27.3: B1.03- Section 2

Last updated
Save as PDF

Page ID: 51771

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

To solve equations, write a simpler equivalent equation using the following rules. Keep writing simpler equivalent equations until you have simplified it enough that one side has the variable alone and the other has a number. That gives the solution.

Add or subtract the same number or expression to both sides of the equation.
Multiply or divide both sides of the equation by the same nonzero number or expression.

You can check the solution to an equation by plugging the number into the original equation and seeing if the resulting statement is true.

For additional explanation and examples, see the course web page for links. These include links to instructional materials and a link to an equation solver, where you can enter any equation and have it find the solution. You can use that to check your work on as many problems as you have the time and energy to work.

Example 1

Solve $3x-4=11$ .

[reveal-answer q=”765770″]Show Answer[/reveal-answer]
[hidden-answer a=”765770″]

Solution: x $\begin{align}&\,\,\,\,\,\,\,3x-4=11\\&3x-4+4=11+4\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,3x=15\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3x}{3}=\frac{15}{3}\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=5\\\end{align}$ .

Check: $\begin{align}&\,\,\,\,\,3x-4=11\\&\\&3\cdot5-4?=?11\\&\,\,\,15-4?=?11\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,11=11\\\end{align}$

[/hidden-answer]

Example 2

Find a formula for y (that is, solve for y in terms of x): $y-7=-4(x-2)+12$

[reveal-answer q=”731148″]Show Answer[/reveal-answer]
[hidden-answer a=”731148″]

Discussion: This is more complicated than a problem like Example 1. However, it uses exactly the same techniques. Problems like this arise when we begin to investigate graphing lines. Probably you have done problems like these in a previous algebra course, but maybe you haven’t done many of them. We will practice many problems like these in the first few weeks of the course. Don’t practice these until you can do problems like Example 1 fairly easily.

Solution:

$\begin{align}&\,\,\,\,\,\,\,y-7=-4(x-2)+12\\&\,\,\,\,\,\,\,y-7=-4x+8+12\\&\,\,\,\,\,\,\,y-7=-4x+20\\&y-7+7=-4x+20+7\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=-4x+27\\\end{align}$

Check by plugging in the answer for y:

$\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y-7=-4(x-2)+12\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\,\,\,)-7=-4(x-2)+12\\&(-4x+27)-7\,\,?=?\,\,\,-4(x-2)+12\\&-4x+27-7\,\,\,?=?\,\,\,-4x+8+12\\&\,\,\,\,\,\,\,\,\,\,-4x+20\,\,\,=\,\,-4x+20\,\,\,\,\,\,\,\,\\\end{align}$

[/hidden-answer]

Example 3

Solve $0.75-0.08t=1.22$ .

[reveal-answer q=”606202″]Show Answer[/reveal-answer]
[hidden-answer a=”606202″]

Discussion: This is basically like Example 1, except that the coefficients are decimals and the variable is t rather than x. In your previous algebra classes, you may not have done much work with decimal numbers, but we will use them often in this class. We will practice many problems like these in the first few weeks of the course. Don’t practice these until you can do problems like Example 1 fairly easily.

Solution: $\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.75-0.08t=1.22\\&0.75-0.08t-0.75=1.22-0.75\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-0.08t=0.47\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{-0.08t}{-0.08}=\frac{0.47}{-0.08}\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=-5.875\\\end{align}$

Check: $\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.75-0.08t=1.22\\&\\&0.75-0.08\cdot(-5.875)\,\,?=?\,\,1.22\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.75+0.47\,\,?=?\,\,1.22\\&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1.22=1.22\\\end{align}$

[/hidden-answer]

Mathematics for Modeling. Authored by: Mary Parker and Hunter Ellinger. License: CC BY: Attribution