Linear equations
Method: solving equations
The general steps to solve equations are:
Expand (Remove) all brackets.
"Move" all terms with the variable to the left hand side of the equation, and
all constant terms (the numbers) to the right hand side of the equals sign.Bearing in mind that the sign of the terms will change from (
$+$ ) to (
$-$ ) or vice
versa, as they "cross over" the equals sign.
Group all like terms together and simplify as much as possible.
Factorise if necessary.
Find the solution.
Substitute solution into
original equation to check answer.
Khan academy video on equations - 1
Determine what is given and what is required
We are given
$4-x=4$ and are required to solve for
$x$ .
Determine how to approach the problem
Since there are no brackets, we can start with grouping like terms and then
simplifying.
Solve the problem
$$\begin{array}{ccc}\hfill 4-x& =& 4\hfill \\ \hfill -x& =& 4-4\phantom{\rule{1.em}{0ex}}\left(\mathrm{move}\mathrm{all}\mathrm{constant}\mathrm{terms}\right(\mathrm{numbers}\left)\mathrm{to}\mathrm{the}\mathrm{RHS}\right(\mathrm{right}\mathrm{hand}\mathrm{side}\left)\right)\hfill \\ \hfill -x& =& 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{group}\mathrm{like}\mathrm{terms}\mathrm{together}\right)\hfill \\ \hfill -x& =& 0\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill -x& =& 0\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}x& =& 0\hfill \end{array}$$
Check the answer
Substitute solution into original equation:
$$4-0=4$$
$$4=4$$
Since both sides are equal, the answer is correct.
Write the final answer
The solution of
$4-x=4$ is
$x=0$ .
Solve for
$x$ :
$4(2x-9)-4x=4-6x$
Determine what is given and what is required
We are given
$4(2x-9)-4x=4-6x$ and are required to solve for
$x$ .
Determine how to approach the problem
We start with expanding the brackets, then grouping like terms and then
simplifying.
Solve the problem
$$\begin{array}{ccc}\hfill 4(2x-9)-4x& =& 4-6x\hfill \\ \hfill 8x-36-4x& =& 4-6x\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{expand}\mathrm{the}\mathrm{brackets}\right)\hfill \\ \hfill 8x-4x+6x& =& 4+36\phantom{\rule{1.em}{0ex}}\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{with}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\mathrm{to}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{the}=\\ \hfill (8x-4x+6x)& =& (4+36)\phantom{\rule{1.em}{0ex}}\left(\mathrm{group}\mathrm{like}\mathrm{terms}\mathrm{together}\right)\hfill \\ \hfill 10x& =& 40\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill \frac{10}{10}x& =& \frac{40}{10}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}10\right)\hfill \\ \hfill x& =& 4\hfill \end{array}$$
Check the answer
Substitute solution into original equation:
$$\begin{array}{ccc}\hfill 4\left(2\right(4)-9)-4\left(4\right)& =& 4-6\left(4\right)\hfill \\ \hfill 4(8-9)-16& =& 4-24\hfill \\ \hfill 4(-1)-16& =& -20\hfill \\ \hfill -4-16& =& -20\hfill \\ \hfill -20& =& -20\hfill \end{array}$$
Since both sides are equal to
$-20$ , the answer is correct.
Write the final answer
The solution of
$4(2x-9)-4x=4-6x$ is
$x=4$ .
Solve for
$x$ :
$\frac{2-x}{3x+1}=2$
Determine what is given and what is required
We are given
$\frac{2-x}{3x+1}=2$ and are required to solve for
$x$ .
Determine how to approach the problem
Since there is a denominator of (
$3x+1$ ), we can start by multiplying both sides
of the equation by (
$3x+1$ ). But because division by 0 is not permissible, there
is a restriction on a value for x. (
$x\ne \frac{-1}{3}$ )
Solve the problem
$$\begin{array}{ccc}\hfill \frac{2-x}{3x+1}& =& 2\hfill \\ \hfill (2-x)& =& 2(3x+1)\hfill \\ \hfill 2-x& =& 6x+2\phantom{\rule{1.em}{0ex}}(\mathrm{remove}/\mathrm{expand}\mathrm{brackets})\hfill \\ \hfill -x-6x& =& 2-2\phantom{\rule{1.em}{0ex}}\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{containing}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\left(\mathrm{numbers}\right)\mathrm{to}\mathrm{the}\mathrm{RHS}.\\ \hfill -7x& =& 0\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill x& =& 0\xf7(-7)\hfill \\ \hfill therefore\phantom{\rule{1.em}{0ex}}x& =& 0\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\mathrm{zero}\mathrm{divided}\mathrm{by}\mathrm{any}\mathrm{number}\mathrm{is}0\hfill \end{array}$$
Check the answer
Substitute solution into original equation:
$$\begin{array}{ccc}\hfill \frac{2-\left(0\right)}{3\left(0\right)+1}& =& 2\hfill \\ \hfill \frac{2}{1}& =& 2\hfill \end{array}$$
Since both sides are equal to 2, the answer is correct.
Write the final answer
The solution of
$\frac{2-x}{3x+1}=2$ is
$x=0$ .
Solve for
$x$ :
$\frac{4}{3}x-6=7x+2$
Determine what is given and what is required
We are given
$\frac{4}{3}x-6=7x+2$ and are required to solve for
$x$ .
Determine how to approach the problem
We start with multiplying each of the terms in the equation by 3, then
grouping like terms and then simplifying.
Solve the problem
$$\begin{array}{ccc}\hfill \frac{4}{3}x-6& =& 7x+2\hfill \\ \hfill 4x-18& =& 21x+6\phantom{\rule{1.em}{0ex}}\left(\mathrm{each}\mathrm{term}\mathrm{is}\mathrm{multiplied}\mathrm{by}3\right)\hfill \\ \hfill 4x-21x& =& 6+18\phantom{\rule{1.em}{0ex}}(\mathrm{move}\mathrm{all}\mathrm{terms}\mathrm{with}\mathrm{x}\mathrm{to}\mathrm{the}\mathrm{LHS}\hfill \\ \hfill \mathrm{and}\mathrm{all}\mathrm{constant}\mathrm{terms}\mathrm{to}\mathrm{the}\mathrm{RHS}\mathrm{of}\mathrm{the}=)\\ \hfill -17x& =& 24\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}\left(\mathrm{simplify}\mathrm{grouped}\mathrm{terms}\right)\hfill \\ \hfill \frac{-17}{-17}x& =& \frac{24}{-17}\phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}(\mathrm{divide}\mathrm{both}\mathrm{sides}\mathrm{by}-17)\hfill \\ \hfill x& =& \frac{-24}{17}\hfill \end{array}$$
Check the answer
Substitute solution into original equation:
$$\begin{array}{ccc}\hfill \frac{4}{3}\times \frac{-24}{17}-6& =& 7\times \frac{-24}{17}+2\hfill \\ \hfill \frac{4\times (-8)}{\left(17\right)}-6& =& \frac{7\times (-24)}{17}+2\hfill \\ \hfill \frac{(-32)}{17}-6& =& \frac{-168}{17}+2\hfill \\ \hfill \frac{-32-102}{17}& =& \frac{(-168)+34}{17}\hfill \\ \hfill \frac{-134}{17}& =& \frac{-134}{17}\hfill \end{array}$$
Since both sides are equal to
$\frac{-134}{17}$ , the answer is correct.
Write the final answer
The solution of
$\frac{4}{3}x-6=7x+2$ is,
$x=\frac{-24}{17}$ .
Solving linear equations
Solve for
$y$ :
$2y-3=7$
Solve for
$w$ :
$-3w=0$
Solve for
$z$ :
$4z=16$
Solve for
$t$ :
$12t+0=144$
Solve for
$x$ :
$7+5x=62$
Solve for
$y$ :
$55=5y+\frac{3}{4}$
Solve for
$z$ :
$5z=3z+45$
Solve for
$a$ :
$23a-12=6+2a$
Solve for
$b$ :
$12-6b+34b=2b-24-64$
Solve for
$c$ :
$6c+3c=4-5(2c-3)$
Solve for
$p$ :
$18-2p=p+9$
Solve for
$q$ :
$\frac{4}{q}=\frac{16}{24}$
Solve for
$q$ :
$\frac{4}{1}=\frac{q}{2}$
Solve for
$r$ :
$-(-16-r)=13r-1$
Solve for
$d$ :
$6d-2+2d=-2+4d+8$
Solve for
$f$ :
$3f-10=10$
Solve for
$v$ :
$3v+16=4v-10$
Solve for
$k$ :
$10k+5+0=-2k+-3k+80$
Solve for
$j$ :
$8(j-4)=5(j-4)$
Solve for
$m$ :
$6=6(m+7)+5m$
Questions & Answers
Introduction about quantum dots in nanotechnology
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is system testing?
AMJAD
how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:
OpenStax, Linear equations. OpenStax CNX. Jun 15, 2015 Download for free at https://legacy.cnx.org/content/col11828/1.1
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