Make w the subject of the formula y = 3w - 10
w =
Rearranging (harder) — Notes & Worked Examples Show / Hide
Explanation:
When the letter you want to make the subject appears more than once in a formula, follow this standard 4-step method: (1) multiply out any fractions and expand brackets; (2) move all terms containing the target letter to one side and all other terms to the opposite side; (3) factorise out the target letter as a single common factor; (4) divide by the bracket.
If the target letter is trapped inside a square root or power, isolate the root or power term before squaring or taking roots of both sides.
Example 1: Make x the subject of y = 3x + 2x − 4
Solution:
Multiply by (x − 4): y(x − 4) = 3x + 2
Expand bracket: xy − 4y = 3x + 2
Collect x terms on left: xy − 3x = 4y + 2
Factorise out x: x(y − 3) = 4y + 2
Answer: x = 4y + 2y − 3
Example 2: Make m the subject of T = 2√(m + 1m − 3)
Solution:
Square both sides: T2 = 4(m + 1m − 3)
Multiply & expand: T2m − 3T2 = 4m + 4
Collect m terms: m(T2 − 4) = 3T2 + 4
Answer: m = 3T2 + 4T2 − 4
Make d the subject of the formula c = 8d + 5
d =
Worked Solution & Answer
Make x the subject of the formula y = 2x + 4
x =
Worked Solution & Answer
Rearrange g = 3h - 1 to make h the subject.
Answer
Worked Solution & Answer
Make k the subject of the formula y = √(2m - k)
k =
Worked Solution & Answer
Square both sides: y2 = 2m - k ⇒ k = 2m - y2 M1 A1
Answer: k = k = 2m - y2
Make a the subject of the formula p = 3a - 9
a =
Worked Solution & Answer
Make n the subject of the formula x = 3p + n3n - 4
n =
Worked Solution & Answer
x(3n − 4) = 3p + n M1
⇒ 3nx − 4x = 3p + n
⇒ 3nx − n = 3p + 4x M1
⇒ n(3x − 1) = 3p + 4x
⇒ n = (3p + 4x)/(3x − 1). A1
⇒ 3nx − 4x = 3p + n
⇒ 3nx − n = 3p + 4x M1
⇒ n(3x − 1) = 3p + 4x
⇒ n = (3p + 4x)/(3x − 1). A1
Answer: n = n = 3p + 4x3x - 1
Make m the subject of k = p + 2m5
m =
Worked Solution & Answer
k – p = 2m/5 M1
⇒ 5(k – p) = 2m M1
⇒ m = 5(k – p)/2 A1
⇒ 5(k – p) = 2m M1
⇒ m = 5(k – p)/2 A1
Answer: m = m = 5(k - p)2
Make p the subject of the formula t = 2(2p - 3)5 - 2p
p =
Worked Solution & Answer
t(5 – 2p) = 4p – 6 M1
⇒ 5t – 2pt = 4p – 6 M1
⇒ 5t + 6 = 4p + 2pt M1
⇒ p(4 + 2t) = 5t + 6
⇒ p = (5t + 6)/(4 + 2t) A1
⇒ 5t – 2pt = 4p – 6 M1
⇒ 5t + 6 = 4p + 2pt M1
⇒ p(4 + 2t) = 5t + 6
⇒ p = (5t + 6)/(4 + 2t) A1
Answer: p = p = 5t + 64 + 2t
Make t the subject of the formula c = t2 + 37 - 8t2
t =
Worked Solution & Answer
c(7 – 8t²) = t² + 3 M1
⇒ 7c – 3 = t² + 8ct² M1
⇒ t²(1 + 8c) = 7c – 3 M1
⇒ t = ±√((7c – 3)/(1 + 8c)) A1
⇒ 7c – 3 = t² + 8ct² M1
⇒ t²(1 + 8c) = 7c – 3 M1
⇒ t = ±√((7c – 3)/(1 + 8c)) A1
Answer: t = t = ±√(<span class="opg-frac"><span class="opg-num">7c - 3</span><span class="opg-den">1 + 8c</span></span>)