Question

My friend Challenged me to solve this.

Help! I don't wanna lose!​

Answer 1

Answer:

x = 1 ,1

Explain:-

x = 1/2-1/2-1/2-x

x = 1/2-x (x = 1/2-1/2-1/2-x)

x (2-x) = 1

2x - x^2 = 1

2x -x^2-1 = 0

-x^2 +2x -1=0

x^2 -2x +1=0

x^2 - x - x +1=0

x(x-1) - 1(x-1)=0

(x-1) (x-1)=0

either or

x - 1 = 0 x-1=0

x=0 x=1

Hope it will help you.

Answer 2

${\red{\maltese\;}} {\underline{\underline{\textbf{\textsf{Given that :- }}}}}$

$\bigstar \; {\underline{\boxed{\tt {x = \dfrac{1}{2- \dfrac{1}{2-\dfrac{1}{\;2-x} } } \;\; ,x\neq 2}}}}$

${\red{\maltese\;}} {\underline{\underline{\textbf{\textsf{To Find :- }}}}}$

• The value of the variable x in the equation

${\red{\maltese\;}} {\underline{\underline{\textbf{\textsf{Required Solution :- }}}}}$

• The solution to the equation is 1

${\red{\maltese\;}} {\underline{\underline{\textbf{\textsf{Full Solution :- }}}}}$

• Here we have been provided with a complex equation with multiple mixed fractions in the denominator of a fraction , now in order to solve it let's simplify it from thee bottom

ㅤㅤNow it is been observed that the bottom part of the fraction is in the form $\tt a - \dfrac{b}{c}$ so , let's split it into the form $\tt a \times c - b$ simplifying it

$\rightarrow \tt \qquad x = \dfrac{1}{2-\dfrac{1}{2-\dfrac{1}{\; 2 - x} } } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{2-\dfrac{1}{\;\;2(2 -x ) -1 } } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{2-\dfrac{1}{\; \; 4 - 2x -1} } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{2-\dfrac{1}{\; \; 3 - 2x} }$

→ Now let's use the same concept and repeat the procedure in the rest of the solution

$\rightarrow \tt \qquad x = \dfrac{1}{2-\dfrac{1}{\; \; 3 - 2x} } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{\;\;2(3- 2x)-1 } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{\;\;6 - 4x -1 } \\ \\ \\ \rightarrow \tt \qquad x = \dfrac{1}{\;-4x + 5}$

→  Now let's transpose the term in denominator of the fraction to the other hand side of the equation

$\longrightarrow \tt x( - 4x + 5 ) = 1$

$\longrightarrow \tt - 4x^2 + 5x - 1 = 0$

→ As now we have framed a quadratic equation of the form ax² + bx + c

so, let's use middle term method to find out the zeros of the equation to find the solution of the equation

$\longrightarrow \tt - 4x^2 + 5x - 1 = 0$

$\longrightarrow \tt - 4x^2 + 4x + 1x- 1 = 0$

$\longrightarrow \tt 4x ( - x + 1 ) - 1( - x + 1 )= 0$

$\longrightarrow \tt (4x- 1) ( - x + 1 ) = 0$

→ Now let's find out the zeros of the equation

$\longrightarrow \tt 4x - 1 = 0$

$\longrightarrow \tt 4x = 1$

$\longrightarrow \tt x = 1/4$

Or that,

$\longrightarrow \tt - x + 1 = 0$

$\longrightarrow \tt -x = - 1$

$\longrightarrow \tt x = 1$

→ Now as we have found the zeros and answer to the question will either 1/4 or 1 , In order to find the answer we'll have to substitute the value of  1/4 and 1 and see whether which number would satisfy the condition

ㅤㅤSince the term 1/4 is a fraction and a bit complicated to simplify so, let's first put the value of 1 in the equation and see whether it is thee rigth answer or not, if it isn't thee correct one then 1/4 would be the solution

$\rightarrow \qquad \tt x = \dfrac{1}{2- \dfrac{1}{2-\dfrac{1}{\;2-x} } }$

$\rightarrow \qquad \tt 1 = \dfrac{1}{2- \dfrac{1}{2-\dfrac{1}{\;2-1} } }$

$\rightarrow \qquad \tt 1 = \dfrac{1}{2- \dfrac{1}{2-\dfrac{1}{\;1} } }$

$\rightarrow \qquad \tt 1 = \dfrac{1}{2- \dfrac{1}{1} }$

$\rightarrow \qquad \tt 1 = \dfrac{1}{\dfrac{1}{1} }$

$\rightarrow \qquad {\pink{\boxed{\frak{ 1 = \frac{1}{1} }}}\purple\bigstar}$

• Hence verified..!!!

${\red{\maltese\;}} {\underline{\underline{\textbf{\textsf{Therefore :- }}}}}$

• The value of the variable x is 1 which is the solution to the equation

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