# How do you solve -3/11=(5-h)/(h+1.4)?

May 16, 2017

See a solution process below:

#### Explanation:

First, multiply each side of the equation by $\textcolor{red}{11} \textcolor{b l u e}{\left(h + 1.4\right)}$ to eliminate the fractions while keeping the equation balanced. $\textcolor{red}{11} \textcolor{b l u e}{\left(h + 1.4\right)}$ is the Lowest Common Denominator of the two fractions:

$\textcolor{red}{11} \textcolor{b l u e}{\left(h + 1.4\right)} \times - \frac{3}{11} = \textcolor{red}{11} \textcolor{b l u e}{\left(h + 1.4\right)} \times \frac{5 - h}{h + 14.4}$

$\cancel{\textcolor{red}{11}} \textcolor{b l u e}{\left(h + 1.4\right)} \times - \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{11}}}} = \textcolor{red}{11} \cancel{\textcolor{b l u e}{\left(h + 1.4\right)}} \times \frac{5 - h}{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{h + 14.4}}}}$

$\textcolor{b l u e}{\left(h + 1.4\right)} \left(- 3\right) = \textcolor{red}{11} \left(5 - h\right)$

Next, expand the terms in parenthesis on both sides of the equation by multiplying each term within the parenthesis by the term outside the parethesis:

$\left(- 3 \times \textcolor{b l u e}{h}\right) + \left(- 3 \times \textcolor{b l u e}{1.4}\right) = \left(\textcolor{red}{11} \times 5\right) - \left(\textcolor{red}{11} \times h\right)$

$- 3 h - 4.2 = 55 - 11 h$

Then, add $\textcolor{red}{4.2}$ and $\textcolor{b l u e}{11 h}$ to each side of the equation to isolate the $h$ term while keeping the equation balanced:

$- 3 h - 4.2 + \textcolor{red}{4.2} + \textcolor{b l u e}{11 h} = 55 - 11 h + \textcolor{red}{4.2} + \textcolor{b l u e}{11 h}$

$- 3 h + \textcolor{b l u e}{11 h} - 4.2 + \textcolor{red}{4.2} = 55 + \textcolor{red}{4.2} - 11 h + \textcolor{b l u e}{11 h}$

$\left(- 3 + \textcolor{b l u e}{11}\right) h - 0 = 59.2 - 0$

$8 h = 59.2$

Now, divide each side of the equation by $\textcolor{red}{8}$ to solve for $h$ while keeping the equation balanced:

$\frac{8 h}{\textcolor{red}{8}} = \frac{59.2}{\textcolor{red}{8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} h}{\cancel{\textcolor{red}{8}}} = 7.4$

$h = 7.4$

May 16, 2017

color(purple)(h=7.4

#### Explanation:

$- \frac{3}{11} = \frac{5 - h}{h + 1.4}$

$\therefore - \frac{3}{11} = \frac{5 - h}{h + \frac{14}{10}}$

multiply L.H.S. and R.H.S. by $\left(h + \frac{14}{10}\right)$

$\therefore - \frac{3}{11} \left(h + \frac{14}{10}\right) = 5 - h$

$\therefore - \frac{3}{11} h - \left(\frac{14}{10}\right) \left(\frac{3}{11}\right) = 5 - h$

$\therefore - \frac{3}{11} h - \frac{42}{110} = 5 - h$

$\therefore - \frac{3}{11} h + h = 5 + \frac{42}{110}$

$\therefore - \frac{3}{11} h + \frac{11}{11} h = 5 + \frac{42}{110}$

$\therefore \frac{8}{11} h = 5 \frac{42}{110}$

$\therefore \frac{8}{11} h = \frac{592}{110}$

:.$\frac{80 h = 592}{110}$

multiply L.H.S. and R.H.S. by $110$

$\therefore 80 h = 592$

$\therefore h = \frac{592}{80}$

:.color(purple)(h=7.4

check:

substitute color(purple)(h=7.4

$\therefore - \frac{3}{11} = \frac{5 - 7.4}{7.4 + 1.4}$

$\therefore - \frac{3}{11} = - \frac{2.4}{8.8}$

$\therefore - \frac{3}{11} = - 0.272727272$

:.color(purple)(-0.272727272=-0.272727272