SOLUTION: c=x+8 b=x a=x+7 find x i tried everything i could..i cudnt find any solutions and it drove me crazy...help would be soo appreciated.

Algebra ->  Pythagorean-theorem -> SOLUTION: c=x+8 b=x a=x+7 find x i tried everything i could..i cudnt find any solutions and it drove me crazy...help would be soo appreciated.      Log On


   



Question 174426: c=x+8
b=x
a=x+7
find x
i tried everything i could..i cudnt find any solutions and it drove me crazy...help would be soo appreciated.

Found 4 solutions by josmiceli, stanbon, Electrified_Levi, Edwin McCravy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Usually, in order to be solvable, the number of equations
has to equal (or be greater than) the number of unknowns
You have 3 equations and 4 unknowns
c+=+x+%2B+8
b+=+x
a+=+x+%2B+7
As written, there is no way to solve this
Suppose I say x+=+5, then
c+=+5+%2B+8
b+=+5
a+=+5+%2B+7
c=13,b=5,anda=12
But suppose I say x+=+11, then
c+=+11+%2B+8
b+=+11
a+=+11+%2B+7
The point is, there is no single value of x that makes
the solution the one and only correct one.
Any x that you pick will make a solution

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
c=x+8
b=x
a=x+7
find x
------------
x = c-8
x = a-7
=========
c-8 = a-7
c = a + 1
--------------
Solutions: All ordered pairs (a,c) where c=a+1
Therefore : x = c-8 = (a+1)-8 = a-7
-------------
Final solution: x = a-7
and b = a-7 since x = b.
=============================
Cheers,
Stan H.


Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
.
If this is a Pythagorean Theorem, this is how you would find "x"
.
c=x+8
.
b=x
.
a=x+7
.
find x
.
Pythagorean Theorem is +a%5E2+%2B+b%5E2+=+c%5E2+
.
Now replace the letters "a", "b", and "c" with what they equal in "x" terms
.
Replace "a" with (x+7)
.
Replace "b" with (x)
.
Replace "c" with (x+8)
.
+a%5E2+%2B+b%5E2+=+c%5E2+ = +%28x%2B7%29%5E2+%2B+%28x%29%5E2+=+%28x%2B8%29%5E2+, now you would just solve for "x"
.
+%28x%2B7%29%5E2+%2B+%28x%29%5E2+=+%28x%2B8%29%5E2+ = +%28x%2B7%29%28x%2B7%29+%2B+x%5E2+=+%28x%2B8%29%28x%2B8%29+, you would now use the FOIL meathod to multiply
.
+%28x%2B7%29%28x%2B7%29+%2B+x%5E2+=+%28x%2B8%29%28x%2B8%29+ = = = =
.
+%28x%5E2%2B7x%2B7x%2B49%29+%2B+x%5E2+=+%28x%5E2+%2B+8x+%2B+8x+%2B+64%29+ = +%28x%5E2%2B14x%2B49%29+%2B+x%5E2+=+%28x%5E2+%2B+16x+%2B+64%29+ = +x%5E2%2B14x%2B49+%2B+x%5E2+=+x%5E2+%2B+16x+%2B+64+ , Now we will combine like terms
.
+x%5E2%2B14x%2B49+%2B+x%5E2+=+x%5E2+%2B+16x+%2B+64+ = +2x%5E2%2B14x%2B49+=+x%5E2+%2B+16x+%2B+64+, we will now move everything on the right side to the left side
.
+2x%5E2%2B14x%2B49+=+x%5E2+%2B+16x+%2B+64+ = +2x%5E2+-+x%5E2+%2B14x+-+16x+%2B49+-+64+=+x%5E2+-x%5E2+%2B+16x+-+16x+%2B+64+-+64+ = +x%5E2+-2x++-+15+=+0+, now we have a quadratic equation
.
We will solve using the quadratic equation
.
Quadratic equation = x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
Quadratic equations are in the form +ax%5E2+%2B+bx+%2B+c+=+0+
.
+x%5E2+-2x++-+15+=+0+, +ax%5E2+%2B+bx+%2B+c+=+0+
.
In our equation
.
a = 1
.
b = (-2)
.
c = (-15)
.
Now replace these letters with the numbers in the quadratic equation
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ = = x+=+%282+%2B-+sqrt%28+4-%28-60%29+%29%29%2F2+ = x+=+%282+%2B-+sqrt%28+4%2B60+%29%29%2F2+ = x+=+%282+%2B-+sqrt%28+64+%29%29%2F2+ = x+=+%282+%2B-+8%29%2F2+
.
x+=+%282+%2B+8%29%2F2+ = x+=+%2810%29%2F2+ = x+=+5+
.
x+=+%282+-+8%29%2F2+ = x+=+%28-6%29%2F2+ = x+=+%28-3%29+
.
You can check by replacing these numbers into the original quadratic equation
.
(5), +x%5E2+-2x++-+15+=+0+ = +%285%29%5E2+-2%285%29++-+15+=+0+ = +25+-10++-+15+=+0+ = +0+=+0+ (True)
.
(-3), +x%5E2+-2x++-+15+=+0+ = +%28-3%29%5E2+-2%28-3%29++-+15+=+0+ = +9+%2B+6++-+15+=+0+ = +0+=+0+ ( True )
.
We could have used factoring to solve, but this only works sometimes
.
+x%5E2+-2x++-+15+=+0+
.
We could have listed all of the factors of (-15), the two factors would have to equal (-2)
.
We would start like this
.
( x )( x )
.
(1, -15) ( added = (-14))
.
( 5, -3) ( added = 2 )
.
(+highlight%28-5%29+ , +highlight%283%29+) ( added = (-2))
.
The factors would then be
.
( x + 3 )( x - 5) = 0
.
You would then set each factor equal to "0"
.
+%28+x+%2B+3+%29+=+0+ = ++x+%2B+3+=+0+ = ++x+%2B+3+-+3+=+0+-+3+ = +x+=+%28-3%29+
.
+%28+x+-+5%29+=+0+ = +x+-+5+=+0+ = +x+-+5+%2B+5+=+0+%2B+5+ = +x+=+5+
.
These are our two answers/solutions
.
You can check even further if we replaced "x" in our very original equation
.
(5), +%28x%2B7%29%5E2+%2B+%28x%29%5E2+=+%28x%2B8%29%5E2+ = +%28%285%29%2B7%29%5E2+%2B+%28%285%29%29%5E2+=+%28%285%29%2B8%29%5E2+ = +12%5E2+%2B+5%5E2+=+13%5E2+ = +144+%2B+25+=+169+ = +169+=+169+ (True)
.
(-3), +%28%28-3%29%2B7%29%5E2+%2B+%28%28-3%29%29%5E2+=+%28%28-3%29%2B8%29%5E2+ = +4%5E2+%2B+%28-3%29%5E2+=+5%5E2+ =
+16+%2B+9+=+25+ = +25+=+25+ ( True )
.
Don't be fooled with your two answers, since this is a triangle, your answer cannot be negative ( you cannot have negative sides ), so (-3) is not an answer
.
The only answer that works is +5+
.
+x+=+5+
.
Hope I helped, Levi

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
c=x+8
b=x
a=x+7
find x
i tried everything i could..i cudnt find any solutions and it drove me crazy...help would be soo appreciated.


There are infinitely many solutions.

1. Pick any number you want for x to be.
2. Add 8 to what you picked for x and you have c.
3. b is that same number you picked for x.
4. Add 7 to what you picked for x and you have a.

Suppose you pick x=4.
Then add 8 and you have c=12
Then b is the same as x, so b=4
Then add 7 to the 4 you picked for x and so a=11.

So a=11, b=4, c=12, x=4 

Lets check by substituting in

c=x+8
b=x
a=x+7

12=4+8 that checks
4=4 that checks
11=4+7 that checks

------

Suppose you pick x=100.
Then add 8 and you have c=108
Then b is the same as x, so b=100
Then add 7 to the 100 you picked for x and so a=107.

So a=107, b=100, c=108, x=100 

Lets check by substituting in

c=x+8
b=x
a=x+7

108=100+8 that checks
100=100 that checks
107=100+7 that checks

-------

Suppose you pick x=-7.3689
Then add 8 and you have c=.6311
Then b is the same as x, so b=-7.3689
Then add 7 to the -7.3689 you picked for x and so a=-.3689

So a=-.3689, b=-7.3689, c=.6311, x=-7.3689 

Lets check by substituting in

c=x+8
b=x
a=x+7

.6311=-7.3689+8 that checks
-7.3689=-7.3689 that checks
-.3689=-7.3689+7 that checks

There are infinitely many solution sets for a,b,c,x.

There's not just one.

Edwin