In many applied problems, an equation may involve several variables, but you need to solve for just one of them in terms of the others. This process is essential in science, engineering, and everyday life. In general, the approach depends on whether the equation is linear or quadratic (in the variable of interest), or if it can be rearranged to match one of those types.

Example 1. Ideal Gas Law (Chemistry/Physics) 

Equation: \[PV = nRT,\] where

  • \(P\) is the pressure,
  • \(V\) is the volume,
  • \(n\) is the number of moles of gas,
  • \(R\) is the universal gas constant,
  • \(T\) is the temperature.

Goal: Solve for \(T\) in terms of \(P\) , \(V\) , \(n\) , and \(R\) .

 

Solution

 

  • Since the equation is linear in \(T\) , isolate \(T\) by dividing both sides by \(nR\) : \[T = \frac{PV}{nR}.\] 

 

Example 2. Perimeter of a Rectangle (Everyday Geometry) 

Equation: \[P = 2L + 2W,\] where

  • \(P\) is the perimeter of a rectangle,
  • \(L\) is the length,
  • \(W\) is the width.

Goal: Solve for \(L\) in terms of \(P\) and \(W\) .

 

Solution

 

  • Subtract \(2W\) from both sides: \[P - 2W = 2L.\]
  • Divide both sides by \(2\) to isolate \(L\) : \[L = \frac{P - 2W}{2}.\] 

 

Example 3. Projectile Motion (Physics/Engineering) 

Equation (vertical position): \[y = -16t^2 + v_0 t + y_0,\] where

  • \(t\) is the time (in seconds),
  • \(v_0\) is the initial velocity,
  • \(y_0\) is the initial height,
  • \(-16\) is the constant reflecting gravitational acceleration (ft/s \(^2\) ).

Goal: Solve for \(t\) in terms of \(y\) , \(v_0\) , and \(y_0\) .

 

Solution

 

  • Rewrite the equation to set it equal to zero: \[-16t^2 + v_0 t + y_0 - y = 0.\]
  • Use the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) , letting \(a = -16\) , \(b = v_0\) , and \(c = y_0 - y\) : \[t = \frac{-v_0 \pm \sqrt{(v_0)^2 - 4(-16)(y_0 - y)}}{2(-16)}.\]
  • Simplify: \[t = \frac{-v_0 \pm \sqrt{v_0^2 + 64(y_0 - y)}}{-32}.\] 

 

Example 4. Lens Formula (Optics) 

Equation: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i},\] where

  • \(f\) is the focal length of the lens,
  • \(d_o\) is the distance from the lens to the object,
  • \(d_i\) is the distance from the lens to the image.

Goal: Solve for \(d_o\) in terms of \(f\) and \(d_i\) .

 

Solution

 

  • Isolate \(\frac{1}{d_o}\) : \[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}.\]
  • Combine the right-hand side: \[\frac{1}{f} - \frac{1}{d_i} = \frac{d_i - f}{f\,d_i}.\]
  • Therefore: \[\frac{1}{d_o} = \frac{d_i - f}{f\,d_i}.\]
  • Take the reciprocal: \[d_o = \frac{f\,d_i}{d_i - f}.\] 

 

Example 5. Period of a Pendulum (Physics) 

Equation (approximate): \[T = 2\pi \sqrt{\frac{L}{g}},\] where

  • \(T\) is the period (in seconds),
  • \(L\) is the length of the pendulum (in meters),
  • \(g\) is the acceleration due to gravity (in m/s \(^2\) ).

Goals: 

  • A: Solve for \(L\) in terms of \(T\) and \(g\) .
  • B: Solve for \(g\) in terms of \(T\) and \(L\) .

 

Solution

 

  • Goal A: Square both sides and solve for \(L\) . \[T^2 = 4\pi^2 \frac{L}{g} \quad \Longrightarrow \quad L = \frac{g\,T^2}{4 \pi^2}.\]
  • Goal B: From \(T^2 = 4\pi^2 \frac{L}{g}\) , rearrange to solve for \(g\) . \[g = 4\pi^2 \frac{L}{T^2}.\] 

 

Example 6. Pythagorean Theorem (Geometry/Engineering) 

Equation: \[a^2 + b^2 = c^2,\] where

  • \(a\) and \(b\) are the legs of a right triangle,
  • \(c\) is the hypotenuse.

Goal: Solve for \(b\) in terms of \(a\) and \(c\) .

 

Solution

 

  • Subtract \(a^2\) from both sides: \[b^2 = c^2 - a^2.\]
  • Take the positive square root (assuming \(b>0\) ): \[b = \sqrt{c^2 - a^2}.\]