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
x-2y+3z=-3
2x-y+z=7
-x+3y-z=6
Need help solving this problem (2/7)^-2
what is the coefficient of -4×
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2
then, f(3) when 5f(x+1)
5((3-2)+1)
5(1+1)
5(2)
10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
ninjadapaul
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2
so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
is it a question of log
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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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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