Substitution resulting in an equation

In Using Your Formulae, we looked at how to substitute values into common equations to evaluate the subject of the formulae. Sometimes we don't know all of the values, and we just want to substitute the ones that we do know. This leaves us with a simpler equation that involves less variables.

Examples

Question 1

In Physics, Ohm's Law states that $V=I\times R$V=I×R for a conductor, where $V$V is the voltage (in Volts), $I$I is the current (in Amps) and $R$R is the resistance (in Ohms). If a particular object in a circuit has a resistance of $100$100 Ohms, write the voltage in terms of the current.

Think: We want to substitute the given value of $R$R, so that we are left with an equation using only $V$V and $I$I.

Do: We have that $R=100$R=100, so $V=I\times100$V=I×100, or $V=100I$V=100I.

 

Question 2

The area $A$A of a trapezium is given by the equation $A=\frac{1}{2}h\left(a+b\right)$A=12​h(a+b), where $a$a and $b$b are the lengths of the parallel sides and $h$h is the perpendicular distance between them. If the perpendicular height of a trapezium is $12$12 cm, write an equation for the area in terms of the lengths $a$a and $b$b.

Think: We want to substitute the given value of $h$h, so that we are left with an equation for $A$A using only $a$a and $b$b. We will also be able to simplify this time.

Do: We have that $h=12$h=12, so $A=\frac{1}{2}\times12\left(a+b\right)$A=12​×12(a+b), which simplifies to $A=6\left(a+b\right)$A=6(a+b).

Question 3

Question 4

Question 5