# Elementary algebra: solving linear equations in one variable  (Page 3/3)

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Solve for a: $-4a+2-a=3+5a-2$ The solution set is $\left\{\frac{1}{10}\right\}$

## Simplifying expressions first

When solving linear equations the goal is to determine what value, if any, will solve the equation. A general guideline is to use the order of operations to simplify the expressions on both sides first.

Solve for x: $5\left(3x+2\right)-2=-2\left(1-7x\right)$ The solution set is $\left\{-10\right\}$ .

## Conditional equations, identities, and contradictions

There are three different kinds of equations defined as follows.

Conditional Equation
A conditional equation is true for particular values of the variable.
Identity
An identity is an equation that is true for all possible values of the variable. For example, x = x has a solution set consisting of all real numbers, $\Re$ .
A contradiction is an equation that is never true and thus has no solutions. For example, x + 1 = x has no solution. No solution can be expressed as the empty set .

So far we have seen only conditional linear equations which had one value in the solution set. If when solving an equation and the end result is an identity, like say 0 = 0, then any value will solve the equation. If when solving an equation the end result is a contradiction, like say 0 = 1, then there is no solution.

Solve for x: $4\left(x+5\right)+6=2\left(2x+3\right)$ $\begin{array}{cccc}4\left(x+5\right)+6& =& 2\left(2x+3\right)& \mathit{\text{Distribute}}\hfill \\ 4x{+}{20}{+}{6}& =& 4x+6& \mathit{\text{Add same side like terms}}\text{.}\hfill \\ 4x+26& =& 4x+6& \\ 4x+26{-}{4}{x}& =& 4x+6{-}{4}{x}& \mathit{\text{Subtract 4x on both sides.}}\hfill \\ 26& =& 6& \mathit{\text{False}}\hfill \end{array}$ There is no solution, $\varnothing$ .

Solve for y: $3\left(3y+5\right)+5=10\left(y+2\right)-y$ The equation is an identity, the solution set consists of all real numbers, $\Re$ .

## Linear literal equations

Literal equations, or formulas, usually have more than one variable. Since the letters are placeholders for values, the steps for solving them are the same. Use the properties of equality to isolate the indicated variable.

Solve for a: $P=2a+b$ Solution: $a=\frac{P-b}{2}$

Solve for x: $z=\frac{x+y}{2}$ Solution $x=2z-y$

Yes

No

Yes

Yes

No

## Solving in one step

$x=-3$

$y=-5$

$x=\frac{5}{6}$

$x=\frac{5}{6}$

$x=-11$

$a=10$

$y=3$

$x=-\frac{3}{2}$

$t=-3$

$x=-\frac{3}{14}$

## Solve in two steps

$a=10$

$y=5$

$x=0$

$x=\frac{2}{15}$

$y=7$

$x=-\frac{5}{2}$

$a=\frac{1}{8}$

$x=1$

$y=\frac{1}{3}$

$x=-3$

## Solve in multiple steps

$x=-12$

$y=-4$

$a=-\frac{26}{15}$

$x=-1$

No Solution, $\varnothing$

All Reals, $\Re$

$y=\frac{1}{2}$

$x=2$

$a=-\frac{10}{9}$

All Reals, $\Re$

## Literal equations

$w=\frac{P-2l}{2}$

$b=P-a-c$

$C=\frac{5F-160}{9}$

$r=\frac{C}{2\pi }$

$y=-5z+x$

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yes that's correct
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I think
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