Changing the Subject of a Formula
The aim here is to isolate whatever we want to be the subject of the formula.
Change the subject of the formula \[ V = IR \] to make \( I \) the subject.
Step 1: Start with the original formula
\[ V = IR \]
Step 2: Divide both sides by \( R \)
To get \( I \) on its own, divide both sides of the equation by \( R \).
\[ \frac{V}{R} = I \]
Final Answer
\[ I = \frac{V}{R} \]
Change the Subject of the Formula \[ A = BC - 3D \] to make \( C \) the subject.C
Step 1: Add \( 3D \) to both sides
Move the \(-3D\) term to the left by adding \( 3D \) to both sides.
\[ A + 3D = BC \]
Step 2: Divide both sides by \( B \)
To leave \( C \) on its own, divide both sides of the equation by \( B \).
\[ \frac{A + 3D}{B} = C \]
Final Answer
\[ C = \frac{A + 3D}{B} \]
Change the Subject of the Formula \[ l = \tfrac14 yu + k \] to make \( u \) the subject.
Problem:
Change the subject of the formula
.
Step 1: Subtract \( k \) from both sides
Move the constant term to the left-hand side.
\[ l - k = \tfrac14 yu \]
Step 2: Multiply both sides by 4
This removes the fraction.
\[ 4(l - k) = yu \]
Step 3: Divide both sides by \( y \)
This leaves \( u \) on its own.
\[ u = \frac{4(l - k)}{y} \]
Final Answer
\[ u = \frac{4(l - k)}{y} \]
Change the Subject of the Formula \[ s = ut + \tfrac12 a t^2 \] to make \( u \) the subject.
Step 1: Move the \( \tfrac12 a t^2 \) term
Subtract \( \tfrac12 a t^2 \) from both sides to isolate the \( ut \) term.
\[ s - \tfrac12 a t^2 = ut \]
Step 2: Divide by \( t \)
To leave \( u \) on its own, divide both sides by \( t \).
\[ \frac{s - \tfrac12 a t^2}{t} = u \]
Step 3: Split the fraction
Separate the expression into two simpler terms.
\[ u = \frac{s}{t} - \tfrac12 a t \] \[ u = t\left( \frac{s}{t^{2}} - \frac12 a \right) \]
Final Answer
\[ u = t\left( \frac{s}{t^{2}} - \frac12 a \right) \]
Change the Subject — Long vs Short Method
Problem:
Solve the equation for \( A \):
\[
\frac{3(A+8)}{20} = \frac{4(A-16)}{5}
\]
Long Method
\[ \frac{3(A+8)}{20} = \frac{4(A-16)}{5} \]
Multiply both sides by 20: \[ 3(A+8) = \frac{4(A-16)}{5} \times 20 \]
\[ 3(A+8) = 4(A-16) \times 4 \]
\[ 3(A+8) = 16(A-16) \]
Expand: \[ 3A + 24 = 16A - 384 \]
Collect like terms: \[ 24 = 19A - 384 \] \[ 408 = 19A \]
Divide by 19: \[ A = \frac{408}{19} \] \[ A = 21\frac{9}{19} \]
Short Method
Start with: \[ \frac{3(A+8)}{20} = \frac{4(A-16)}{5} \]
Cross‑multiply: \[ 3(A+8) \times 5 = 20 \times 4(A-16) \]
Simplify: \[ 15(A+8) = 80(A-16) \]
Expand: \[ 15A + 120 = 80A - 1280 \]
Collect like terms: \[ 1400 = 65A \]
Divide: \[ A = \frac{1400}{65} \] \[ A = \frac{408}{19} \]
Final Answer
\[ A = \frac{408}{19} = 21\frac{9}{19} \]
