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Algebra - Changing the subject of an equation

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The subject of a formula is the variable that stands alone on one side of the equals sign (usually the left). For example, in y=mx+cy = mx + c, yy is the subject.

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To change the subject, use inverse operations to isolate the desired variable. If a variable is added, subtract it; if it is multiplied, divide it.

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Order of operations in reverse: When rearranging, it is often easiest to 'undo' addition and subtraction first, followed by multiplication and division, and finally powers or roots.

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If the new subject appears more than once in the equation (e.g., ax+b=cxax + b = cx), you must move all terms containing that variable to one side and factorise it out.

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When dealing with fractions, multiply the entire equation by the denominator to 'clear' the fraction before attempting to isolate the subject.

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Squaring and Square Roots: If the subject is squared (x2x^2), the inverse is taking the square root (x\sqrt{x}). Remember that x2=±x\sqrt{x^2} = \pm x in some algebraic contexts, though often only the positive root is required in geometry.

📐Formulae

y=x+a  ⟹  x=y−ay = x + a \implies x = y - a

y=ax  ⟹  x=yay = ax \implies x = \frac{y}{a}

y=x2  ⟹  x=yy = x^2 \implies x = \sqrt{y}

y=ax  ⟹  x=ayy = \frac{a}{x} \implies x = \frac{a}{y}

V=13Ah  ⟹  h=3VAV = \frac{1}{3}Ah \implies h = \frac{3V}{A}

💡Examples

Problem 1:

Given the formula for the final velocity v=u+atv = u + at, make tt the subject.

Solution:

t=v−uat = \frac{v - u}{a}

Explanation:

First, subtract uu from both sides to get v−u=atv - u = at. Next, divide both sides by aa to isolate tt.

Problem 2:

Make rr the subject of the formula for the area of a circle: A=πr2A = \pi r^2.

Solution:

r=Aπr = \sqrt{\frac{A}{\pi}}

Explanation:

Divide both sides by π\pi to get Aπ=r2\frac{A}{\pi} = r^2. Then, take the square root of both sides to solve for rr.

Problem 3:

Make xx the subject of the equation y=x+ax−by = \frac{x + a}{x - b}.

Solution:

x=by+ay−1x = \frac{by + a}{y - 1}

Explanation:

  1. Multiply both sides by (x−b)(x - b) to clear the fraction: y(x−b)=x+ay(x - b) = x + a.
  2. Expand the brackets: xy−by=x+axy - by = x + a.
  3. Move all terms with xx to one side and others to the other: xy−x=by+axy - x = by + a.
  4. Factorise xx: x(y−1)=by+ax(y - 1) = by + a.
  5. Divide by (y−1)(y - 1) to isolate xx.

Problem 4:

Rearrange P=2l+2wP = 2l + 2w to make ww the subject.

Solution:

w=P−2l2w = \frac{P - 2l}{2}

Explanation:

Subtract 2l2l from both sides: P−2l=2wP - 2l = 2w. Then divide the entire left side by 22.