The numbers from $1$ to $8$ with a variable $x$ is formed a fraction in algebraic form by the combination of both reciprocal and addition. It is given that the value of the algebraic expression is equal to $9$ by $10$.
$1+ \dfrac{2}{3+\dfrac{4}{5+\dfrac{6}{7+\dfrac{8}{x}}}}$ $\,=\,$ $\dfrac{9}{10}$
The value of variable $x$ should be calculated by solving the above algebraic equation in this math problem.
It is not an appropriate way to find the sum of the fractions from top to bottom on the left hand side of the equation. However, it can be evaluated from the bottom to top. Look at the bottom of the algebraic expression, two fractions $7$ and $8$ by $x$ are added. So, let’s calculate the sum of them to minimize the complexity of the expression as per the addition rule of the fractions.
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+\dfrac{6}{\dfrac{7\times x+8}{x}}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+\dfrac{6}{\dfrac{7x+8}{x}}}}$ $\,=\,$ $\dfrac{9}{10}$
Observe the bottom of the mathematical expression. Two fractions $5$ and $6$ by quotient of $7x+8$ by $x$ are connected by a plus symbol but a fraction $7x+8$ by $x$ is in reciprocal form in the second term and it creates a problem for adding it to the number $5$.
It is better to convert the reciprocal of a fraction into a fraction form purely. So, let’s split the fraction as a product of two fractions firstly as per the multiplication rule of the fractions.
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+\dfrac{6 \times 1}{\dfrac{7x+8}{x}}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+6 \times \dfrac{1}{\dfrac{7x+8}{x}}}}$ $\,=\,$ $\dfrac{9}{10}$
According to the reciprocal rule of a fraction, the reciprocal of $7x+8$ by $x$ is equal to the $x$ by $7x+8$.
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+6 \times \dfrac{x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+ \dfrac{2}{3+\dfrac{4}{5+\dfrac{6 \times x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{5+\dfrac{6x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
Now, use the same procedure for adding the algebraic fractions mathematically.
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{\dfrac{5 \times (7x+8)+6x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
Use the distributive property for distributing the number $5$ over the addition basis binomial $7x+8$.
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{\dfrac{5 \times 7x+5 \times 8+6x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{\dfrac{35x+40+6x}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
It’s time to simplify the algebraic expression at the bottom on the left-hand side of the equation for finding the sum of the terms.
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{\dfrac{35x+6x+40}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{4}{\dfrac{41x+40}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
Just like in the above steps, repeat the concepts of both reciprocal property and sum rule of the fractions for simplifying the algebraic expression on the left-hand side of the equation.
$\implies$ $1+\dfrac{2}{3+\dfrac{4 \times 1}{\dfrac{41x+40}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+4 \times \dfrac{1}{\dfrac{41x+40}{7x+8}}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+4 \times \dfrac{7x+8}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{4 \times (7x+8)}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{4 \times 7x+4 \times 8}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{3+\dfrac{28x+32}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{\dfrac{3 \times (41x+40)+28x+32}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{\dfrac{3 \times 41x+3 \times 40+28x+32}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{\dfrac{123x+120+28x+32}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{\dfrac{123x+28x+120+32}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2}{\dfrac{151x+152}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2 \times 1}{\dfrac{151x+152}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+2 \times \dfrac{1}{\dfrac{151x+152}{41x+40}}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+2 \times \dfrac{41x+40}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2 \times (41x+40)}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{2 \times 41x+2 \times 40}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $1+\dfrac{82x+80}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $\dfrac{1\times (151x+152)+82x+80}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $\dfrac{1\times 151x+1\times 152+82x+80}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $\dfrac{151x+152+82x+80}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $\dfrac{151x+82x+152+80}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
$\implies$ $\dfrac{233x+232}{151x+152}$ $\,=\,$ $\dfrac{9}{10}$
Now, transform the equation in fractions into an algebraic equation by the cross-multiplication system.
$\implies$ $10 \times (233x+232)$ $\,=\,$ $9 \times (151x+152)$
According to the distributive property of multiplication over the addition, multiply the terms in both sides of the binomials by their corresponding numeral coefficients.
$\implies$ $10 \times 233x$ $+$ $10 \times 232$ $\,=\,$ $9 \times 151x$ $+$ $9 \times 152$
$\implies$ $2330x$ $+$ $2320$ $\,=\,$ $1359x$ $+$ $1368$
It is a linear equation in one variable and it is time to solve it for finding the solution of the given fraction form basis algebraic equation.
$\implies$ $2330x$ $-$ $1359x$ $\,=\,$ $1368$ $-$ $2320$
$\implies$ $971x$ $\,=\,$ $-$ $952$
$\implies$ $971 \times x$ $\,=\,$ $-$ $952$
$\,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $-$ $\dfrac{952}{971}$
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