Factorisation would help
⎩ ⎪ ⎨ ⎪ ⎧ x y + 2 x + y = 2 8 x z + 3 x + z = 3 7 y z + 3 y + 2 z = 4 2 Positive reals x , y , and z satisfy the equation above. Find x + y + z .
The answer is 11.
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2 solutions
@Chew-Seong Cheong , why did you delete your solution?
(everything copied from one of your solutions and only changed values)
for x in range(0,100):
for y in range(0, 100):
for z in range(0, 100):
if (x*y+2*x+y==28):
if (x*z+3*x+z==37):
if(y*z+3*y+2*z==42):
print(x,y,z)
Output:
4 4 5
(and this is how I "wrote" my first python code)
The set of equations are equal to
⎩
⎪
⎨
⎪
⎧
x
y
+
2
x
+
y
+
2
=
(
x
+
1
)
(
y
+
2
)
=
3
0
x
z
+
3
x
+
z
+
3
=
(
x
+
1
)
(
z
+
3
)
=
4
0
y
z
+
3
y
+
2
z
+
6
=
(
y
+
2
)
(
z
+
3
)
=
4
8
.
Substitute
a
,
b
,
c
into
x
+
1
,
y
+
2
,
z
+
3
. With some reasoning, we get
(
a
b
c
)
2
=
3
0
×
4
0
×
4
8
=
5
7
6
0
0
,
a
b
c
=
2
4
0
,
So
a
=
b
c
a
b
c
=
5
,
e
t
c
.
and we get
a
=
5
,
b
=
6
,
c
=
8
,
i.e.
x
=
4
,
y
=
4
,
z
=
5
.
So the answer is
4
+
4
+
5
=
4
+
2
+
5
=
1
1
.
This method can be used in general for different values and more unknowns, and might give multiple sets of answers.
Nice! @Jeff Giff More of these type of questions - I've got your back!
P.S Can I also submit my idea for the next question (of this type)?
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Thanks! Feel free to give any suggestions! :)
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Can we do like 4 equations in the format above but the final answer is... imaginary! What do you think?
The first half of your solution is same as mine, the latter differs. Nice Problem!
\[\begin{cases} xy + 2x + y {\color{red}{ +2}} = 28 {\color{red}{ +2}} \\ xz + 3x + z {\color{green}{ +3}} = 37 {\color{green}{ +3}} \\ yz + 3y + 2z {\color{blue}{ +6}} = 42 {\color{blue}{ +6}} \end{cases}
\implies
\begin{cases} (x+1)(y+2) = 30 &\ldots {\color{red}{(1)}} \\ (x+1)(z+3) = 40 &\ldots {\color{green}{(2)}} \\ (y+2)(z+3) = 48 &\ldots {\color{blue}{(3)}} \end{cases} \]
( y + 2 ) ( z + 3 ) ( x + 1 ) ( y + 2 ) ⋅ ( x + 1 ) ( z + 3 ) ( x + 1 ) 2 ( x + 1 ) 2 x = ( 3 ) ( 1 ) ( 2 ) = 4 8 3 0 ⋅ 4 0 = 2 5 = 4
Similarly, all the values can be found by either substituting or by doing the method above. We get y = 4 , z = 5