 1times 209dots$$
Treat the others the same, and see what drops out.
$(2b+5)(2c+5)=297=11times 27$ (and various other possible factorisations) and $(2a+5)(2c+5)=513=27times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
answered yesterday
Mark Bennet
76.3k772169
If you are looking for solutions in integers, try this.
Multiply the first equation by $2$ to obtain $$4ab+10a+10b=184$$ and rewrite as $$(2a+5)(2b+5)=209=11times 19=-11times -19= 1times 209dots$$
Treat the others the same, and see what drops out.
$(2b+5)(2c+5)=297=11times 27$ (and various other possible factorisations) and $(2a+5)(2c+5)=513=27times 19$
For positive integers $2b+5=11, 2c+5=27, 2a+5=19$
answered yesterday
Mark Bennet
76.3k772169
edited yesterday
answered yesterday
Mark Bennet
76.3k772169
answered yesterday
Mark Bennet
76.3k772169
answered yesterday
Mark Bennet
76.3k772169
76.3k772169
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
add a comment |Â
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
1
1
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Yes... and you can eliminate some of the factorizations immediately by the requirement $a,b,c in mathbbZ^+$, and specifics such as no factor can be $1$.
– David G. Stork
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
Because $prodlimits_cyc(2a+5)^2=5643^2.$ Nice! +1
– Michael Rozenberg
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
@DavidG.Stork Indeed, factors must be at least $7$
– Mark Bennet
yesterday
add a comment |Â
up vote
3
down vote
In the same sprit as David G. Stork, consider
$$ 5a + 5b + 2ab = 92tag 1$$ $$ 5b+5c+2bc=136tag 2$$ $$5c+5a+2ac=244tag3$$
From $(1)$ elimintate $b$ to get
$$b=frac92-5 a2 a+5$$ Plug in $(2)$ and eliminate $c$ to get
$$c=frac27 a+20119 $$ Plug in $(3)$ and simplify to get
$$frac5419 (a-7) (a+12)=0$$
answered yesterday
Claude Leibovici
111k1054126
add a comment |Â
up vote
3
down vote
In the same sprit as David G. Stork, consider
$$ 5a + 5b + 2ab = 92tag 1$$ $$ 5b+5c+2bc=136tag 2$$ $$5c+5a+2ac=244tag3$$
From $(1)$ elimintate $b$ to get
$$b=frac92-5 a2 a+5$$ Plug in $(2)$ and eliminate $c$ to get
$$c=frac27 a+20119 $$ Plug in $(3)$ and simplify to get
$$frac5419 (a-7) (a+12)=0$$
answered yesterday
Claude Leibovici
111k1054126
add a comment |Â
up vote
3
down vote
up vote
3
down vote
In the same sprit as David G. Stork, consider
$$ 5a + 5b + 2ab = 92tag 1$$ $$ 5b+5c+2bc=136tag 2$$ $$5c+5a+2ac=244tag3$$
From $(1)$ elimintate $b$ to get
$$b=frac92-5 a2 a+5$$ Plug in $(2)$ and eliminate $c$ to get
$$c=frac27 a+20119 $$ Plug in $(3)$ and simplify to get
$$frac5419 (a-7) (a+12)=0$$
answered yesterday
Claude Leibovici
111k1054126
In the same sprit as David G. Stork, consider
$$ 5a + 5b + 2ab = 92tag 1$$ $$ 5b+5c+2bc=136tag 2$$ $$5c+5a+2ac=244tag3$$
From $(1)$ elimintate $b$ to get
$$b=frac92-5 a2 a+5$$ Plug in $(2)$ and eliminate $c$ to get
$$c=frac27 a+20119 $$ Plug in $(3)$ and simplify to get
$$frac5419 (a-7) (a+12)=0$$
answered yesterday
Claude Leibovici
111k1054126
answered yesterday
Claude Leibovici
111k1054126
answered yesterday
Claude Leibovici
111k1054126
answered yesterday
Claude Leibovici
111k1054126
111k1054126
add a comment |Â
add a comment |Â
up vote
2
down vote
We can write the system of equations as following $$(2a+5)(2b+5)=209\(2b+5)(2c+5)=297\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
answered yesterday
Mostafa Ayaz
8,5203530
add a comment |Â
up vote
2
down vote
We can write the system of equations as following $$(2a+5)(2b+5)=209\(2b+5)(2c+5)=297\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
answered yesterday
Mostafa Ayaz
8,5203530
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We can write the system of equations as following $$(2a+5)(2b+5)=209\(2b+5)(2c+5)=297\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
answered yesterday
Mostafa Ayaz
8,5203530
We can write the system of equations as following $$(2a+5)(2b+5)=209\(2b+5)(2c+5)=297\(2a+5)(2c+5)=513$$the first equation has two answers $$a=3,b=7\a=7,b=3$$the first case is impossible since by substitution is 2nd equation we must have $19|297$ which is impossible therefore $$a=7,b=3$$and we obtain $$c=11$$so $$7a+8b+9c=49+24+99=172$$
answered yesterday
Mostafa Ayaz
8,5203530
answered yesterday
Mostafa Ayaz
8,5203530
answered yesterday
Mostafa Ayaz
8,5203530
answered yesterday
Mostafa Ayaz
8,5203530
8,5203530
add a comment |Â
add a comment |Â
up vote
0
down vote
Two solutions (found by elimination of variables), only one with all positive values:
a -> 7, b -> 3, c -> 11
so $7 a + 8 b + 9 c = 172$
answered yesterday
David G. Stork
7,3102728
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
add a comment |Â
up vote
0
down vote
Two solutions (found by elimination of variables), only one with all positive values:
a -> 7, b -> 3, c -> 11
so $7 a + 8 b + 9 c = 172$
answered yesterday
David G. Stork
7,3102728
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Two solutions (found by elimination of variables), only one with all positive values:
a -> 7, b -> 3, c -> 11
so $7 a + 8 b + 9 c = 172$
answered yesterday
David G. Stork
7,3102728
Two solutions (found by elimination of variables), only one with all positive values:
a -> 7, b -> 3, c -> 11
so $7 a + 8 b + 9 c = 172$
answered yesterday
David G. Stork
7,3102728
answered yesterday
David G. Stork
7,3102728
answered yesterday
David G. Stork
7,3102728
answered yesterday
David G. Stork
7,3102728
7,3102728
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
add a comment |Â
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
Sir, can you explain how you have solved this ? I tried using elimination method. But didn't work for me. So please explain.
– Identicon
yesterday
add a comment |Â
up vote
0
down vote
From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1le a,ble 12 textand\
a+b=frac92-2ab5=18+frac2(1-ab)5 Rightarrow abequiv 1pmod5.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation.
From the second and third equations one can easily find the complete set.
answered yesterday
farruhota
13.4k2632
add a comment |Â
up vote
0
down vote
From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1le a,ble 12 textand\
a+b=frac92-2ab5=18+frac2(1-ab)5 Rightarrow abequiv 1pmod5.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation.
From the second and third equations one can easily find the complete set.
answered yesterday
farruhota
13.4k2632
add a comment |Â
up vote
0
down vote
up vote
0
down vote
From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1le a,ble 12 textand\
a+b=frac92-2ab5=18+frac2(1-ab)5 Rightarrow abequiv 1pmod5.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation.
From the second and third equations one can easily find the complete set.
answered yesterday
farruhota
13.4k2632
From the first equation $5a + 5b + 2ab = 92$, we can find:
$$1le a,ble 12 textand\
a+b=frac92-2ab5=18+frac2(1-ab)5 Rightarrow abequiv 1pmod5.$$
The possible candidates for $(a,b)$ or $(b,a)$ are :
$$(1,6);(1,11);(2,3);(2,8);(3,7);(3,12);(4,9).$$
Only $(3,7)$ or $(7,3)$ satisfy the first equation.
From the second and third equations one can easily find the complete set.
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
answered yesterday
farruhota
13.4k2632
13.4k2632
add a comment |Â
add a comment |Â
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Answering the question literally, the brute-force computer program quickly shows that $7 cdot 3 + 8 cdot 7 + 9 cdot 11 = 176$.
– Weather Vane
yesterday