Equations and inequalities: solving linear equations
The simplest equation to solve is a linear equation. A linear equation is an
equation where the power of the variable(letter, e.g.
$x$ ) is 1(one). The
following are examples of linear equations.
In this section, we will learn how to find the value of the variable that makes
both sides of the linear equation true. For example, what value of
$x$ makes
both sides of the very simple equation,
$x+1=1$ true.
Since the definition of a linear equation is that if the variable has a highest power of one (1), there is
at most
one solution or
root for the equation.
This section relies on all the methods we have already discussed: multiplying
out expressions, grouping terms and factorisation. Make sure that you arecomfortable with these methods, before trying out the work in the rest of this
chapter.
That is all that there is to solving linear equations.
Solving equations
When you have found the solution to an equation,
substitute the solution into the original equation, to check your answer.
Method: solving linear equations
The general steps to solve linear equations are:
Expand (Remove) all brackets that are in the equation.
"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.
If necessary factorise.
Find the solution and write down the answer(s).
Substitute solution into
original equation to check answer.
Solve for
$x$ :
$4-x=4$
We are given
$4-x=4$ and are required to solve for
$x$ .
Since there are no brackets, we can start with rearranging and then grouping like terms.
We are given
$\frac{2-x}{3x+1}=2$ and are required to solve for
$x$ .
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. (
$x\xe2\u2030\frac{-1}{3}$ )
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