## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$z=0.8$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $0.91-0.2z=1.23-0.6z ,$ remove first the decimal numbers by multiplying both sides by a power of $10.$ Then use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ Since the largest number of decimal places a term has is $2,$ multiply both sides by $100.$ This results to \begin{array}{l}\require{cancel} 0.91-0.2z=1.23-0.6z \\\\ 100(0.91-0.2z)=100(1.23-0.6z) \\\\ 91-20z=123-60z .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 91-20z=123-60z \\\\ -20z+60z=123-91 \\\\ 40z=32 \\\\ z=\dfrac{32}{40} \\\\ z=\dfrac{\cancel8(4)}{\cancel8(5)} \\\\ z=\dfrac{4}{5} \\\\ z=0.8 .\end{array} Checking: If $z=0.8,$ then \begin{array}{l}\require{cancel} 0.91-0.2z=1.23-0.6z \\\\ 0.91-0.2(0.8)=1.23-0.6(0.8) \\\\ 0.91-0.16=1.23-0.48 \\\\ 0.75=0.75 \text{ (TRUE) } .\end{array} Hence, the solution is $z=0.8 .$