Changing the Subject
Formula rearrangement — 48 items
Prepared by Miss Clarissa Ng
www.clartutors.com
www.clartutors.com
Attempt each part first — the working space is there for your own line of reasoning. The full worked solutions follow on a new page, one line per step, so the questions can be printed on their own.
Q1(a) y + w = c
Q1(b) y − p = m
Q1(c) m + y = s
Q1(d) y − 2g = n
Q1(e) 3y = c
Q1(f) ay = w
Q1(g) yc = w
Q1(h) ya = 2c
Q1(i) a = y + p
Q1(j) c = y − k
Q1(k) y² = s
Q1(l) y³ = x
Q1(m) √y = g
Q1(n) πy = c
Q1(o) n − y = t
Q1(p) ry = c
Q1(q) 4πy = b
Q1(r) y + 7t = c + r
Q1(s) ry = w
Q1(t) y² = k + x
Q1(u) A = xy
Q2(a) 4x + c = w
Q2(b) dx − t = 8
Q2(c) x² + 3 = h
Q2(d) 2x + 2y = P
Q2(e) s = x² − 3
Q2(f) y = xz + s
Q2(g) xn + 2 = w
Q2(h) x6 − 5 = w
Q2(i) x + 3c = h
Q2(j) 3y = 4x + 1
Q2(k) x² + a = v
Q2(l) x³ − 4 = 5y
Q2(m) x + tm = 2c
Q2(n) w + xu = 3z
Q2(o) A = πx²
Q2(p) A = ½bx
Q2(q) V = abx
Q2(r) v² = u² + 2ax
Q2(s) a + bx = r
Q2(t) 5cxb = a
Q2(u) ∛xk = w
Q3(a) (a + c)² = t
Q3(b) v = u + ac
Q3(c) v = πc²h
Apply 1 The circumference of a circle is given as c = 2πr. Make the radius, r, the subject.
Apply 2 The formula to convert degrees Fahrenheit to degrees Celsius is 59(F − 32) = C. Find the formula to convert from degrees Celsius to degrees Fahrenheit by making F the subject.
Apply 3 Can you spot any mistakes below?
Every line names the operation and what it gives, ending in the finished formula. The rearrangements were re-checked by solving each original equation for its subject.
| Q1(a) | subtract w y = c − w |
| Q1(b) | add p y = m + p |
| Q1(c) | subtract m y = s − m |
| Q1(d) | add 2g y = n + 2g |
| Q1(e) | divide by 3 y = c3 |
| Q1(f) | divide by a y = wa |
| Q1(g) | ×c y = cw |
| Q1(h) | ×a y = 2ac |
| Q1(i) | subtract p y = a − p |
| Q1(j) | add k y = c + k |
| Q1(k) | square root both sides y = √s |
| Q1(l) | cube root both sides y = ∛x |
| Q1(m) | square both sides y = g² |
| Q1(n) | divide by π y = cπ |
| Q1(o) | subtract n: −y = t − n → ×(−1) y = n − t |
| Q1(p) | divide by r y = cr |
| Q1(q) | divide by 4π y = b4π |
| Q1(r) | subtract 7t y = c + r − 7t |
| Q1(s) | ×y: r = wy → divide by w y = rw |
| Q1(t) | square root both sides y = √(k + x) |
| Q1(u) | divide by x y = Ax |
| Q2(a) | subtract c: 4x = w − c → divide by 4 x = w − c4 |
| Q2(b) | add t: dx = 8 + t → divide by d x = 8 + td |
| Q2(c) | subtract 3 → square root x = √(h − 3) |
| Q2(d) | subtract 2y: 2x = P − 2y → divide by 2 x = P − 2y2 |
| Q2(e) | add 3 → square root x = √(s + 3) |
| Q2(f) | subtract s: y − s = xz → divide by z x = y − sz |
| Q2(g) | subtract 2: xn = w − 2 → ×n x = n(w − 2) |
| Q2(h) | add 5: x6 = w + 5 → ×6 x = 6(w + 5) |
| Q2(i) | ×c: x + 3 = ch → subtract 3 x = ch − 3 |
| Q2(j) | subtract 1: 3y − 1 = 4x → divide by 4 x = 3y − 14 |
| Q2(k) | subtract a → square root x = √(v − a) |
| Q2(l) | add 4: x³ = 5y + 4 → cube root x = ∛(5y + 4) |
| Q2(m) | ×m: x + t = 2cm → subtract t x = 2cm − t |
| Q2(n) | ×u: w + x = 3uz → subtract w x = 3uz − w |
| Q2(o) | divide by π: Aπ = x² → square root x = √Aπ |
| Q2(p) | ×2: 2A = bx → divide by b x = 2Ab |
| Q2(q) | divide by ab x = Vab |
| Q2(r) | subtract u²: v² − u² = 2ax → divide by 2a x = v² − u²2a |
| Q2(s) | ×x: a + b = rx → divide by r x = a + br |
| Q2(t) | ×b: 5cx = ab → divide by 5c x = ab5c |
| Q2(u) | cube both sides: xk = w³ → ×k x = kw³ |
| Q3(a) | square root: a + c = √t → subtract a c = √t − a |
| Q3(b) | subtract u: v − u = ac → divide by a c = v − ua |
| Q3(c) | divide by πh: vπh = c² → square root c = √vπh |
| Apply 1 | divide by 2π r = c2π |
| Apply 2 | ×9: 5(F − 32) = 9C → divide by 5: F − 32 = 95C → add 32 F = 95C + 32 |
| (i) | The a must come off before the square root: k − a = y² gives y = √(k − a). You cannot square-root the two terms separately. |
| (ii) | The first line is right (t − 1 = v ÷ 4), but then both sides must be multiplied by 4, not divided: v = 4(t − 1). |