y9. Algebra. Changing Subject of Formula - Maths with David - Free Printable

Let’s solve each part step by step.

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Question 1: Make y the subject

We need to rearrange each equation so that y = … (i.e., isolate y on one side).

(a) $ y + w = c $
→ Subtract $ w $ from both sides:
$ y = c - w $

(b) $ y - p = m $
→ Add $ p $ to both sides:
$ y = m + p $

(c) $ m + y = s $
→ Subtract $ m $:
$ y = s - m $

(d) $ y - 2g = n $
→ Add $ 2g $:
$ y = n + 2g $

(e) $ 3y = c $
→ Divide both sides by 3:
$ y = \frac{c}{3} $

(f) $ ay = w $
→ Divide both sides by $ a $ (assuming $ a \ne 0 $):
$ y = \frac{w}{a} $

(g) $ \frac{y}{c} = w $
→ Multiply both sides by $ c $:
$ y = wc $

(h) $ \frac{y}{a} = 2c $
→ Multiply both sides by $ a $:
$ y = 2ac $

(i) $ a = y + p $
→ Subtract $ p $:
$ y = a - p $

(j) $ c = y - k $
→ Add $ k $:
$ y = c + k $

(k) $ y^2 = s $
→ Take square root of both sides:
$ y = \pm\sqrt{s} $
(But unless specified, usually we give both roots. However, in many school contexts, they may expect just $ y = \sqrt{s} $ if only real positive is considered — but strictly, it's ±. Let’s keep both for accuracy.)

(l) $ y^3 = x $
→ Cube root both sides:
$ y = \sqrt[3]{x} $

(m) $ \sqrt{y} = g $
→ Square both sides:
$ y = g^2 $

(n) $ \pi y = c $
→ Divide by $ \pi $:
$ y = \frac{c}{\pi} $

(o) $ n - y = t $
→ Subtract $ n $: $ -y = t - n $
→ Multiply by –1:
$ y = n - t $

(p) $ ry = c $
→ Divide by $ r $:
$ y = \frac{c}{r} $

(q) $ 4\pi y = b $
→ Divide by $ 4\pi $:
$ y = \frac{b}{4\pi} $

(r) $ y + 7t = c + r $
→ Subtract $ 7t $:
$ y = c + r - 7t $

(s) $ \frac{r}{y} = w $
→ Multiply both sides by $ y $: $ r = wy $
→ Then divide by $ w $:
$ y = \frac{r}{w} $ (assuming $ w \ne 0 $)

(t) $ y^2 = k + x $
→ Take square root:
$ y = \pm\sqrt{k + x} $

(u) $ A = xy $
→ Divide both sides by $ x $:
$ y = \frac{A}{x} $ (assuming $ x \ne 0 $)

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Question 2: Make x the subject

Now isolate x.

(a) $ 4x + c = w $
→ Subtract $ c $: $ 4x = w - c $
→ Divide by 4:
$ x = \frac{w - c}{4} $

(b) $ dx - t = 8 $
→ Add $ t $: $ dx = 8 + t $
→ Divide by $ d $:
$ x = \frac{8 + t}{d} $

(c) $ x^2 + 3 = h $
→ Subtract 3: $ x^2 = h - 3 $
→ Square root:
$ x = \pm\sqrt{h - 3} $

(d) $ 2x + 2y = P $
→ Subtract $ 2y $: $ 2x = P - 2y $
→ Divide by 2:
$ x = \frac{P - 2y}{2} = \frac{P}{2} - y $

(e) $ s = x^2 - 3 $
→ Add 3: $ x^2 = s + 3 $
→ Square root:
$ x = \pm\sqrt{s + 3} $

(f) $ y = xz + s $
→ Subtract $ s $: $ y - s = xz $
→ Divide by $ z $:
$ x = \frac{y - s}{z} $

(g) $ \frac{x}{n} + 2 = w $
→ Subtract 2: $ \frac{x}{n} = w - 2 $
→ Multiply by $ n $:
$ x = n(w - 2) $

(h) $ \frac{x}{6} - 5 = w $
→ Add 5: $ \frac{x}{6} = w + 5 $
→ Multiply by 6:
$ x = 6(w + 5) $

(i) $ \frac{x + 3}{c} = h $
→ Multiply by $ c $: $ x + 3 = ch $
→ Subtract 3:
$ x = ch - 3 $

(j) $ 3y = 4x + 1 $
→ Subtract 1: $ 3y - 1 = 4x $
→ Divide by 4:
$ x = \frac{3y - 1}{4} $

(k) $ x^2 + a = v $
→ Subtract $ a $: $ x^2 = v - a $
→ Square root:
$ x = \pm\sqrt{v - a} $

(l) $ x^3 - 4 = 5y $
→ Add 4: $ x^3 = 5y + 4 $
→ Cube root:
$ x = \sqrt[3]{5y + 4} $

(m) $ \frac{x + t}{m} = 2c $
→ Multiply by $ m $: $ x + t = 2cm $
→ Subtract $ t $:
$ x = 2cm - t $

(n) $ \frac{w + x}{u} = 3z $
→ Multiply by $ u $: $ w + x = 3zu $
→ Subtract $ w $:
$ x = 3zu - w $

(o) $ A = \pi x^2 $
→ Divide by $ \pi $: $ x^2 = \frac{A}{\pi} $
→ Square root:
$ x = \pm\sqrt{\frac{A}{\pi}} $

(p) $ A = \frac{1}{2}bx $
→ Multiply both sides by 2: $ 2A = bx $
→ Divide by $ b $:
$ x = \frac{2A}{b} $

(q) $ V = abx $
→ Divide by $ ab $:
$ x = \frac{V}{ab} $

(r) $ v^2 = u^2 + 2ax $
→ Subtract $ u^2 $: $ v^2 - u^2 = 2ax $
→ Divide by $ 2a $:
$ x = \frac{v^2 - u^2}{2a} $

(s) $ \frac{a + b}{x} = r $
→ Multiply both sides by $ x $: $ a + b = rx $
→ Divide by $ r $:
$ x = \frac{a + b}{r} $

(t) $ \frac{5cx}{b} = a $
→ Multiply both sides by $ b $: $ 5cx = ab $
→ Divide by $ 5c $:
$ x = \frac{ab}{5c} $

(u) $ \sqrt[3]{\frac{x}{k}} = w $
→ Cube both sides: $ \frac{x}{k} = w^3 $
→ Multiply by $ k $:
$ x = kw^3 $

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Now double-check a few tricky ones:

- (k) $ y^2 = s $ → $ y = \pm\sqrt{s} $ ✔️
- (t) $ y^2 = k + x $ → $ y = \pm\sqrt{k+x} $ ✔️
- (s) $ \frac{r}{y} = w $ → $ y = \frac{r}{w} $ ✔️
- (u) $ A = xy $ → $ y = \frac{A}{x} $, correct
- (l) $ x^3 = 5y + 4 $ → cube root ✔️
- (u) cube root equation: yes, cube both sides first.

All steps look solid.

Final Answer will list all answers clearly.

Since the user asked only for the final answer (and not explanation), and ExplainAnswer = false, we output only the final answers in order.

But per instructions: *Complete ALL reasoning and explanations BEFORE writing the Final Answer.* So now we write the Final Answer section with all results.

We’ll format as:

Question 1:
(a) y = c − w
(b) y = m + p
… etc.

Question 2:
(a) x = (w − c)/4
… etc.

Let’s compile cleanly.

Final Answer:
Question 1:
(a) $ y = c - w $
(b) $ y = m + p $
(c) $ y = s - m $
(d) $ y = n + 2g $
(e) $ y = \dfrac{c}{3} $
(f) $ y = \dfrac{w}{a} $
(g) $ y = wc $
(h) $ y = 2ac $
(i) $ y = a - p $
(j) $ y = c + k $
(k) $ y = \pm\sqrt{s} $
(l) $ y = \sqrt[3]{x} $
(m) $ y = g^2 $
(n) $ y = \dfrac{c}{\pi} $
(o) $ y = n - t $
(p) $ y = \dfrac{c}{r} $
(q) $ y = \dfrac{b}{4\pi} $
(r) $ y = c + r - 7t $
(s) $ y = \dfrac{r}{w} $
(t) $ y = \pm\sqrt{k + x} $
(u) $ y = \dfrac{A}{x} $

Question 2:
(a) $ x = \dfrac{w - c}{4} $
(b) $ x = \dfrac{8 + t}{d} $
(c) $ x = \pm\sqrt{h - 3} $
(d) $ x = \dfrac{P - 2y}{2} $
(e) $ x = \pm\sqrt{s + 3} $
(f) $ x = \dfrac{y - s}{z} $
(g) $ x = n(w - 2) $
(h) $ x = 6(w + 5) $
(i) $ x = ch - 3 $
(j) $ x = \dfrac{3y - 1}{4} $
(k) $ x = \pm\sqrt{v - a} $
(l) $ x = \sqrt[3]{5y + 4} $
(m) $ x = 2cm - t $
(n) $ x = 3zu - w $
(o) $ x = \pm\sqrt{\dfrac{A}{\pi}} $
(p) $ x = \dfrac{2A}{b} $
(q) $ x = \dfrac{V}{ab} $
(r) $ x = \dfrac{v^2 - u^2}{2a} $
(s) $ x = \dfrac{a + b}{r} $
(t) $ x = \dfrac{ab}{5c} $
(u) $ x = kw^3 $