Linear Equation in One Variable
p.1. Number of Roots of Linear Equation in One Variable
A linear equation in one variable x is an equation of the form ax=b, where a and b are real numbers.
a is called a coefficient of the variable, and b is a free term .
When solving a linear equation, the following three cases occur.
a ≠ 0
$b \in \Bbb R$ - любой
One root
a = 0
b = 0
$x \in \Bbb R$ - любой
Infinitely many roots
a = 0
b ≠ 0
$x \in \Bbb \varnothing $
No roots
p.2. Examples of Problem Solving
Example 1. Solve the equation 6-5x = 8(3,5-2x)
Solution:
$ 6-5x = 8(3,5-2x) \iff 6-5x = 28-16x \iff -5x+16x = 28-6 \iff $
$ \iff 11x = 22 \iff x = 2 $
Answer: x=2
Example 2. Solve the equation $\frac{2}{3} x-\frac{4}{5} = 0,6x$
Solution:
$ \frac{2}{3}x-\frac{4}{5} = 0,6x | ×15 \iff 2x∙5-4∙3 = 0,6x∙15 \iff 10x-12=9x \iff $
$ \iff 10x-9x = 12 \iff x = 12 $
Answer: x = 12
Example 3. Solve the equation 8(x+7)-7(2x-3) = 2(5x-11)
Solution:
$ 8(x+7)-7(2x-3) = 2(5x-11) \iff 8x+56-14x+21 = 10x-22 \iff$
$ \iff -6x+77 = 10x-22 \iff -6x-10x = -22-77 \iff -16x=-99 \iff $
$ \iff x = \frac{-99}{-16} = 6\frac{3}{16}$
Answer: x = $6\frac{3}{16}$
Example 4. Find all of the values of the coefficient a such that the root of the equation ax=-6– is an integer.
Solution:
$$ax = -6 \Rightarrow {\left\{ \begin{array}{c} a ≠ 0 \\ x=- \frac{6}{a} \end{array} \right.}$$
x is an integer for a = $\pm$6, $\pm$3, $\pm$2,$\pm$1
Answer: a = $\pm$6, $\pm$3, $\pm$2, $\pm$1
Example 5*. Solve the equation $ ax = a^2 -3a $
Solution:
$$ ax = a^2-3a \iff \left[ \begin{array}{cc} {\left\{ \begin{array}{c} a≠0 \\ x = \frac{(a^2-3a)}{a} = \frac{a(a-3)}{a} = a-3 \end{array} \right.} \\ {\left\{ \begin{array}{c} a = 0 \\ 0x = 0 \end{array} \right.} \end{array} \right. \iff \left[ \begin{array}{cc} {\left\{ \begin{array}{c} a≠0 \\ x = a-3 \end{array} \right.} \\ {\left\{ \begin{array}{c} a = 0 \\ x \in \Bbb R \end{array} \right.} \end{array} \right. $$
Answer: при a ≠ 0,x = a-3; при a = 0, $x \in \Bbb R$ - any
Example 6*. Solve the equation (k+1)x = k
Solution:
$$ (k+1)x = k \iff \left[ \begin{array}{cc} {\left\{ \begin{array}{c} k+1 ≠ 0 \\ x = \frac{k}{k+1} \end{array} \right.} \\ {\left\{ \begin{array}{c} k+1 = 0 \\ 0x = -1 \end{array} \right.} \end{array} \right. \iff \left[ \begin{array}{cc} {\left\{ \begin{array}{c} k ≠ -1 \\ x = \frac{k}{k+1} \end{array} \right.} \\ {\left\{ \begin{array}{c} k = -1 \\ x \in \Bbb \varnothing - no \quad roots \end{array} \right.} \end{array} \right. $$
Answer: при k ≠ -1, $ x = \frac{k}{k+1} $, при k = -1 no roots
Example 7*. Solve the equation ax+b = cx+d
Solution:
$$ ax+b = cx+d \iff ax-cx = d-b \iff (a-c)x = d-b \iff $$
$$ \left[ \begin{array}{cc} {\left\{ \begin{array}{c} a-c ≠ 0 \\ x = \frac{d-b}{a-c} \end{array} \right.} \\ {\left\{ \begin{array}{c} a-c = 0 \\ d-b = 0 \\ 0x = 0 \end{array} \right.} \\ {\left\{ \begin{array}{c} a-c = 0 \\ d-b ≠ 0 \\ 0x ≠ 0 \end{array} \right.} \end{array} \right. \iff \left[ \begin{array}{cc} {\left\{ \begin{array}{c} a ≠ c \\ x = \frac{d-b}{a-c} \end{array} \right.} \\ {\left\{ \begin{array}{c} a = c \\ d = b \\ x \in \Bbb R - any \end{array} \right.} \\ {\left\{ \begin{array}{c} a = c \\ d ≠ b \\ x \in \Bbb \varnothing - no \quad roots \end{array} \right.} \end{array} \right. $$
