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Learning how to solve various algebraic equations is one of our main goals in algebra. This module introduces the basic techniques for solving linear equations in one variable. (Prerequisites: Working knowledge of real numbers and their operations.)
We begin by establishing some definitions.
Following are some examples of linear equations in one variable, all of which will be solved in the course of this module.
The variable in the linear equation $\mathrm{2x}+3=\text{13}$ is $x$ . Values that can replace the variable to make a true statement compose the solution set. Linear equations have at most one solution. After some thought, you might deduce that $x=5$ is a solution to $\mathrm{2x}+3=\text{13}$ . To verify this we substitute the value 5 in for $x$ and see that we get a true statement, $2\left(\mathbf{5}\right)+3=\text{10}+3=\text{13}$ .
Is $x=3$ a solution to $-\mathrm{2x}-3=-9$ ?
Yes, because $-2\left({3}\right)-3=-6-3=-9$
Is $a=-\frac{1}{2}$ a solution to $-\text{10}a+5=\text{25}$ ?
No, because $-\text{10}\left({-}\frac{{1}}{{2}}\right)+5=5+5=\text{10}\ne \text{25}$
When evaluating expressions, it is a good practice to replace all variables with parenthesis first, then substitute in the appropriate values. By making use of parenthesis we could avoid some common errors using the order of operations.
Is $y=-3$ a solution to $\mathrm{2y}-5=-y-\text{14}$ ? $$\begin{array}{cccc}2\left(\text{}\right)-5& =& -\left(\text{}\right)-14& \mathit{\text{Replacevariableswithparenthesis}}\text{.}\hfill \\ 2\left({-}{3}\right)-5& =& -\left({-}{3}\right)-14& \mathit{\text{Substitutetheappropriatevalue}}\text{.}\hfill \\ -6-5& =& 3-14& \mathit{\text{Simplify}}\text{.}\hfill \\ -11& =& -11& \mathit{\text{True}}\text{.}\hfill \end{array}$$ Yes because $y=-3$ produces a true mathematical statement.
When the coefficients of linear equations are numbers other than nice easy integers, guessing at solutions becomes an unreasonable prospect. We begin to develop an algebraic technique for solving by first looking at the properties of equality.
Given algebraic expressions A and B where c is a real number:
$$\text{If}A=B\text{then}A+c=B+c$$
$$\text{If}A=B\text{then}A-c=B-c$$
$$\text{If}A=B\text{and}c\ne 0\text{thenc}A=cB$$
$$\text{If}A=B\text{and}c\ne 0\text{then}\frac{A}{c}=\frac{B}{c}$$
To summarize, the equality is retained if we add, subtract, multiply and divide both sides of an equation by any nonzero real number. The central technique for solving linear equations involves applying these properties in order to isolate the variable on one side of the equation.
Use the properties of equality to solve: $x+3=-5$ $$\begin{array}{cccc}x+3& =& -5& \\ x+3{-}{3}& =& -5{-}{3}& \mathit{\text{Subtract3onbothsides}}\text{.}\hfill \\ x& =& -8& \mathit{\text{Simplify}}\hfill \end{array}$$ The solution set is $\left\{-8\right\}$ .
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