Equations with One Variable

From the Introduction you know the concept of an equation.

Now we turn to the question of how to solve equations. For this purpose we first study the general rules for transforming and solving equations, together with some examples. In a second part more specific techniques for two of the most important cases - linear and quadratic equations - are presented.

In the chapter Equations - Additional Aspects we will study some additional categories of equations such as

• Equations with fractional expressions
• Radical equations
• Exponential equations
  

 

Concepts

Examples

$\begin{align} & x\,\,=\,\,2 \\ & y-3\,\,=\,\,5 \\ & 10\,\,=\,\,{{x}^{2}}-3x \\ \end{align}$

Examples

$\begin{align} & x\,\,=\,\,2 \\ & y-3\,\,=\,\,5 \\ & 10\,\,=\,\,{{x}^{2}}-3x \\ \end{align} $	The domain G of all of these three equations ist the set of real numbers: $G=\mathbb{R}$
$\sqrt{x}\,\,=\,\,2$	The domain of this equation is the set of all positive real numbers: $G={{\mathbb{R}}^{+}}$ (the square root of negative numbers is undefined)
$\large \frac{2}{x}\,\normalsize =\,\,1$	Zero doesn't belong to the domain G, because division by zero is not possible: $G=\mathbb{R}\backslash \left\{ 0 \right\}$

 

Examples

$x\,\,=\,\,2$	$L\,\,=\,\,\left\{ 2 \right\}$
$ y-3\,\,=\,\,5$	$L\,\,=\,\,\left\{ 8 \right\}$
$10\,\,=\,\,{{x}^{2}}-3x$	$L\,\,=\,\,\left\{ 5;-2 \right\}$

 

Equivalence of Equations

The general approach to finding the solution to equations is to change the equation into simpler equations. Of course, these manipulations must preserve the solution set of the equation; if this is the case, we say that the two equations are equivalent.

The two most important types of manipulations with this equivalence preserving property are presented first: there is an additive and a multiplicative version of it, and we call them E1 and E2, respectively.

 

Example 1

$\begin{align} & y-3\,\,=\,\,5 \\ & y-3+3\,\,=\,\,5+3 \\ & y\,\,=\,\,8 \end{align}$

We write:

$\ y-3\,\,=\,\,5 \quad \left| \,\,+3 \right.$

y can be isolated by adding 3 to both sides of the equation, thereby getting the most simple form of the equation:

$y = 8$.

The solution set consist of the (unique) element 8:

$L\,\,=\,\,\left\{ 8 \right\}$

Replacing x by 8 in the original equation turns it into a true statement: $8-3=5$

Furthermore, 8 is the only solution of the original equation. So "Adding 3 to both sides of the equation" is an operation that preserves equivalence.

Example 2

$ 2x-10\,\,=\,\,x+2 \quad \left| \,\,-x \right. $

Subtracting x from both sides of the equation has the effect that x vanishes on the right side of the equation:

$ x – 10 = 2 \quad \left| \,\,+10 \right. $

Adding 10 then isolates x :

$x = 12$.

 

Example 3

$\begin{align} &0.5x\,\,=\,\,2 \quad \,\,\,\,\left| \,\,\cdot \,\,2 \right. \quad \quad G\,\,=\,\,\mathbb{R} \\ &x\,\,=\,\,4 \end{align}$

Example 4

$\begin{align} &4x\,\,=\,\,8 \quad \quad \left| \,\,\div \,\,4 \right. \quad \quad G\,\,=\,\,\mathbb{R} \\ &\,x\,\,=\,\,2 \end{align}$

The next example shows an equation with fractional expressions: the variable appears in the denominator of the fraction. Therefore, when determining the domain of the equation we have to avoid division by zero.

Example 5

$ \large \frac{2}{x} \small \,\,=\,\,1 \quad \quad \left| \,\,\cdot x \right. \quad \quad G\,\,=\,\,\mathbb{R}\backslash \left\{ 0 \right\}$    (all reals without zero)
$\,x\,\,=\,\,2$

Note that multiplication by x can be considered as an equivalence preserving manipulation only due the fact, that 0 is not part of the domain, so we are sure not to multiply by zero.

We will come back to solution methods for fractional equations in the chapter Equations - Additional Aspects.

 

Linear Equations

All examples discussed so far were linear equations (as 1 to 4) or it was easily possible to rearrange the equation to a linear one (as in example 5).

The solution method for linear equations (as shown in the examples above) can be described as follows: by use of equivalence preserving manipulations we try to isolate the variable on one side of the equals sign.

The following example 6 shows an equation that looks more complicated than the previous ones; besides the manipulations E1 and E2 that we have introduced above (and that have to be applied on both sides of the equation) we will additionally make use of rearrangements of terms, as described in the earlier paragraphs of this tutorial. Note that you are allowed to use these on each side separatly, if useful.

Example 6

$- x + 1 + {(3x - 2)^2} = 2x(5x - 6) - {(x + 4)^2}\quad \quad G\,=\,\mathbb{R} $

In a first step, we simplify the two sides of the equation separately; you can find more information about the used techniques in the chapters Addition, Subtraction, Brackets, Multiplication and Expanding and Binomial Formulas.

$\begin{eqnarray} - x + 1 + (9{x^2} - 12x + 4) &=& 10{x^2} - 12x - ({x^2} + 8x + 16)\\ - x + 1 + 9{x^2} - 12x + 4 &=& 10{x^2} - 12x - {x^2} - 8x - 16\\9{x^2} - 13x + 5 &=& 9{x^2} - 20x - 16 \end{eqnarray}$

Now equivalences E1 and E2 are applied:

$ \begin{eqnarray} 9{x^2} - 13x + 5 &=& 9{x^2} - 20x - 16 \quad \quad &\left| \,\,-9{x^2} \right. \\ - 13x + 5 &=& - 20x - 16 \quad \quad &\left| \,\,+20x-5 \right. \\ 7x&=&-21 \\ x&=&-3 \end{eqnarray}$

 

Quadratic Equations

In the following example the method presented above fails. The equation is no longer linear but contains quadratic terms.

Example 7

$ {x^2} -6x=40 \quad \quad G\,=\,\mathbb{R} $

We cannot solve directly for x even though the variable is isolated on the left side of the equation.

But by manipulation E1 we can complete the left side to be equivalent to the square of a binomial term (see Binomial Formulas):

$\begin{eqnarray} {x^2} -6x &=& 40 &\left| \;\; \color{red}{+9} \right.\\ {x^2} - 6x + \color{red}{9} &=& 40 + \color{red}{9}\\{(x-3)^2} &=& 49 \end{eqnarray}$

You can simplify this equation by taking the square root of both sides:

$x-3=7$

Note, however, that this operation changes the solution set of the equation, i.e. it is not equivalent: if the square of an expression equals 49, this expression is not neccessarily equal to +7, but there is a second possibility -7, since $(-7)^2$ equals 49 as well!
So we have to deal with both:

$x-3=+7 \quad\rightarrow \quad {x_1}=10$

$x-3=-7 \quad \rightarrow \quad {x_2}=-4$

The solution set of the equation consists of two values: $L\,=\,\left\{ 10;\,-4 \right\}$

The method we used in example 7 is known as "completing the square"; it can be applied as an approach to find the solutions of a general quadratic equation of the form

$a{x^2}+bx+c=0$

 

Quadratic Formula:

\(\qquad {x_{1,2}} =\large \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

For details concerning the proof of the formula see here.

In example 7:

$ \begin{align}{x^2} -6x=40& \quad \left| \;\; -40\right.\\ \color{blue} 1 {x^2} -6x-40=0& \end{align} $

$a=\color{blue}1 \color{black} \quad b=-6 \quad c=-40 \,$:

\({x_{1,2}} =\large \frac{{ - ( - 6) \pm \sqrt {{{( - 6)}^2} - 4 \cdot 1 \cdot ( - 40)} }}{{2 \cdot 1}} \normalsize = \large \frac{{6 \pm \sqrt {196}}}{2} = \frac{{6 \pm 16}}{2}\)

\( x_1=\large \frac{{6 + 16}}{2} \normalsize =10 \)

\( x_2=\large \frac{{6 - 16}}{2} \normalsize =-4 \)

Example 8

$ {x^2} -x-1=0 $    (Equation of the "Golden Ratio")

$a=1 \quad b=-1 \quad c=-1 \,$:

\({x_{1,2}} =\large \frac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \cdot 1 \cdot ( - 1)} }}{{2 \cdot 1}} \normalsize = \large \frac{{1 \pm \sqrt {5}}}{2} \)

\( x_1=\large \frac{{1 + \sqrt{5}}}{2} \small =1.618 \)

\( x_2=\large \frac{{1 - \sqrt{5}}}{2} \small =-0.618 \)

As you see a quadratic equation can have two solutions. Yet there are different cases: the quadratic formula contains a square root that is defined only for non-negative reals. Therefore it is obvious that the expression under the square root sign plays an important role for the number of solutions; its name is derived from the latin word "discriminare" (to decide):

The number of solutions (in the field of real numbers) depends directly on the sign of the discriminant D:

$D>0$	2 solutions	\({x_{1}} = \frac{{ - b + \sqrt {D} }}{{2a}} \; , \; {x_{2}} = \frac{{ - b - \sqrt {D} }}{{2a}}\)	Expl.:  \(2{x^2}+x-1=0 \; \) with solutions\({x_1}= 0.5 \; ,\;{x_2}= -1 \)
$D=0$	1 solution	\( x =\frac{- b}{2a} \)	Expl.:  \({x^2}-2x+1=0 \; \) with the only solution \( x=1 \)
$D<0$	no solution	-	Expl.:  \({x^2}-x+2=0 \;\) doesn't have any solutions

Remark:

Instead of using the quadratic formula you can try to factorise the quadratic expression, i.e. to find a product of bracket expressions \( \,(x-{x_1})(x-{x_2}) \,\) that is equivalent to the given term. However this alternative seems applicable only if the equation has simple (i.e. integer) solutions.
For more detail see Factorization.

 

Equations of Higher Degree

Equations containing powers of three or more are difficult (or impossible) to solve by formal methods: in most cases you will need numerical methods to find their solutions ("Solver" of your pocket calculator or computer).

In special cases you may succed in reducing the complexity by factorizing the equation (see Factorization) and solving the resulting fragments:

Example 9

$ {x^3} = 2{x^2} + 3x \quad \quad G\,=\,\mathbb{R} $

After collecting all terms with the variable on the left side (using E1) you can place x outside the brackets:

$ \quad {x^3}=2{x^2} +3x \quad \quad \left| \,\,-2{x^2} -3x \right.$

$ \begin{eqnarray} {x^3} - 2{x^2} - 3x=0 \\x \cdot ({x^2} - 2x - 3) =0\end{eqnarray}$

To the left of the equals sign we have a product.

If a product is equal to 0 then (at least) one of the two factors must be equal to 0 !

Therefore either: $x=0$  or   ${x^2} - 2x - 3=0$

Alternative 1 provides the first solution:   ${x_1}=0$

Alternative 2 provides, using the quadratic formula for the equation ${x^2}-2x-3=0$, two more solutions of the original equation:

\({x_2} = 3 \quad {x_3} = -1 \)

Thus, the solution set of the original equation $\,{x^3} = 2{x^2} + 3x \,$ is  $L\,=\,\left\{0;3;-1 \right\}$

Please note that the alternative idea of simplifying the original equation by dividing it by x is problematic: a division by zero doesn't lead to an equivalent equation because x=0 is part of the domain and therefore we risk a division by zero. Indeed, with this approach we would loose the solution $\,{x_1}=0\,$ .

Exercises - Linear Equations

1)

Solve the following linear equations (using E1/E2):

a)

\( \begin{equation} 3(x - 1) - [7 - 2\,(2x - 3)] = - x \end{equation}\)

b)

\( \begin{equation}{\frac{{4x}}{5}}+ 3 = 5x - 6\,(1 + {\frac{x}{2}} )\end{equation}\)

c)

\(\begin{equation}{(x - 5)^2} = {(2x + 3)^2} - (x + 7)(3x - 1) + 15\end{equation}\)

d)

\(\begin{equation}\frac{{3(t + 2)}}{4} - \frac{{t - 5}}{6} = 2 + \frac{{t + 1}}{3}\end{equation}\)

e)

\(\begin{equation}{(y+2)^2}- 2\,[7y - 5\,(3y +1)]= {y^2}-11 \end{equation}\)

Solution

 

Exercises - Quadratic Equations

2)

Solve the following quadratic equations by putting the quadratic formula to work (if needed, use E1/E2 to bring the equation to the proper form):

a)

\( \begin{equation} 2{x^2}-x-6=0 \end{equation}\)

b)

\( \begin{equation}(x + 2)(2x + 3) = 4{x^2}\end{equation}\)

c)

\(\begin{equation}(2x +1)^2 = x \end{equation}\)

d)

\(\begin{equation}4y(y - 1) = 7 \end{equation}\)

e)

\(\begin{equation}{z^3}=(z+3)(z-4)(z+6) \end{equation}\)

Solution

More explanations under the following links:

Video: Solving Linear Equations (Example 1, simple)

Video: Solving Linear Equations (Example 2, more complicated)

Video: Solving Quadratic Equations (Factoring and Quadratic Formula)

Study Guides: Linear Equations (Introduction with solved examples)

www.mathcentre.ac.uk  (linear equations; with solved examples and additional exercises)

www.mathcentre.ac.uk  (applying the quadratic formula; with solved examples and additional exercises)

 

Additional exercises with linear and quadratic equations you find here:

math.com: Linear Equations  (interactive exercises)

vitutor.com: Linear Equations  (15 exercises with solutions)

vitutor.com: Quadratic Equations  (16 exercises with solutions)

shmoop.com: Solving Equations  (exercises, different topics)