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This question already has an answer here:

For example

\documentclass[14pt,twoside]{extreport}
\renewcommand{\baselinestretch}{1.5} 
\usepackage[utf8]{inputenc}
\usepackage{graphicx}
\usepackage[a4paper,width=150mm,top=25mm,bottom=25mm,bindingoffset=6mm]{geometry}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhead{}
\fancyhead{}
\fancyhead[RO,LE]{\small\leftmark }
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{amsmath}
\usepackage[english]{babel}

\begin{document}
First, we compute the following products:
\begin{align*}
R_2R_5&=a_2 a_5 t^2+\left(a_5 b_2+a_2 b_5\right) t+b_2 b_5,\\
R_2 R_3 R_5&=a_2 a_3 a_5 t^3+\left(a_3 a_5 b_2+a_2 a_5 b_3+a_2 a_3 b_5\right) t^2+\left(a_5 b_2 b_3+a_2 b_5 b_3+a_3 b_2 b_5\right) t\\
&+b_2 b_3 b_5,\\
R_4 R_6&=a_4 a_6 t^2+\left(a_6 b_4+a_4 b_6\right) t+b_4 b_6,\\
R_3 R_4 R_6&=a_3 a_4 a_6 t^3+\left(a_4 a_6 b_3+a_3 a_6 b_4+a_3 a_4 b_6\right) t^2+\left(a_6 b_3 b_4+a_3 b_6 b_4+a_4 b_3 b_6\right) t\\
&+b_3 b_4 b_6,
\end{align*}
By replacement, ordering by powers of t and identifying, we obtain a system whose solutions describe the number of straight lines situated on the "reliability" hypersurface. We write the system ordering by the coefficients of the powers of degree from zero to five,relative to t:
\begin{align*}
b_7&=b_2 b_5+ b_2 b_3 b_5 - b_2 b_3 b_6 b_5 + b_3 b_6 b_5 - b_2 b_4 b_6 b_5 + b_2 b_3 b_4 b_6 b_5 - b_3 b_4 b_6 b_5 + b_3 b_4 b_6 + b_4 b_6,
\end{align*}
Starting from the last equation, at least one of the numbers $a_i$; $i = 2,\ldots , 6$ must be zero
(number of cases: $C_1^5 + C_2^5 + C_3^5 + C_4^5 = 30)$. So the straight-lines are parallel to some hyperplane of coordinates. The first equation shows that at t = 0, point $(b_2,\ldots, b_7)$ is on the "reliability" hypersurface. This remark requires the following procedure: we choose arbitrarily $b_2, \ldots ,b_6$, and compute $b_7$. We replace the values $(b_2,\ldots, b_7)$ in the remaining equations. If the new system, in unknown $(a_2,\ldots, a_7)$ , it has a solution with at least a nonzero component, then there exists one straight line passing through the point $(b_2,\ldots, b_7)$ 
and lying on the "reliability" hypersurface. Explicitly, after solving the algebraic system, we have the following cases:




\end{document}

The result is this enter image description here

How can I adjust the equation within the limits of the text?

thanks for the help

marked as duplicate by CarLaTeX, user36296, TeXnician, Zarko, Moriambar Jun 7 '17 at 18:00

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    I think the only way to get automatic breaking of equations is the breqn package (e.g. tex.stackexchange.com/questions/27365). You'll need to use the environments provided by that package though. – Torbjørn T. Jun 7 '17 at 16:14
  • @TorbjørnT. Can you do an example? – Emad kareem Jun 7 '17 at 16:21
  • Breaking equations is something TeX doesn't do automatically. You have to do it manually. Do you need an example of how to do it? – Michael Fraiman Jun 7 '17 at 16:27
  • @MichaelFraiman Can you answer the example above – Emad kareem Jun 7 '17 at 16:36
  • @Emadkareem do you mean if I can provide an example using breqn? – Michael Fraiman Jun 7 '17 at 16:37
0

It is much better to break equations by hand in general. Replacing amsmath with its extensionmathtools, here a way to nicely break the equations with themultlined` environment:

\documentclass[14pt,twoside]{extreport}
\renewcommand{\baselinestretch}{1.5}
\usepackage[utf8]{inputenc}
\usepackage{graphicx}
\usepackage[a4paper,width=150mm,top=25mm,bottom=25mm,bindingoffset=6mm, showframe]{geometry}
\usepackage{fancyhdr}
\pagestyle{fancy}
\fancyhead{}
\fancyhead{}
\fancyhead[RO,LE]{\small\leftmark }
\usepackage{amsfonts}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{mathtools}
\usepackage[english]{babel}

\begin{document}
First, we compute the following products:
\begin{align*}
R_2R_5&=a_2 a_5 t^2+\left(a_5 b_2+a_2 b_5\right) t+b_2 b_5,\\
R_2 R_3 R_5&=\begin{multlined}[t] a_2 a_3 a_5 t^3+\left(a_3 a_5 b_2+a_2 a_5 b_3+a_2 a_3 b_5\right) t^2 \\[-1ex]
+\left(a_5 b_2 b_3+a_2 b_5 b_3+a_3 b_2 b_5\right) + b_2 b_3 b_5,t \end{multlined} \\
R_4 R_6&=a_4 a_6 t^2+\left(a_6 b_4+a_4 b_6\right) t + b_4 b_6,\\
R_3 R_4 R_6&=\begin{multlined}[t] a_3 a_4 a_6 t^3+\left(a_4 a_6 b_3+a_3 a_6 b_4+a_3 a_4 b_6\right) t^2 \\[-1ex]
+\left(a_6 b_3 b_4+a_3 b_6 b_4+a_4 b_3 b_6\right) t + b_3 b_4 b_6. \end{multlined}
\end{align*}
By replacement, ordering by powers of $ t $ and identifying, we obtain a system whose solutions describe the number of straight lines situated on the "reliability" hypersurface. We write the system ordering by the coefficients of the powers of degree from zero to five,relative to $ t $:
\begin{multline*}
b_7=b_2 b_5+ b_2 b_3 b_5 - b_2 b_3 b_6 b_5 + b_3 b_6 b_5 - b_2 b_4 b_6 b_5 \\[-1ex]
+ b_2 b_3 b_4 b_6 b_5 - b_3 b_4 b_6 b_5 + b_3 b_4 b_6 + b_4 b_6.
\end{multline*}
Starting from the last equation, at least one of the numbers $a_i$; $i = 2,\ldots , 6$ must be zero
(number of cases: $C_1^5 + C_2^5 + C_3^5 + C_4^5 = 30)$. So the straight-lines are parallel to some hyperplane of coordinates. The first equation shows that at $ t = 0 $, point $(b_2,\ldots, b_7)$ is on the "reliability" hypersurface. This remark requires the following procedure: we choose arbitrarily $b_2, \ldots ,b_6$, and compute $b_7$. We replace the values $(b_2,\ldots, b_7)$ in the remaining equations. If the new system, in unknown $(a_2,\ldots, a_7)$ , it has a solution with at least a nonzero component, then there exists one straight line passing through the point $(b_2,\ldots, b_7)$ and lying on the "reliability" hypersurface. Explicitly, after solving the algebraic system, we have the following cases:

\end{document} 

enter image description here

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