Solve for G.

[latex]\Large 27+ \text{G} = 43+49[/latex]

Step 1: Isolate the variable you are trying to find. In this case “G.” In order to do this, we must remove the 27 from the left side of the equation. This is done by subtracting 27 from the left side and then also from the right side.

[latex]\Large \begin{array}{c}27+\text{G} = 43+49 \\ \mathbf{(27-27)}+\text{G} = 43+49 - \mathbf{27}\end{array}[/latex]

Step 2: Work through the equation.

[latex]\Large \begin{array}{c}(27-27) +\text{G} = 43+49-27 \\ 0 + \text{G} = 92-27 \\ \text{G}=65\end{array}[/latex]

Step 3: Check your answer.

[latex]\Large \begin{array}{rl}\text{Replace G with 65} &\\ \downarrow & \\ 27 + \text{G} & = 43 + 49 \\ 27 + 65 & = 43 + 49 \\ 92 & =92\end{array}[/latex]

Solve for H.

[latex]\Large \text{H}-16=13+19[/latex]

Step 1: Isolate the variable you are trying to find. In this case “H.” In order to do this we must remove the 16 from the left side of
the equation. This is done by adding 16 to the left hand side of the equation and then also adding 16 to the right hand side of
the equation.

[latex]\Large \begin{array}{c}\text{H}-16 = 13+19 \\ \text{H} + \mathbf{(-16+16)}=13 + 19 \mathbf{ +16}\end{array}[/latex]

Step 2: Work through the equation.

[latex]\Large \begin{array}{c}\text{H}+(-16+16)=13+19+16 \\ \text{H}+0=32+16 \\ \text{H}=48 \end{array}[/latex]

Step 3: Check your answer.

[latex]\Large \begin{array}{l}\text{Replace H with 48} \\ \downarrow \\ \text{H}-16 = 13+19 \end{array}[/latex]

[latex]\Large \begin{array}{c} 48-16=13+19 \\32=32 \end{array}[/latex]

We’ll try another example, but this time we’ll deal with fractions that will have to be moved around. We’ll also start to do this in steps as we have done with previous questions.

Solve for L.

[latex]\Large \dfrac{4}{9} \times \text{L} = 12 \times 12[/latex]

Step 1:  Isolate L. In this case, we have to move the 4/9 from one side to the other. In order to do this we have to multiply both sides by the reciprocal of 4/9. This will essentially remove 4/9 from the left hand side of the equation. The reciprocal is of 4/9 is 9/4.

[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} = \dfrac{36}{36} = 1[/latex]

Therefore we get:

[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4}[/latex]

Step 2: Solve the equation

[latex]\Large \begin{array}{c} \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4} \\ 1 \times \text{L} = 144 \times \dfrac{9}{4} \\ \text{L} = \dfrac{1296}{4} \\ \text{L}=324 \end{array}[/latex]

Step 3: Confirm your answer

[latex]\Large \begin{array}{rl}\text{Replace L with 324} & \\ \downarrow & \\ \dfrac{4}{9} \times \text{L} & = 12 \times 12 \\ \dfrac{4}{9} \times 324 & = 12 \times 12 \\ \dfrac{1296}{9} & = 144 \\ 144 & = 144 \end{array}[/latex]

This one is a bit more challenging. You’ll note in the question that there are no numbers, only letters. Take a look at the question, and see if you can come up with any ideas on how you would solve this equation.

Solve for “D” in the following equation.

[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]

Step 1:  Determine which variable you must isolate. In this case it’s given for us and it’s “D.”

[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\textbf{D}}[/latex]

The challenge here is that “D” is in the denominator (bottom) of the fraction. If we were to isolate it but it was still in the denominator of a fraction we would not have something that we could work with.

When “D” is isolated it must not only be by itself on one side of the equation but also appear as a whole number and not as the denominator of a fraction.

Step 2: To make this process easier break the equation down to look like the following.

[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}}[/latex]

Remember that:

[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{A}}{\text{B}}[/latex]

We haven’t really changed anything mathematically we have just made the equation into something that is easier to work with.

Step 3:  Isolate “D.” In order to do this we must remove “C” from the right hand side of the equation. Do this by multiplying both the right and left hand side by 1/C.

[latex]\Large \mathbf{\dfrac{1}{\text{C}}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}} \times \mathbf{\dfrac{1}{\text{C}}}[/latex]

Now this might start looking a little bit complicated but if you follow it through you’ll start to see the answer form. Look at the right hand side of the equation. It now has both and C/1 and a 1/C. Multiplying those reciprocals together you get 1 and effectively take out the “C” on the right hand side.

[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{1}{\text{C}}= \dfrac{\text{C}}{\text{C}}=1[/latex]

So we end up with:

[latex]\Large \dfrac{1}{\text{C}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= 1 \times \dfrac{1}{\text{D}}[/latex]

If we wanted to simplify this we could just do the following:

[latex]\Large \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}}[/latex]

Step 4: Get D into the numerator of the equation. What we need to do will take a little bit of math and some patience. We need to follow the same rules that we have been following.

The idea here is to multiply each side by D/1. This will make the right hand side of the equation equal to 1 but now the left hand side of the equation will have D in the numerator.

[latex]\Large \begin{array}{c}\dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}} \times \dfrac{\text{D}}{1} \\ \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C}\times \text{B}} = 1 \end{array}[/latex]

What you’ll notice is that it actually creates more work but we have managed to get “D” into the numerator of the equation. The only problem is that we also have a bunch of other stuff on the same side as “D” and now our job becomes getting rid of all that stuff.

Now all we have to do is follow the rules and get the A, B and C over to the right hand side and isolate D. I’ll do this simply and in one quick calculation.

[latex]\Large \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} \times \dfrac{\text{C} \times \text{B}}{\text{A}} = \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]

What we end up with is:

[latex]\Large \text{D}= \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]

There is a lot of math involved there but if you follow the rules and go one step at a time you’ll eventually get there. My suggestion here is that you go over what we just went through a couple times before moving on. Understanding the math involved here is quite important when it comes to transposing formulas and equations.

At this point you might be wondering if there was a shortcut and as luck would have it there is for this procedure which I’ll go through now.. We’ll start with an example.

Take the following equation:

[latex]\Large \dfrac{3}{4}=\dfrac{3}{4}[/latex]

This equation is true. Now take each of the fractions and flip them around.

[latex]\Large \dfrac{4}{3} = \dfrac{4}{3}[/latex]

It seems that if you flip both fractions around then the equation is still true. Mathematically you are doing the same thing to one side as you are doing to the other. There is a lot of math involved here but the main point is that if you were to do all the math you would end up with the same answer.

How we can use this to help us simplify our question is to just do the following. Remember we started with:

[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]

We needed to isolate D but the main problem (and the thing that causes us to do a lot of work) was that D was in the denominator and we needed it in the numerator.

Through our math wizardry we can get D into the numerator by just flipping around the left hand fraction. Whatever we do to that side we then must do to the right side. What we end up with is:

[latex]\Large \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}}[/latex]

All we need to do now is get C out of the left hand side by multiplying it by C/1 and then doing the same thing to the left hand side.

[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}} \times \dfrac{\text{D}}{1}[/latex]

We end up with…

[latex]\Large \dfrac{\text{C} \times \text{B}}{\text{A}} = \text{D}[/latex]

Now to take all that in in one shot might be a little much so you may want to go back and reread that whole explanation. But always keep in mind the math that goes along with that.

In this example, we are going to use the formula for the area of a circle. Calculating the area of a circle is actually something we are going to do further into this volume, but for now we are just going to deal with the formula itself.

[latex]\Large \begin{array}{c} \text{Area of a circle} \\ \text{A} = {\text{D}}^{2} \times 0.7854 \end{array}[/latex]

[latex]\Large \begin{array}{cl} \text{Where:} & \text{A = area of the circle} \\ & \text{D = diameter of the circle} \end{array}[/latex]

Solving for the area of the circle would be pretty straight forward but as we are dealing with transposing equations in this section what we’ll do is solve for the diameter (D).

Step 1: Identify the variable you are trying to solve for.

[latex]\Large \text{A} = \mathbf{{\text{D}}^{2}} \times 0.7854[/latex]

From this you can see we have a couple problems we have to deal with. The first one is that we have to isolate D. The second one is that D has the exponent 2 attached to it so we have to somehow eliminate that.

Step 2: Isolate D.

For this we have to move the 0.7854 from the right hand side of the equation to the left hand side. We simply follow the rules we have used up to this point.

[latex]\Large \begin{array}{c} \mathbf{\dfrac{1}{0.7854}} \times \text{A} = {\text{D}}^{2} \times 0.7854 \times \mathbf{\dfrac{1}{0.7854}} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times \dfrac{0.7854}{0.7854} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times 1 \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \end{array}[/latex]

That takes care of the first problem. Now we have to tackle the D² issue.

Step 3: Remove the exponent from D.

Once again we go back to our original rule. Whatever you do to one side you must to the same thing to the other side.

We have to refer to the video that you watched earlier in this section regarding exponents and square roots. We’ll do a quick refresher before we solve for D.

Remember that:

[latex]\Large \begin{array}{c} {\text{D}}^{2} = \text{D} \times \text{D} \\ \text{and... } \\ \sqrt{\text{D} \times \text{D}} = \text{D} \end{array}[/latex]

So what we see is that if you square root a number that is squared (has an exponent of 2) you end up with the number itself. I’ll quickly show you with numbers.

[latex]\Large \begin{array}{c} {4}^{2} = 4 \times 4 \\ {4}^{2} = 16 \\ \sqrt{16} = \sqrt{4 \times 4} \\ \sqrt{16} = 4 \end{array}[/latex]

Having gone through all that we can now solve the problem.

[latex]\Large \begin{array}{c} \sqrt{\dfrac{\text{A}}{0.7854}}= \sqrt{{\text{D}}^{2}} \\ \sqrt{\dfrac{\text{A}}{0.7854}} = \text{D} \end{array}[/latex]