Applications of Linear Equations

This section this textbook explains how to translate the situations described in word problems to equations and provides a variety of examples. Read the chapter and work through the problems. Some examples involved the geometric facts you have learned in Unit 2.

Example 107.

The second angle of a triangle is double the first. The third angle is $40$ less than the first. Find the three angles.

First $x$	With nothing given about the first we make that $x$
Second $2x$	The second is double the first,
Third $x − 40$	The third is $40$ less than the first
$F + S + T = 180$	All three angles add to $180$
$(x) +(2x)+ (x − 40)= 180$	Replace $F$ , $S$ , and $T$ with the labeled values
$x +2x + x − 40 = 180$	Here the parenthesis are not needed.
$4x − 40 = 180$	Combine like terms, $x +2x + x$
$\underline {+ 40 + 40}$	Add $40$ to both sides
$\underline {4x = 220}$	The variable is multiplied by $4$
$4 \quad \quad 4$	Divide both sides by $4$
$x = 55$	Our solution for $x$
First $55$	Replace $x$ with $55$ in the original list of angles
Second $2(55)= 110$ Third $(55) − 40 = 15$	Our angles are $55$ , $110$ , and $15$

Another geometry problem involves perimeter or the distance around an object. For example, consider a rectangle has a length of 8 and a width of 3. There are two lengths and two widths in a rectangle (opposite sides) so we add $8 + 8 + 3 + 3 = 22$ . As there are two lengths and two widths in a rectangle an alternative to find the perimeter of a rectangle is to use the formula $P = 2L + 2W$ . So for the rectangle of length 8 and width 3 the formula would give, $P = 2(8) + 2(3) = 16 + 6 = 22$ . With problems that we will consider here the formula $P = 2L + 2W$ will be used.