Equation Lab #7 – Order of Operations

Yesterday, I went over the last of the types of additional steps when solving linear equations, which was about variables on both sides of the equation. Now we’re at the end of the road in regards to multi-step equations. Today’s lesson is on the order of how to solve equations.

Multi-Step Equations – Order of Operations:

So over the week, I went over addition and subtraction, multiplication and division, combining like terms, clearing fractions, and variables on both sides. What if all five steps are present in one equation? This would have to be the order of operations.

1. Clear out the fractions.
2. Combine like terms.
3. Move variables to one side.
4. Add/subtract the constants.
5. Divide the coefficient.

If at least one of these steps is absent, remember that clearing out fractions has the highest priority. Then it goes to combining like terms, then moving variables to one side.

Let’s do one with four steps. The equation is $\frac{3}{5} t+\frac{3}{4} t-3-1=50$. First, we multiply the whole equation by the least common denominator, which is the least common multiple of $5$ and $4$. Since they have no common factors, we’ll use $20$. $\frac{3}{5} t\cdot 20=12t$, $\frac{3}{4} t\cdot 20=15t$, $-3\cdot 20=-60$, $-1\cdot20=-20$, and $50\cdot 20=1000$. We now have $12t+15t-60-20=1000$. The next step to do is to combine like terms. $12t+15t=27t$, and $-60-20=-80$. The equation is $27t-80=1000$. $27t-80+80=27t$, and $1000+80=1080$. We are down to $27t=1080$. $27t/27=t$, and $1080/27=40$. The solution is $t=40$.

1. $\frac{3}{5} t+\frac{3}{4} t-3-1=50$
2. $(\frac{3}{5} t+\frac{3}{4} t-3-1)\cdot20=50\cdot 20$
3. $12t+15t-60-20=1000$
4. $(12t+15t)+(-60-20)=1000$
5. $27t-80=1000$
6. $27t-80+80=1000+80$
7. $27t=1080$
8. $27t/27=1080/27$
9. $t=40$

We got one more equation to go, and this time, we’re going to use all five steps. The epic equation is:

$\frac{1}{3} t+\frac{3}{4} t+6+3=\frac{7}{6} t+\frac{3}{2} t-20-9$

First, we clear out the denominators by multiplying the entire equation by all denominators’ least common multiple. Between $2$, $3$, $4$, and $6$, the smallest number to be divisible by all four of them is $12$, so we’ll have to multiply both sides by $12$. Clearing the denominators, $\frac{1}{3} t\cdot 12=4t$, $\frac{3}{4} t\cdot 12=9t$, $6\cdot 12=72$, $3\cdot 12=36$, $\frac{7}{6} t\cdot 12=14t$, $\frac{3}{2} t\cdot 12=18t$, $-20\cdot 12=-240$, and $-9\cdot 12=-108$. The equation is $4t+9t+72+36=14t+18t-240-108$. Next, we have to combine like terms. Make sure not to combine terms from different sides of the equation. On the left side, the like terms are $4t$ and $9t$, and $72$ and $36$. $4t+9t=13t$, and $72+36=108$. On the right side, the like terms are $14t$ and $18t$, and $-240$ and $-108$. $14t+18t=32t$, and $-240-108=-348$. The equation is now $13t+108=32t-348$. The fractions are cleared, and the like terms are combined, but there’s still variables on both sides. Therefore, the next step is to move all variables to one side. $13t-13t+108=108$, and $32t-13t-348=19t-348$. We are down to $19t-348=108$, which is a two-step equation. $19t-348+348=19t$, and $108+348=456$. The last equation is $19t=456$. $19t/19=t$, and $456/19=24$. Our solution to this epic equation is $t=24$.

 1. $\frac{1}{3}t+\frac{3}{4}t+6+3=\frac{7}{6}t+\frac{3}{2}t-20-9$
 2. $(\frac{1}{3}t+\frac{3}{4} t+6+3)\cdot 12=(\frac{7}{6} t+\frac{3}{2} t-20-9)\cdot 12$
 3. $4t+9t+72+36=14t+18t-240-108$
 4. $(4t+9t)+(72+36)=(14t+18t)+(-240-108)$
 5. $13t+108=32t-348$
 6. $13t-13t+108=32t-13t-348$
 7. $19t-348=108$
 8. $19t-348+348=108+348$
 9. $19t=456$
10. $19t/19=456/19$
11. $t=24$

Normally, I would have a word problem after showing two equations, but since this is a big one, let’s leave it as this.

Quiz #7:

As the last Equation Lab lesson of the month, here is the last quiz. There are two questions that involve combining and clearing, two questions that involve combining and moving, two questions that involve clearing and moving, and two questions that involve all three steps. Whether you put down the missing value or the variable with the missing value is fine for as long as you use the correct variable and correct answer. For instance, when the question is $b+3=6$, you can use $3$ or $b=3$, but not $a=3$.