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

Let's solve both Question 1 and Question 2 step by step, making the required variable the subject of each equation.

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✔ Question 1: Make $ y $ the subject

We'll isolate $ y $ in each equation.

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

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

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

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

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

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(f) $ ay = w $
Divide by $ a $:
$$
y = \frac{w}{a}
$$

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

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

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

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

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(k) $ y^2 = s $
Take square root:
$$
y = \pm\sqrt{s}
$$
*(Note: Two solutions due to squaring)*

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(l) $ y^3 = x $
Take cube root:
$$
y = \sqrt[3]{x}
$$

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

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

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(o) $ n - y = t $
Rearrange:
$$
-y = t - n \Rightarrow y = n - t
$$

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

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

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

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(s) $ \frac{r}{y} = w $
Multiply both sides by $ y $:
$$
r = wy \Rightarrow y = \frac{r}{w}
$$

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

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(u) $ A = xy $
Divide by $ x $:
$$
y = \frac{A}{x}
$$

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✔ Question 1 Summary (All answers):

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

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✔ Question 2: Make $ x $ the subject

Now we isolate $ x $ in each formula.

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

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

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

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

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

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

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

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

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(i) $ \frac{x + 3}{c} = h $
Multiply by $ c $:
$$
x + 3 = hc \Rightarrow x = hc - 3
$$

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

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

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

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

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

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

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

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

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

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(s) $ \frac{a + b}{x} = r $
Multiply by $ x $:
$$
a + b = rx \Rightarrow x = \frac{a + b}{r}
$$

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(t) $ \frac{5cx}{b} = a $
Multiply by $ b $:
$$
5cx = ab \Rightarrow x = \frac{ab}{5c}
$$

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

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✔ Question 2 Summary (All answers):

| Equation | Answer |
|--------|--------|
| (a) $ 4x + c = w $ | $ x = \frac{w - c}{4} $ |
| (b) $ dx - t = 8 $ | $ x = \frac{8 + t}{d} $ |
| (c) $ x^2 + 3 = h $ | $ x = \pm\sqrt{h - 3} $ |
| (d) $ 2x + 2y = P $ | $ x = \frac{P - 2y}{2} $ |
| (e) $ s = x^2 - 3 $ | $ x = \pm\sqrt{s + 3} $ |
| (f) $ y = xz + s $ | $ x = \frac{y - s}{z} $ |
| (g) $ \frac{x}{n} + 2 = w $ | $ x = n(w - 2) $ |
| (h) $ \frac{x}{6} - 5 = w $ | $ x = 6(w + 5) $ |
| (i) $ \frac{x + 3}{c} = h $ | $ x = hc - 3 $ |
| (j) $ 3y = 4x + 1 $ | $ x = \frac{3y - 1}{4} $ |
| (k) $ x^2 + a = v $ | $ x = \pm\sqrt{v - a} $ |
| (l) $ x^3 - 4 = 5y $ | $ x = \sqrt[3]{5y + 4} $ |
| (m) $ \frac{x + t}{m} = 2c $ | $ x = 2cm - t $ |
| (n) $ \frac{w + x}{u} = 3z $ | $ x = 3zu - w $ |
| (o) $ A = \pi x^2 $ | $ x = \pm\sqrt{\frac{A}{\pi}} $ |
| (p) $ A = \frac{1}{2}bx $ | $ x = \frac{2A}{b} $ |
| (q) $ V = abx $ | $ x = \frac{V}{ab} $ |
| (r) $ v^2 = u^2 + 2ax $ | $ x = \frac{v^2 - u^2}{2a} $ |
| (s) $ \frac{a + b}{x} = r $ | $ x = \frac{a + b}{r} $ |
| (t) $ \frac{5cx}{b} = a $ | $ x = \frac{ab}{5c} $ |
| (u) $ \sqrt[3]{\frac{x}{k}} = w $ | $ x = kw^3 $ |

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✔ Final Notes:

- When solving equations involving squares or cubes, remember to include ± for even roots.
- Always check for division by zero if variables are in denominators.
- The process is about rearranging algebraically, using inverse operations.

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