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Rewriting Formulas and Equations

- Calculus
- Absolute Maxima and Minima
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Perhaps you need to find the width of your pool, you've been given the volume, height, and length. These are all variables that are used in the formula to find the volume of a rectangular prism. You can still find the width of the pool, you just need to rewrite your **formula**! Continue reading to find out how you can do this.

**Formulas** are used to calculate something specific, they contain an equals sign and at least two variables. A formula is considered to always be true.

Let's look at some examples of formulas.

Some examples of formulas that you may be familiar with include:

- Area of a circle = \(\pi r^2\)
- Area of a triangle = \(\frac{1}{2}bh\)
- Distance = \(rt\)

Formulas involve two or more quantities, and variables and are used to find different things based on the formulas themselves.

An **equation** involves an equals sign and can contain missing values. It shows you that what is on one side of the equal sign is equal to what is on the right.

Here are some examples.

Some examples of equations include:

- \(3x+7=13\)
- \(x-6=2\)

Since equations do not have to always be true, the above equations will only be true if the missing variable allows for the calculation to be correct. For example in the equation \(3x + 7 = 13\), \(x\) must equal 2 in order for the equation to be true.

It can be helpful to know how to rearrange formulas and equations. For example you may be given the area of a circle and asked to find the radius.

It will be helpful to know how to rewrite formulas so that it is possible to solve for a variable. When you rewrite a formula the aim is to create an equation that is equivalent to the formula but with the missing variable by itself. There are two main rules to follow to ensure that you are rewriting the formula correctly:

Whether you are adding, subtracting, multiplying, or dividing make sure that you do it to

**both sides**of the equation, this means both sides of the \(=\) sign.When multiplying or dividing, make sure you apply it to

**all of the terms**in the equation, don't leave any out.

Solve for \(H\) in the equation, \(H+3=b\).

Answer:

To do this you can subtract 3 from both sides of the equation:

\[H+3-3=b-3.\]

Then you can simplify to get

\[H=b-3.\]

Rewriting formulas and equations can be helpful when you are trying to find specific answers.

Find the length of the rectangle with an area of \(20\,cm^2\)^{ }and width of \(8\,cm \).

Answer:

To solve this you would first need to think about the correct formula to use. It can be helpful to first list all of the information you know:

- the shape is a rectangle
- the area is \(20\,cm^2\)
- the width is \(8\,cm \) .

Now you can tell that you are going to need the formula for the area of a rectangle, which is

\[A=lw\]

where

\[ \begin{align} A &= \text{ Area of rectangle} \\ l &= \text{ length of rectangle} \\ w &= \text{ width of rectangle} \end{align} \]

There are two ways you can go from here. Either one will give you exactly the same answer.

**Method one**: solve for the variable you need in the formula, then plug in values.

Lets have a look how to do this, you need to get \(l\) by itself. First, the formula for the area is

\[A=lw.\]

Since you want to isolate the length, divide both sides by \(w\) to get:

\[\frac{A}{w}=\frac{lw}{w}.\]

Then you can cancel on the right hand side to get:

\[\frac{A}{w}=l.\]

Now that you have rewritten the formula, you can input the variables and find the value of \(l\):

\[\frac{20}{8}=l\]

so including the units \(l=2.5\,cm\).

**Method two**: first plug in the information you have, then solve for the variable you need.

Plugging in the numbers you found above into the formula \(A=lw\),

\[ 20 = l\cdot 8.\]

Then you can solve for \(l\) by dividing both sides by \(8\) and simplifying:

\[ \frac{20}{8} = l \]

so

\[l = 2.5.\]

Don't forget to include the units! The length of the rectangle is \(l=2.5\,cm\).

When rewriting formulas, it is important to remember, if you divide on one side, you need to do the same to the other side of the equation.

You can rewrite algebraic equations too to help you find values of \(x\) or \(y\).

Solve \(5x+y=18\) for \(y\) when \(x=2\).

Answer:

To begin with, you need to rewrite the algebraic equation to make \(y\) the subject, to do this you can subtract \(5x\) from both sides of the equation

\[5x+y=18\]

giving you

\[-5x+5x+y=18-5x.\]

Then cancelling on the left side gives you

\[y=18-5x.\]

Now you can substitute the value of \(x = 2\) to solve for \(y\), which gives you

\[\begin{align} y&=18-5(2)\\ &=8. \end{align}\]

In algebra, the standard form is a way of writing out an equation, for a line it follows the form \(Ax+By=C\).

To rewrite an equation using standard form, you can simply move the terms of the equation around. Let's take a look at an example.

Rewrite the equation \(y=12-2x\).

Answer:

You can start by looking at the standard form:

\[Ax+By=C\]

To rewrite this in the standard form you need to move the \(x\) value to the same side of the equation as the \(y\) term. So adding \(2x\) to both sides of the equation gives you

\[2x+y=12.\]

This is now written in standard form!

It is important to remember that the form \(Ax+By=C\) is the standard form for a line, things like circles will have a different standard form.

A function takes an input and creates an output.

Let's look at some examples.

Functions are often written as:

- \(f(x)=x^2\)
- \(f(x)=2x^3\)

If you want to rewrite a function and turn it into an equation, you can simply replace the \(f(x)\) with a \(y\). There are lots of other properties of functions and things you can do with them. Take a look at the Functions article for more details.

Find the radius of the circle with a circumference of \(35\, in\).

Answer:

Let's begin by looking at an appropriate formula;

\[C=2 \pi r\]

where

\[ \begin{align} C &= \text{ circumference of the circle} \\ r &= \text{ radius of the circle}.\end{align}\]

Now you can rewrite the formula to help you find the radius. To do this you can divide both sides by \(2\pi\);

\[\frac{C}{2 \pi}=r.\]

Now you can input your variables into the formula to find the radius;

\[r=\frac{35}{2 \pi},\]

so

\[r \approx 5.57\]

Don't forget units! The radius of the circle is approximately \(5.57\, in\).

Let's take a look at another example.

Solve \(2x+2y=22\) for \(y\) when \(x=4\).

Answer:

Let's start by rewriting the formula in terms of \(y\). Since

\[2x+2y=22,\]

you get

\[2y=22-2x,\]

and then dividing both sides by \(2\)

\[\begin{align} y &=\frac{22-2x}{2} \\ &= 11-x. \end{align}\]

Now you can substitute in the value of \(x=4\), which gives you

\[\begin{align} y&=11 - 4 \\ &=7. \end{align}\]

- You can rewrite formulas and equations to help you find missing variables.
- It is important to remember that when you are subtracting, adding, multiplying or dividing, you need to do it to both sides of the equation.
- When you are multiplying or dividing, you need to remember to do it to all of the terms!

When rewriting an equation or a formula there are two important things to remember;

- When adding, subtracting, multiplying or dividing, you need to make sure you do it to both sides of the = sign.
- When multiplying or dividing, you need to remember to do it to all of the terms.

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