# Rearranging formulae 2

Last updated: September 28, 2021

## Why do you need to know how to rearrange formulae?

Rearranging complicated formulae can be considered the peak of algebra skills. If you manipulate an equation and make your required variable the subject before adding the numeric values, solving for variables becomes easy. Only, solving a variable for several different values is necessary. There are some rules to follow while rearranging formulae and manipulating equations. In this article, we will learn about rearranging complicated formulae and the related examples.

## E2.1: Construct and rearrange complicated formulae and equations

Some important rules:

• You must keep both sides of an equation equal. Whatever you do to one side, you must do to the other, whether it is addition, subtraction, multiplication or division.

• If you are multiplying or dividing, you must do it to every term in the equation.

Solving equations involves rearranging and manipulating the terms in the equation until we have isolated the variable for which you are solving the equation. To manipulate equations, you must perform the same operations on both sides of the equation.

For example, consider an equation $$3x=12$$. If we divide the left-hand side of the equation by $$3$$, we must also divide the right-hand side by $$3$$ for the equation to remain true. Thus, $$3x=12$$ is equivalent to $$\frac{3x}{3}=\frac{12}{3}$$, which is equivalent to $$x=4$$. We have now isolated the variable $$x$$ and we can see that the solution is $$x=4$$.

### Worked examples of rearranging formulae

Example 1: Make $$a$$ the subject and find the formula for $$a$$ by rearranging $$t=\frac{1}{4}\sqrt{\frac{5a}{h}}$$.

Step 1: Consider the given formula.

The given formula is $$t=\frac{1}{4}\sqrt{\frac{5a}{h}}$$.

Step 2: Multiply the given formula by $$4$$ on both sides.

$$4t=4\times \frac{1}{4}\sqrt{\frac{5a}{h}}$$

Simplify the above equation.

$$4t=\sqrt{\frac{5a}{h}}$$

Step 3: Square both sides.

$$4t=\sqrt{\frac{5a}{h}}$$

Now, the square root on LHS gets cancelled by squaring both sides.

$$16t^2=\frac{15a}{h}$$

Step 4: Multiply by $$h$$.

Now, multiply by $$h$$ on both sides.

$$16t^2h=5a$$

Step 5: Divide by $$5$$ on both sides.

$$\frac{16t^2h}{5}=a$$

Hence, the final solution is of the form $$a=\frac{16t^2h}{5}$$.

Example 2: Make $$t$$ as the subject of the formula $$a-xt=b+yt$$. Find the formula for $$t$$.

Step 1: Consider the given formula.

The given formula is $$a-xt=b+yt$$.

Step 2: Add $$xt$$ on both sides.

Now start by collecting all the $$t$$ terms on the right-hand side.

Add $$xt$$ on both sides.

$$a-xt+xt=b+yt+xt$$

This gives $$a=b+yt+xt$$.

Step 3: Now shift the terms without $$t$$ on the left hand side by subtracting $$b$$ from both sides.

$$a-b=b+yt+xt-b$$

This gives $$a-b=yt+xt$$.

Step 4: Factorise the $$RHS$$ side.

Now factorise the right-hand side of the equation.

$$a-b=t(y+x)$$

Step 5: Divide by $$\left (y-x\right)$$ on both sides.

$$\frac{a-b}{y+x}=t\frac{y+x}{y+x}$$

This gives $$\frac{a-b}{y+x}=t$$

Hence, the required formula for $$t$$ is of the form $$t=\frac{a-b}{y+x}$$

Example 3: Make $$y$$ the subject of the formula $$x=\frac{y+z}{y-z}$$.

Step 1: Consider the given formula.

The given formula is of the form $$x=\frac{y+z}{y-z}$$

Step 2: Multiply both sides by $$\left (y-z\right)$$.

Now multiply both sides by $$\left (y-z\right)$$ in the given formula.

$$x\left (y-z\right)=\frac{y+z}{y-z}\times \left (y-z\right )$$

This gives $$x\left (y-z\right)={y+z}$$.

Step 3: Simplify the equation.

Now simplify the equation by expanding the bracket.

$$x\left (y-z\right)={y+z}$$

$$xy-xz=y+z$$

This implies that $$xy-y=z+zx$$

Step 4: Factorise the last equation.

$$y\left (x-1\right)=z\left (x+1\right)$$

Step 5: Divide by $$\left (x-1\right)$$ on both sides.

$$\frac{y\left (x-1\right)}{x-1}=\frac{z\left (x+1\right)}{x-1}$$

This implies that $$y=\frac{z\left (x+1\right)}{x-1}$$.