Proving System of Equations for Algebra Challenge

In summary, the system of equations given is used to prove that the expressions $\dfrac{a^2}{1-x^2}$, $\dfrac{b^2}{1-y^2}$, and $\dfrac{c^2}{1-z^2}$ are all equal to each other. The approach of using what needs to be proved in the argument is recommended. A solution by another individual is also shared.
  • #1
anemone
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Given the system of equation below:

$a = zb + yc$

$b = xc + za$

$c = you + xb$

Prove that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}$.
 
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  • #2
anemone said:
Given the system of equation below:

$a = zb + yc---(1)$

$b = xc + za---(2)$

$c = you + xb---(3)$
Prove that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}---(4)$.
if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same
 
  • #3
Albert said:
if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same

$(4) => (5),(6),(7)$
but that does not mean that
$(5),(6),(7) =>(4)$
 
  • #4
kaliprasad said:
$(4) => (5),(6),(7)$
but that does not mean that
$(5),(6),(7) =>(4)$
if $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
then $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
and if $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
then $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
of course all the denominators $\neq 0$
and more I got (5)(6)(7) from(1)(2)(3) not from (4)
 
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  • #5
Albert said:
if $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
then $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
and if $\dfrac {a_1+a_2}{b_1+b_2}=\dfrac {a_1-a_2}{b_1-b_2}=------=k$
then $\dfrac {a_1}{b_1}=\dfrac {a_2}{b_2}=\dfrac {a_3}{b_3}=k$
of course all the denominators $\neq 0$
and more I got (5)(6)(7) from(1)(2)(3) not from (4)

I understand you proved (5),(6),(7) but that does not mean that (4) is true,
 
  • #6
Albert said:
if (4) is true , then (5)(6)(7) also hold:
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}---(5)$
$\dfrac {b^2-c^2}{z^2-y^2}=\dfrac{a^2}{1-x^2}---(6)$
$\dfrac {a^2-c^2}{z^2-x^2}=\dfrac{b^2}{1-y^2}---(7)$
proof of (5)
(1)+(2) :$a+b=\dfrac {c(x+y)}{1-z}---(8)$
(1)-(2) :$a-b=\dfrac {-c(x-y)}{1+z}---(9)$
$(8)\times (9) :a^2-b^2=\dfrac {-c^2(x^2-y^2)}{1-z^2}$
and (5) holds
the proof of (6) and (7) are the same
from (5)
proportion by addition and substraction
$\dfrac {a^2-b^2}{y^2-x^2}=\dfrac{c^2}{1-z^2}=\dfrac {a^2}{p-x^2}=\dfrac{b^2}{p-y^2}=\dfrac{c^2}{1-z^2}$
compare(6) and (7) it is easy to see p=1
 
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  • #7
Hi Albert,

Thanks for participating but I don't think your proof is complete.:( It's easier to use what we're asked to prove in our approach and then prove everything holds but more often than not, it's easy to lose sight of what we need to prove as we might have lost ourselves in our argument.

Here is the brilliant and neat solution of other that I wanted to share with the folks of MHB:

Express $x,\,y,\,z$ in terms of $a,\,b,\,c$, we see that we've:

$a = zb + yc \implies z=\dfrac{a-yc}{b}$

$b = xc + za\implies z=\dfrac{b-xc}{a}$

So equating the two we get:

$\dfrac{a-yc}{b}=\dfrac{b-xc}{a}$---(1)

But from $c = you + xb$, we can make $y$ the subject and then replace it into (1):

$y=\dfrac{c-xb}{a}$

$\therefore \dfrac{a-\left(\dfrac{c-xb}{a}\right)c}{b}=\dfrac{b-xc}{a}$

$x=\dfrac{b^2+c^2-a^2}{2bc}$

By the same token, we also get:

$y=\dfrac{a^2+c^2-b^2}{2ac}$; $z=\dfrac{a^2+b^2-c^2}{2ab}$

Now, we can find, separately, the expressions for $\dfrac{a^2}{1-x^2},\,\dfrac{b^2}{1-y^2},\,\dfrac{c^2}{1-z^2}$ all in terms of $a,\,b,\,c$:

$\begin{align*}\dfrac{a^2}{1-x^2}&=\dfrac{a^2}{1-\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2}\\&=\dfrac{4a^2b^2c^2}{(2bc)^2-(b^2+c^2-a^2)^2}\\&=\dfrac{4a^2b^2c^2}{(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)}\\&=\dfrac{4a^2b^2c^2}{(a^2-(b-c)^2)((b+c)^2-a^2)}\\&=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}\end{align*}$

Similarly,

$\dfrac{b^2}{1-y^2}=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}$

$\dfrac{c^2}{1-z^2}=\dfrac{4a^2b^2c^2}{(a-b+c)(a+b-c)(b+c-a)(b+c+a)}$

Therefore we can conclude by now that $\dfrac{a^2}{1-x^2}=\dfrac{b^2}{1-y^2}=\dfrac{c^2}{1-z^2}$.
 
  • #8
Hi anemone,

Here is my solution:

Since $a = zb + yc$ and $b = xc + za$, then $a = z(xc + za) + yc = (zx + y)c + z^2 a$, and thus

$$(1 - z^2)a = (zx + y)c.\tag{1}$$

As $c = you + xb$, substituting the expression $xc + za$ for $b$ yields $c = (y + xz)a + x^2 c$, or

$$(1 - x^2)c = (y + xz)a.\tag{2}$$

Using $(1)$ and $(2)$, the following proportion holds:

$$\frac{(1 - z^2)a}{(1 - x^2)c} = \frac{c}{a}.$$

This is equivalent to the proportion

$$\frac{a^2}{1 - x^2} = \frac{c^2}{1 - z^2}.\tag{3}$$

Substituting $zb + yc$ for $a$ in the equation $b = xc + za$ results in $(1 - z^2)b = (x + zy)c$, and using the same substitution in the equation $c = you + xb$ yields $(1 - y^2)c = (x + yz)b$. So the following proportion holds:

$$\frac{(1 - z^2)b}{(1 - y^2)c} = \frac{c}{b}.$$

Equivalently,

$$\frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.\tag{4}$$

Equations $(3)$ and $(4)$ together yield

$$\frac{a^2}{1 - x^2} = \frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.$$
 
  • #9
Euge said:
Hi anemone,

Here is my solution:

Since $a = zb + yc$ and $b = xc + za$, then $a = z(xc + za) + yc = (zx + y)c + z^2 a$, and thus

$$(1 - z^2)a = (zx + y)c.\tag{1}$$

As $c = you + xb$, substituting the expression $xc + za$ for $b$ yields $c = (y + xz)a + x^2 c$, or

$$(1 - x^2)c = (y + xz)a.\tag{2}$$

Using $(1)$ and $(2)$, the following proportion holds:

$$\frac{(1 - z^2)a}{(1 - x^2)c} = \frac{c}{a}.$$

This is equivalent to the proportion

$$\frac{a^2}{1 - x^2} = \frac{c^2}{1 - z^2}.\tag{3}$$

Substituting $zb + yc$ for $a$ in the equation $b = xc + za$ results in $(1 - z^2)b = (x + zy)c$, and using the same substitution in the equation $c = you + xb$ yields $(1 - y^2)c = (x + yz)b$. So the following proportion holds:

$$\frac{(1 - z^2)b}{(1 - y^2)c} = \frac{c}{b}.$$

Equivalently,

$$\frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.\tag{4}$$

Equations $(3)$ and $(4)$ together yield

$$\frac{a^2}{1 - x^2} = \frac{b^2}{1 - y^2} = \frac{c^2}{1 - z^2}.$$

Awesome, Euge! And thanks for participating!:cool:
 

FAQ: Proving System of Equations for Algebra Challenge

1. How do I prove a system of equations in algebra?

To prove a system of equations in algebra, you must show that all of the equations in the system are true when you substitute the given values into them. This can be done through various methods, such as substitution, elimination, or graphing.

2. What is the purpose of proving a system of equations in algebra?

The purpose of proving a system of equations is to confirm that the given equations are consistent and that all of the equations are satisfied by the same set of values. This helps to determine if a solution exists for the system or if it is inconsistent.

3. Can a system of equations have more than one solution?

Yes, a system of equations can have more than one solution. This occurs when the equations in the system intersect at multiple points, resulting in different sets of values that satisfy all of the equations.

4. What happens if a system of equations has no solution?

If a system of equations has no solution, it means that there is no set of values that satisfies all of the equations in the system. This could be due to the equations being inconsistent or contradictory.

5. How do I know if my proof for a system of equations is correct?

To ensure the accuracy of your proof, you should double-check your substitutions and calculations to ensure that they are correct. You can also plug in your solutions to the original equations and confirm that they satisfy all of the equations in the system.

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