Change of Subject Solver
Change the subject from y to x (or vice versa) for y=ax+b, y=(ax+b)/(cx+d) and y=ax²+bx+c, with steps.
Input Data
Results
At a glance:Changing the subject isolates one variable on one side. DSE Paper 1 often asks to make x the subject of y=(ax+b)/(cx+d): clear denominators, expand, gather x terms, factor x, then divide. Quadratic forms give a ± pair.
Formula
Linear: y = a x + b ⇒ x = (y − b)/a.
Fractional: y = (a x + b)/(c x + d) ⇒ x = (b − d y)/(c y − a).
Quadratic: y = a x² + b x + c ⇒ x = [−b ± √(b² − 4a(c − y))]/(2a).
$$x=\frac{y-b}{a}$$$$x=\frac{b-dy}{cy-a}$$$$x=\frac{-b\pm\sqrt{b^{2}-4a(c-y)}}{2a}$$How to Use
- Pick a form (linear / fractional / quadratic).
- Pick the target subject (x or y).
- Enter parameters a, b, c (and d for fractional) and the known variable.
- The tool returns the subject value, the rearranged formula and the steps.
Common change-of-subject forms
| Original | Subject x | Note |
|---|---|---|
| y = a x + b | x = (y−b)/a | linear |
| y = (a x + b)/(c x + d) | x = (b − d y)/(c y − a) | fractional |
| y = a x² + b x + c | x = [−b ± √(b²−4a(c−y))]/(2a) | quadratic |
Fractional needs c y − a ≠ 0; quadratic needs the radicand ≥ 0 for real roots.
Case Studies
y = (2x+3)/(x+4) → x
Clear denom: y(x+4)=2x+3.
xy+4y=2x+3 → x(y−2)=3−4y → x=(3−4y)/(y−2).
y = 2x+3 → x
2x = y−3 ⇒ x=(y−3)/2.
FAQ
What is the subject?
The subject is the variable isolated on its own, e.g. y in y=ax+b. Changing subject rewrites it with x isolated.
How to clear the denominator in a fraction?
Multiply both sides by (cx+d), expand, then collect x terms, factor x, and divide by its coefficient.
Why ± for quadratic?
x²=k has two real roots ±√k, so a square root appears and generally gives two solutions.
What if c y − a = 0?
The denominator c y − a becomes 0, so x has no unique value (the original degenerates).
Related Tools
References
Content review: Calculatorism Science Team. The three change-of-subject formulas and their steps have been verified. Results are for reference only.