# Contain linear equations with constants in denominators. Solve each equation \displaystyle\frac{{{5}+{x}-{2}}}{{3}}=\frac{{{x}+{3}}}{{8}}

Question
Contain linear equations with constants in denominators. Solve each equation
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\frac{{{\left\lbrace{\left\lbrace{5}\right\rbrace}+{\left\lbrace{x}\right\rbrace}-{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{3}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{8}\right\rbrace}}}}$$

2021-02-13
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\frac{{{\left\lbrace{\left\lbrace{5}\right\rbrace}+{\left\lbrace{x}\right\rbrace}-{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{3}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{8}\right\rbrace}}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\frac{{{\left\lbrace{\left\lbrace{5}\right\rbrace}+{\left\lbrace{x}\right\rbrace}-{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{3}\right\rbrace}}}}={\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{8}\right\rbrace}}}}$$
Multiply both sides by $$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{3}\right\rbrace}\times{\left\lbrace{8}\right\rbrace}={\left\lbrace{24}\right\rbrace}$$
Note that 24 is a product of 3 and 8, Which are the denominators of the fractions in LHS and RHS
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{24}\right\rbrace}\times{\left\lbrace{5}\right\rbrace}+{\left\lbrace{24}\right\rbrace}{\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}-{\left\lbrace{2}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{3}\right\rbrace}}}}={\left\lbrace{24}\right\rbrace}\times{\frac{{{\left\lbrace{\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{8}\right\rbrace}}}}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{120}\right\rbrace}+{\left\lbrace{8}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}-{\left\lbrace{2}\right\rbrace}\right)}\right\rbrace}={\left\lbrace{3}\right\rbrace}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{120}\right\rbrace}+{\left\lbrace{8}\right\rbrace}{\left\lbrace{x}\right\rbrace}-{\left\lbrace{16}\right\rbrace}={\left\lbrace{3}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{9}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{8}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{104}\right\rbrace}={\left\lbrace{3}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{9}\right\rbrace}$$
Subtract 3x from both sides, To get
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{5}\right\rbrace}{\left\lbrace{x}\right\rbrace}+{\left\lbrace{104}\right\rbrace}={\left\lbrace{9}\right\rbrace}$$
Subtract 104 from both sides, To get
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{5}\right\rbrace}{\left\lbrace{x}\right\rbrace}={\left\lbrace{9}\right\rbrace}-{\left\lbrace{104}\right\rbrace}$$
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{b}\right\rbrace}{\left\lbrace{z}\right\rbrace}=-{\left\lbrace{95}\right\rbrace}$$
Divide both sides by 5, To get
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}={\frac{{{\left\lbrace-{\left\lbrace{95}\right\rbrace}\right\rbrace}}}{{{\left\lbrace{5}\right\rbrace}}}}=-{\left\lbrace{19}\right\rbrace}$$
We get the finally result
$$\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le{\left\lbrace{x}\right\rbrace}=-{\left\lbrace{19}\right\rbrace}$$

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