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Linear equations
Linear equations
Equations and inequalities:
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
how can chip be made from sand
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
is there industrial application of fullrenes.
What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest.
Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.?
How this robot is carried to required site of body cell.?
what will be the carrier material and how can be detected that correct delivery of drug is done
Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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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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