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

BenoguigoliB 2021-02-12 Answered

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

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Mitchel Aguirre
Answered 2021-02-13 Author has 25927 answers

\(\displaystyle\frac{{{5}+{x}-{2}}}{{3}}=\frac{{{x}+{3}}}{{8}}\)
\(\displaystyle\frac{{{5}+{x}-{2}}}{{3}}=\frac{{{x}+{3}}}{{8}}\)
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
\(\displaystyle{x}=-{19}\)

Not exactly what you’re looking for?
Ask My Question
28
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-03-07

Solve the correct answer linear equations by considering y as a function of x, that is, \(\displaystyle{y}={y}{\left({x}\right)}.\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{y}= \cos{{x}}\)

asked 2020-10-28

Solve the linear equations by considering y as a function of x, that is, \(\displaystyle{y}={y}{\left({x}\right)}.{x}{y}'+{\left({1}+{x}\right)}{y}={e}^{{-{x}}} \sin{{2}}{x}\)

asked 2021-01-13

Solve the linear equations by considering y as a function of x, that is, y = y(x).
\(\displaystyle{y}'+{y} \tan{{x}}= \sec{{x}},\)
\(\displaystyle{y}{\left(\pi\right)}={1}\)

asked 2020-12-07

Solve the linear equations by considering y as a function of x, that is, \(y = y(x)\).
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+\frac{y}{{x}}=\frac{{ \cos{{x}}}}{{x}},{y}{\left(\frac{\pi}{{2}}\right)}=\frac{4}{\pi},{x}>{0}\)

asked 2021-02-24

Solve the linear equations by considering y as a function of x, that is, \(y = y(x).\)
\(\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{\left(\frac{1}{{x}}\right)}{y}={x}\)

asked 2021-03-11

Solve the linear equations by considering y as a function of x, that is, \(y = y(x)\).
\(y'+2y=4\)

asked 2021-01-27

Cosider the system of differential equations
\(x_1'=2x_1+x_2\)
\(x_2'=4x_1-x_2\)
Convert this system to a second order differential equations and solve this second order differential equations

...