Rearranging Formulae

Multiply both sides by 3 first, then divide by whatever is multiplying h.

1. Multiply both sides by 3: \;3V=x^2h
2. Divide both sides by x^2
3. Answer: h=\dfrac{3V}{x^2}

Clear the \frac{1}{2} and the t first, so that the bracket is left on its own.

1. Multiply both sides by 2: \;2s=(u+v)t
2. Divide both sides by t: \;\dfrac{2s}{t}=u+v
3. Subtract v from both sides
4. Answer: u=\dfrac{2s}{t}-v

Write down how much wire the square uses and how much the triangle uses: together they make 100 cm.

1. The square uses 4x cm of wire and the triangle uses 3y cm
2. So 4x+3y=100
3. Subtract 4x: \;3y=100-4x
4. Divide by 3
5. Answer: y=\dfrac{100-4x}{3}

Multiply both sides by d^2 so that d is no longer on the bottom of the fraction.

1. Multiply both sides by d^2: \;Bd^2=\dfrac{W}{4\pi}
2. Divide both sides by B: \;d^2=\dfrac{W}{4\pi B}
3. Square root both sides (d is a distance, so it is positive)
4. Answer: d=\sqrt{\dfrac{W}{4\pi B}}

Get t^3 on its own first, and only then take the cube root.

1. Add 5 to both sides: \;M+5=2t^3
2. Divide both sides by 2: \;t^3=\dfrac{M+5}{2}
3. Cube root both sides (the root covers the whole fraction)
4. Answer: t=\sqrt[3]{\dfrac{M+5}{2}}

Multiply both sides by a, then collect both of the a terms on the same side.

1. Multiply both sides by a: \;ha=2a-3b
2. Collect the a terms: \;2a-ha=3b
3. Factorise: \;a(2-h)=3b
4. So a=\dfrac{3b}{2-h}
5. \dfrac{3b}{h-2} has the wrong sign and \dfrac{3b}{h+2} comes from adding ha instead of subtracting it
6. a=\dfrac{2a-3b}{h} still has a on the right, so a is not the subject
7. Answer: a=\dfrac{3b}{2-h}

Square both sides to remove the square root, then make a the subject.

1. Square both sides: \;v^2=u^2+2as
2. So a=\dfrac{v^2-u^2}{2s}
3. a=\dfrac{169-25}{16}=\dfrac{144}{16}
4. Answer: a=9

Get the square root on its own on one side before you square.

1. Add \sqrt{2p+q} and subtract t: \;\sqrt{2p+q}=3-t
2. Square both sides: \;2p+q=(3-t)^2
3. Subtract q: \;2p=(3-t)^2-q
4. Divide by 2
5. Answer: p=\dfrac{(3-t)^2-q}{2}

Undo the operations in reverse order: deal with the -1 and the 3 before you square.

1. Add 1 to both sides: \;A+1=3\sqrt{\dfrac{k}{n}}
2. Divide by 3: \;\dfrac{A+1}{3}=\sqrt{\dfrac{k}{n}}
3. Square both sides: \;\dfrac{(A+1)^2}{9}=\dfrac{k}{n}
4. Multiply by 9n: \;n(A+1)^2=9k
5. Divide by (A+1)^2: \;n=\dfrac{9k}{(A+1)^2}

Get \dfrac{3}{u} on its own, write the other side as a single fraction, then turn both sides upside down.

1. Add \dfrac{1}{w} to both sides: \;\dfrac{3}{u}=2+\dfrac{1}{w}
2. Write as a single fraction: \;\dfrac{3}{u}=\dfrac{2w+1}{w}
3. Turn both sides upside down: \;\dfrac{u}{3}=\dfrac{w}{2w+1}
4. Multiply by 3
5. Answer: u=\dfrac{3w}{2w+1}

Multiply both sides by (r+3) first, then get every term containing r on the same side.

1. Multiply both sides by (r+3): \;p(r+3)=2r-a
2. Expand: \;pr+3p=2r-a
3. Collect the r terms on one side: \;pr-2r=-a-3p
4. Factorise out r: \;r(p-2)=-a-3p
5. Divide by (p-2): \;r=\dfrac{-a-3p}{p-2}
6. Multiply top and bottom by -1: \;r=\dfrac{a+3p}{2-p}

Make y the subject of the second equation and substitute it into the first.

1. From the second equation: \;y=2x-3
2. Substitute: \;x+k(2x-3)=4
3. Expand: \;x+2kx-3k=4
4. Factorise out x: \;x(1+2k)=4+3k
5. Answer: x=\dfrac{3k+4}{2k+1}

Multiply both sides by 2, then collect both t^3 terms on the same side and factorise.

1. Multiply both sides by 2: \;t^3+c=2ct^3
2. Collect the t^3 terms: \;c=2ct^3-t^3
3. Factorise: \;c=t^3(2c-1)
4. Divide by (2c-1): \;t^3=\dfrac{c}{2c-1}
5. Cube root both sides
6. Answer: t=\sqrt[3]{\dfrac{c}{2c-1}}

Expand the bracket, then collect both x^2 terms on the same side and factorise.

1. Expand: \;kx^2+k=3x^2-2
2. Collect the x^2 terms on the right: \;k+2=3x^2-kx^2
3. Factorise: \;k+2=x^2(3-k)
4. Divide: \;x^2=\dfrac{k+2}{3-k}
5. Square root (x>0)
6. Answer: x=\sqrt{\dfrac{k+2}{3-k}}

Write x^2-8x+21 in the form (x-a)^2+b so that x appears only once.

1. Complete the square: \;x^2-8x+21=(x-4)^2-16+21
2. So \;y=(x-4)^2+5
3. Subtract 5: \;(x-4)^2=y-5
4. Square root: \;x-4=\sqrt{y-5} (positive root, as x>4)
5. Answer: x=4+\sqrt{y-5}

The closed cylinder has two circular ends as well as its curved surface. Answer part (a) before attempting part (b).

Part (a)

1. Cylinder: two circles 2\pi x^2 and the curved surface 2\pi xy
2. So \;4\pi r^2=2\pi x^2+2\pi xy
3. Divide by 2\pi: \;2r^2=x^2+xy
4. Divide by 2: \;r^2=\dfrac{x^2+xy}{2}
5. Square root (r>0)
6. Answer: r=\sqrt{\dfrac{x^2+xy}{2}}

Part (b)

1. Square both sides: \;r^2=\dfrac{x^2+xy}{2}
2. Multiply by 2: \;2r^2=x^2+xy
3. Subtract x^2: \;xy=2r^2-x^2
4. Divide by x
5. Answer: y=\dfrac{2r^2-x^2}{x}

Multiply both sides by 8p to clear the fractions, then collect every y^3 term on one side and factorise.

Part (a)

1. Multiply both sides by 8p: \;p(2y^3+1)=8(y^3-p)
2. Expand: \;2py^3+p=8y^3-8p
3. Collect the y^3 terms on the right: \;9p=8y^3-2py^3
4. Factorise: \;9p=y^3(8-2p)
5. Divide: \;y^3=\dfrac{9p}{8-2p}
6. Cube root (the root covers the whole fraction)
7. Answer: y=\sqrt[3]{\dfrac{9p}{8-2p}}

Part (b)

1. y^3=8, so \;8=\dfrac{9p}{8-2p}
2. Multiply by (8-2p): \;64-16p=9p
3. 25p=64
4. Answer: p=\dfrac{64}{25}

Clear the fractions by multiplying through by every denominator, then collect the terms in a.

Part (a)

1. Subtract \dfrac{1}{b}: \;\dfrac{1}{a}=\dfrac{1}{R}-\dfrac{1}{b}
2. Single fraction: \;\dfrac{1}{a}=\dfrac{b-R}{Rb}
3. Turn both sides upside down
4. Answer: a=\dfrac{Rb}{b-R}

Part (b)

1. Using part (a) with R=6 and a=b+4: \;b+4=\dfrac{6b}{b-6}
2. Multiply by (b-6): \;(b+4)(b-6)=6b
3. Expand: \;b^2-2b-24=6b
4. So \;b^2-8b-24=0
5. Quadratic formula: \;b=\dfrac{8\pm\sqrt{64+96}}{2}
6. b=10.32\ldots or b=-2.32\ldots
7. A resistance is positive
8. Answer: b=10.3

Cross-multiply to clear both fractions, expand the brackets, then collect every x^2 term on one side and factorise.

Part (a)

1. Cross-multiply: \;(x^2+4)(y+2)=3x^2y
2. Expand: \;x^2y+2x^2+4y+8=3x^2y
3. Collect the x^2 terms on one side: \;2x^2+x^2y-3x^2y=-4y-8
4. Simplify: \;2x^2-2x^2y=-4y-8
5. Factorise out x^2: \;x^2(2-2y)=-4y-8
6. Divide: \;x^2=\dfrac{-4y-8}{2-2y}=\dfrac{2(y+2)}{y-1}
7. Square root (x>0): \;x=\sqrt{\dfrac{2(y+2)}{y-1}}

Part (b)

1. Substitute y=5 into your formula from part (a): \;x^2=\dfrac{2\times 7}{4}=\dfrac{14}{4}
2. x=\sqrt{\dfrac{14}{4}}=\dfrac{\sqrt{14}}{\sqrt{4}}
3. x=\dfrac{\sqrt{14}}{2}

Write the volume V=x^2h in terms of x only, then differentiate. Answer part (a) before attempting part (b).

Part (a)

1. Base: x^2; each of the four sides: xh
2. So \;x^2+4xh=192
3. Subtract x^2: \;4xh=192-x^2
4. Divide by 4x
5. Answer: h=\dfrac{192-x^2}{4x}

Part (b)

1. V=x^2h=x^2\times\dfrac{192-x^2}{4x}
2. So \;V=48x-\dfrac{x^3}{4}
3. \dfrac{dV}{dx}=48-\dfrac{3x^2}{4}
4. At a maximum \dfrac{dV}{dx}=0: \;x^2=64, so x=8
5. \dfrac{d^2V}{dx^2}=-\dfrac{3x}{2}<0, so this is a maximum
6. V=48\times8-\dfrac{512}{4}=384-128
7. Answer: 256 cm^3