Solving Linear Equations
Let’s practice solving algebra (linear) problems. It seems really difficult at first, but I promise you, after some practice, it will get easier. The thing I love about math is that you have a bunch of rules and all you have to do is apply them (memorize less than say History!).
Let’s solve for \(x\) in the following problems. At this point, don’t worry about where we got the variables and numbers – let’s just go through the mechanics. We’ll do real algebra word problems later.
Again, you want to get the \(\boldsymbol{x}\) by itself – when it’s by itself, we get the answer of what \(\boldsymbol{x}\) is! To do this, you have to move things around by doing the opposite of what you have. For example, if the problem has addition in it, you have to subtract from both sides; if the problem has division, you have to multiply on both sides. Remember also that the sign of a variable or constant is what is in front of it; sometimes it’s an “invisible \(+\)” if it’s at the beginning.
 Step 1: First combine the numbers (called “constants”) and the same letters (variables) on each side if you need to. This is called combining like terms, meaning you can add the same letters together – in this case the \(x\)’s. As we saw earlier, this is because of the Distributive Property, and later we’ll see we can do the same for \({{x}^{2}}\), or actually any term with identical variables. Remember that when we just have an \(x\), it’s the same as \(1x\). We can change the order of our terms using the Commutative Property of Addition. So let’s say we are trying to solve \(4x4x=4+1\) for \(x\):
\(\begin{align}4x4x&=4+1\\4xx4&=4+1\\4x1x4&=5\\3x4&=5\end{align}\)
 Step 2: Add or subtract the “numbers” (“constants”) from each side to get rid of them on the side where the variable (the \(x\)) is. You want to do the opposite or (additive inverse) of what we have to get rid of. If you are subtracting a number, you want to add on both sides, and if you are adding a number, you want to subtract on each side. Line up the letters and numbers vertically. In our case, we are subtracting on each side, and using the Subtraction Property of Equality:
\(\displaystyle \begin{array}{l}3x4=\,5\\\,\,\,\,\,\,\,\,\underline{{+4=+4}}\\3x+0=9\\\,\,\,\,\,\,\,\,\,3x=\,9\end{array}\)
 Step 3: If there’s a number before the \(x\) (which means it’s multiplied by the \(x\)), divide both sides by that number (using the Multiplicative Inverse). If there’s a number below the \(x\) (which means it’s divided by the x), multiply both sides by that number. If there’s a fraction before the \(x\), we need to multiply both sides of the reciprocal of the fraction. (The numbers immediately before the variables are called coefficients.) In our case, we have a number before the \(x\), and we are using the Division Property of Equality.
\(\begin{align}3x&=\,9\\\frac{3}{3}x\,&=\,\frac{9}{3}\\\,1x\,&=\,3\\\,\,\,\,x\,&=\,3\,\,\,\,\,\surd \end{align}\)
Now we have solved to get what \(x\) is: \(x=3\)!
 Step 4: We should always try to check our answer, by plugging the answer we got back in for wherever \(x\) is:
\(\begin{align}\,4x4x&=4+1\\4\left( 3 \right)43&=4+1\\\,\,\,1243&=5\\\,\,\,\,\,83=5\\\,\,\,\,\,\,5&=5\,\,\,\,\,\,\,\surd \end{align}\)
Yeah — we did it!
Here are more examples, since solving these is an important foundation of algebra. Just remember to keep doing the above steps until you get \(x\) by itself.
It doesn’t really matter what order you do the operations, as long as you do the addition and/or subtraction first, and then do the multiplication and/or division. Do you see how this is sort of the opposite of PEMDAS – it’s SADMEP! So first do the addition and subtraction, then the multiplication and division, and then worry about things in parentheses. (We’ll talk about solving with exponents later).
Make sure you can follow each example before you go on to the next section. Do you recognize the algebraic properties we saw earlier?
These will look complicated, but really, you are just doing the same things over and over again — like a puzzle!
Solving Linear Equation  Notes 
\(\displaystyle \begin{array}{l}\,\,\color{#800000}{{5x4\,=3x+2}}\\\,\,\,\,\,\,\,\underline{{\,\,+\,4\,=\,\,\,\,\,\,\,\,+4\,}}\,\\\,\,5x\,\,\,\,\,\,\,\,\,\,=3x+6\\\,\,\,\,\,\underline{{3x\,\,\,=3x\,\,\,\,\,\,\,\,}}\\5x3x=\,\,\,\,\,\,+6\\\,\,\,\,\,\,\,\,\,\,\,\,2x=6\\\,\,\,\,\,\,\,\,\,\,\,\frac{2}{2}x=\frac{6}{2}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=3\,\,\,\,\,\surd \end{array}\)  Back and forth, back and forth….move things around to get all the “\(x\)’s” to one side and all the numbers to the other side.
There is certainly more than one way to do this, but make sure you get the basic idea. 
\(\require{cancel} \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{#800000}{{\frac{{5x}}{4}+3=18}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,\,3=3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{4}x\,\,\,\,\,\,\,\,\,=15\\\left( {\frac{{{}^{1}\cancel{4}}}{{{}_{1}\cancel{5}}}\times \frac{{{{{\cancel{5}}}^{1}}}}{{{{{\cancel{4}}}_{1}}}}} \right)x=\frac{4}{5}\times 15\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1x=\frac{4}{{{}_{1}\cancel{5}}}\times {{\cancel{{15}}}^{3}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=4\times 3\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=12\,\,\,\,\,\surd \end{array}\) 
Note that we can write \(\displaystyle \frac{{5x}}{4}\) as \(\displaystyle \frac{5}{4}x\). Try putting in different numbers for \(x\) to show that this works.
Note that we multiplied both sides by the reciprocal of \(\displaystyle \frac{5}{4}\) (which is \(\displaystyle \frac{4}{5}\)) so that \(x\) could be all by itself on one side (\(1\times x=1\)). This is the Multiplicative Inverse property. 
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\color{#800000}{{\frac{{3\left( {x+5} \right)}}{7}2=\,4}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,+\,2=\,+2}}\\\,\,\,\,\,\,\,\,\,\,\frac{{3\left( {x+5} \right)}}{7}\,\,\,\,\,\,\,\,\,=\,\,6\\\frac{{\cancel{3}\left( {x+5} \right)}}{{\cancel{7}}}\left( {\frac{{\cancel{7}}}{{\cancel{3}}}} \right)=6\times \frac{7}{3}\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x+5} \right)=14\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,\,\,\,5=5}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=9\,\,\,\,\,\surd \end{array}\)  Because of the opposite of PEMDAS, we’ll subtract first, then multiply by the reciprocal, subtract again, and then “open up” the parentheses. Think of \((x+5)\) as a variable until we get it by itself.
Now we have to do subtraction one more to get \(x\) by itself.
Note that we could have “pushed through” the 3 at first, using the Distributive Property; that would have worked too. 
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{#800000}{{\frac{{5+3x}}{2}+5=3x}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,5=\,\,\,\,\,5}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{5+3x}}{2}\,\,\,\,\,\,\,\,\,=3x5\\\,\,\,\,\,\,\frac{{5+3x}}{{{}_{1}\cancel{2}}}\,\,\times \,\,\frac{{{{{\cancel{2}}}^{1}}}}{1}=\left( {3x5} \right)\times \frac{2}{1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5+3x=\left( {3x5} \right)\times 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5+3x=6x10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,\,\,\,\,\,\,\,\,\,3x=3x\,\,\,}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\,\,\,\,\,\,\,\,=3x10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,+10\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,\,\,+10}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,15\,\,\,\,=3x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{15}}{3}=\frac{{{}^{1}\cancel{3}x}}{{{}_{1}\cancel{3}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5=x\,\,\,\,\,\to \,\,\,\,x=5\,\,\,\,\,\surd \end{array}\)  Note that we moved the 5 over first, and then we are left with \(\displaystyle \frac{{5+3x}}{2}\). It’s like the \(5+3x\) has parentheses around it – it’s like one variable.
Since we want to get the \(x\) by itself (which means the “\(5+3x\)” first), we had to multiply both sides by 2.
Note the “pushing through” of the 2 – this is the Distributive Property. We did this to just get it out of the way this time.
Note that it doesn’t matter (at least with equal signs) which side we move the \(x\) to.
We could have also multiplied both sides of the equation by 2 for the first step – we’ll see this later.
I know this isn’t easy, but follow the steps and you’ll get the hang of it! 
Here’s another example with fractions. When we have fractions on both sides, we can always multiply both sides by the least common denominator to get rid of the fractions:
Solving Linear Function  Notes 
\(\require{cancel} \displaystyle \begin{array}{c}\,\color{#800000}{{\,\frac{4}{{15}}x\frac{4}{5}=\frac{8}{3}x+4}}\\\,\,\,\,\,\,\,\,\,\,\,\,15\left( {\frac{4}{{15}}x\frac{4}{5}} \right)=15\left( {\frac{8}{3}x+4} \right)\\{}^{1}\cancel{{15}}\left( {\frac{4}{{{{{\cancel{{15}}}}_{1}}}}x} \right){}^{3}\cancel{{15}}\left( {\frac{4}{{{{{\cancel{5}}}_{1}}}}} \right)={}^{5}\cancel{{15}}\left( {\frac{8}{{{{{\cancel{3}}}_{1}}}}x} \right)+15\left( 4 \right)\\\\\,4x\left( {3\times 4} \right)=\left( {5\times 8} \right)x+60\\\,4x12=40x+60\\\underline{{\,\,\,\,\,\,\,\,\,+12\,\,\,\,\,\,\,\,\,\,\,\,\,\,+12}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,4x=40x+72\\\,\underline{{40x=40x}}\\\,\,\,\,\,\,36x=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,72\\\\\,\,\,\,\,\,\frac{{36}}{{36}}x=\frac{{72}}{{36}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=2\,\,\,\,\,\,\,\,\surd \end{array}\)  Note that we multiply everything by 15, since that’s the least common denominator (LCD) between 15, 3, and 5. This gets rid of fractions. Neat trick!
Also note that we see what’s missing on the bottom (that goes into 15) and multiplying that on the top.
Let’s check our work just to make sure we did it right. Watch the negative numbers! \(\begin{array}{l}\,\,\,\,\,\,\,\,\frac{4}{{15}}x\frac{4}{5}=\frac{8}{3}x+4\\\frac{4}{{15}}(2)\frac{4}{5}=\frac{8}{3}(2)+4\\\,\,\,\,\,\,\,\,\,\frac{{8}}{{15}}\frac{4}{5}=\frac{{16}}{3}+4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{20}}{{15}}=\frac{{4}}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{4}}{3}=\frac{{4}}{3}\,\,\,\,\,\,\,\surd \end{array}\)

Solving Literal Equations (Transforming Formulas)
In this next example, we’ll actually solve for the variable \(x\) “in terms of” some other variables (meaning that we still solve for \(x\), but may not get a number); notice that we treat variables just like numbers!
This is called solving Literal Equations, since sometimes variables are called literals. These are also called Transforming Formulas, or solving Multivariable Equations.
Let’s use the formula that converts Celsius temperatures (the type of temperatures they use in Paris!) to Fahrenheit (the type of temperatures they use here in the United States) to get the formula that converts Fahrenheit to Celsius. elsius is the “metric system” way to figure temperature, and we use Fahrenheit. ee, this stuff can really be useful!
Let’s say we are given \(F\) in terms of \(C\) (meaning \(F\) is by itself). Let’s solve for \(C\) in terms of \(F\) (meaning get \(C\) by itself):
Solving Literal Function  Notes 
\(\displaystyle \begin{array}{l}\,\,\,\,\color{#800000}{{\text{F }\,\,\,\,\,\,\,\,=\text{ }\frac{9}{5}C+\text{32}}}\\\,\,\,\,\,\,\,\underline{{32=\,\,\,\,\,\,\,\,\,\,\,\,\,32}}\\\,F32=\frac{9}{5}C\end{array}\)
\(\displaystyle \require{cancel} \begin{array}{l}\left( {F32} \right)\times \frac{5}{9}=\,\,\left( {\frac{{\cancel{9}}}{{\cancel{5}}}\times \frac{{\cancel{5}}}{{\cancel{9}}}} \right)C\\\left( {F32} \right)\times \frac{5}{9}=\,\,1C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C=\left( {F32} \right)\times \frac{5}{9}\,\,\,\,\,\,\,\surd \end{array}\) 
For example, let’s plug in 30 for \(C\) in the equation that we start with. We see that \(\displaystyle F=\frac{9}{5}\left( {30} \right)+32=86\) degrees. So when it’s 30 degrees Celsius in Paris, it’s 86 degrees Fahrenheit in New York if the two cities have the same temperature.
Now that we know what \(C\) is in terms of \(F\), we can plug in the 86 degrees for \(F\), and get \(\displaystyle C=\left( {8632} \right)\times \frac{5}{9}=30\) degrees back. So again, when it’s 86 degrees Fahrenheit in New York, it’s 30 degrees Celsius in Paris.
See how we would need to use one equation when we’re in Paris, and another when someone from Paris is visiting here!

See – algebra is pretty cool (no pun intended)!
You’ll see later that these two equations are called inverses, since one is sort of the opposite of the other.
Here’s one more example that’s a little bit more difficult, since it has a lot of fractions:
Solving Literal Function  Notes 
\(\require{cancel} \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\color{#800000}{{\frac{q}{s}\,\,\,\,\,\,\,=\,\,\,\,p+\frac{{5+x}}{4}}}\\\,\,\,\,\,\,\underline{{\,\,\,\,p\,=p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\\,\,\,\,\,\frac{q}{s}p\,=\,\,\,\,\,\,\frac{{5+x}}{4}\\\\\left( {\frac{q}{s}p} \right)\times 4\,=\frac{{5+x}}{{\cancel{4}}}\times \frac{{\cancel{4}}}{1}\\\left( {\frac{q}{s}p} \right)\times 4\,=5+x\\\,\underline{{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5=5\,\,}}\\\left( {\frac{q}{s}p} \right)\times 45=x\\\\\,\,\,\,\,\frac{{4q}}{s}4p5=x\\\,\,\,\,\,x=\frac{{4q}}{s}4p5\,\,\,\,\,\surd \end{array}\)  Note that we’re just doing the same thing over and over again; adding and subtracting the numbers on the side with the \(x\) and then multiplying or dividing the numbers on the side with the \(x\).
Note that when we have something like “\(5+x\)” in the numerator, it’s like it has parentheses around it.
Note that we “pushed” the 4 through (distributive property) to get rid of the parentheses. When we pushed it through to a fraction, we put the 4 in the numerator, since \(\displaystyle 4=\frac{4}{1}\)).
Notice that the answer (what \(x\) is) has variables and numbers in it, but we can’t really do anything else (we can put the fraction with the other terms by finding a common denominator, but we haven’t done this yet).

Let’s do the literal equation problem from above again, this time using the multiplythrough fraction approach:
Solving Literal Function  Notes 
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{#800000}{{\frac{q}{s}=\,p+\frac{{5+x}}{4}}}\\\left( {4\cancel{s}} \right)\,\,\frac{q}{{\cancel{s}}}=\left( {4s} \right)\,\left( {p+\frac{{5+x}}{4}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4q=4sp+\cancel{4}s\left( {\frac{{5+x}}{{\cancel{4}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4q=4sp+s\left( {5+x} \right)\\\underline{{\,\,\,\,\,\,\,\,4sp=4sp\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\4q4sp=s\left( {5+x} \right)\\4q4sp=5s+sx\\\underline{{\,\,\,\,\,\,\,\,\,\,\,\,5s=5s}}\\4q4sp5s=sx\\\,\,\,\,\,x=\frac{{\left( {4q4sp5s} \right)}}{s}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{4q}}{s}4p5=x\\\,\,\,\,\,\,x=\frac{{4q}}{s}4p5\,\,\,\,\,\surd \end{array}\)  Let’s multiply everything by \(4s\), which is the LCD of 4 and \(s\).
We have to make sure that we multiply the \(4s\) through every term (push it through).
Remember that we must constantly thrive to get the \(x\) on one side by itself.
It looks like this wasn’t too much easier than doing this the original way; this is a really tough problem!
It is really neat to see a problem worked two different ways and still get the same answer. 
Learn these rules and practice, practice, practice!
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On to Linear Inequalities – you are ready!