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

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

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