Change the Subject of an Equation 1

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Year 8

Equations

Change the Subject of an Equation

Change the Subject of an Equation 1

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Question 1 of 5

1. Question
Write $r$ $r$ as the subject of the equation below

$V = π r^{2} h$ $V=pir^2h$
- 1. $r = \sqrt{\frac{π h}{V}}$ $r=sqrt((pih)/V)$
- 2. $r = \sqrt{\frac{V}{π^{2} h}}$ $r=sqrt(V/(pi^2h))$
- 3. $r = \sqrt{\frac{V}{π h}}$ $r=sqrt(V/(pih))$
- 4. $r = \frac{V^{2}}{π h}$ $r=(V^2)/(pih)$
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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation

Divide both sides of the equation by $π h$ $pih$

$V$ $V$ $=$ $=$ $π r^{2} h$ $pir^2h$

$V$ $V$ $\div π h$ $-:pih$ $=$ $=$ $π r^{2} h$ $pir^2h$ $\div π h$ $-:pih$

$\frac{V}{π h}$ $V/(pih)$ $=$ $=$ $r^{2}$ $r^2$ $π h \div π h$ $pih-:pih$ cancels out

Get the square root of both sides

$\frac{V}{π h}$ $V/(pih)$ $=$ $=$ $r^{2}$ $r^2$

$\sqrt{\frac{V}{π h}}$ $sqrt(V/(pih))$ $=$ $=$ $\sqrt{r^{2}}$ $sqrt(r^2)$

$\sqrt{\frac{V}{π h}}$ $sqrt(V/(pih))$ $=$ $=$ $r$ $r$

$r$ $r$ $=$ $=$ $\sqrt{\frac{V}{π h}}$ $sqrt(V/(pih))$ Interchange the sides

$r = \sqrt{\frac{V}{π h}}$ $r=sqrt(V/(pih))$

Question 2 of 5

2. Question

Write

a

$a$ as the subject of the equation below

b = \frac{a - 3}{a - 4}

$b=(a-3)/(a-4)$

1. $a = \frac{4 b - 3}{b - 1}$ $a=(4b-3)/(b-1)$
2. $a = \frac{b + 3}{b - 1}$ $a=(b+3)/(b-1)$
3. $a = \frac{b - 3}{b - 4}$ $a=(b-3)/(b-4)$
4. $a = \frac{4 b + 5}{b - 3}$ $a=(4b+5)/(b-3)$

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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation

Multiply both sides of the equation to $a - 4$ $a-4$

$b$ $b$	$=$ $=$	$\frac{a - 3}{a - 4}$ $(a-3)/(a-4)$
$b$ $b$ $\times (a - 4)$ $xx(a-4)$	$=$ $=$	$\frac{a - 3}{a - 4}$ $(a-3)/(a-4)$ $\times (a - 4)$ $xx(a-4)$

$b (a - 4)$ $b(a-4)$	$=$ $=$	$a - 3$ $a-3$	$\frac{1}{a - 4} \times (a - 4)$ $1/(a-4)xx(a-4)$ cancels out

$a b - 4 b$ $ab-4b$	$=$ $=$	$a - 3$ $a-3$

Leave only

a

$a$ terms on the left

$a b - 4 b$ $ab-4b$	$=$ $=$	$a - 3$ $a-3$
$a b - 4 b$ $ab-4b$ $- a$ $-a$	$=$ $=$	$a - 3$ $a-3$ $- a$ $-a$	Subtract $a$ $a$ from both sides
$a b - 4 b - a$ $ab-4b-a$	$=$ $=$	$- 3$ $-3$	$a - a$ $a-a$ cancels out
$a b - 4 b - a$ $ab-4b-a$ $+ 4 b$ $+4b$	$=$ $=$	$- 3$ $-3$ $+ 4 b$ $+4b$	Add $4 b$ $4b$ to both sides
$a b - a$ $ab-a$	$=$ $=$	$4 b - 3$ $4b-3$	$- 4 b + 4 b$ $-4b+4b$ cancels out

Factor out

a

$a$

$a b - a$ $ab-a$	$=$ $=$	$4 b - 3$ $4b-3$
$a (b - 1)$ $a(b-1)$	$=$ $=$	$4 b - 3$ $4b-3$

Finally, divide both sides by $b - 1$ $b-1$

$a (b - 1)$ $a(b-1)$	$=$ $=$	$4 b - 3$ $4b-3$
$a (b - 1)$ $a(b-1)$ $\div (b - 1)$ $-:(b-1)$	$=$ $=$	$(4 b - 3)$ $(4b-3)$ $\div (b - 1)$ $-:(b-1)$

$a$ $a$	$=$ $=$	$\frac{4 b - 3}{b - 1}$ $(4b-3)/(b-1)$

a = \frac{4 b - 3}{b - 1}

$a=(4b-3)/(b-1)$

Question 3 of 5

3. Question

Write

x

$x$ as the subject of the equation below

7 - 3 x = m x + t

$7-3x=mx+t$

1. $x = 7 m + 8$ $x=7m+8$
2. $x = \frac{t - 7}{m - 3}$ $x=(t-7)/(m-3)$
3. $x = \frac{7 - t}{m + 3}$ $x=(7-t)/(m+3)$
4. $x = 7 - 3 t$ $x=7-3t$

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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation

Keep only

x

$x$ terms on the right side

$7 - 3 x$ $7-3x$	$=$ $=$	$m x + t$ $mx+t$
$7 - 3 x$ $7-3x$ $+ 3 x$ $+3x$	$=$ $=$	$m x + t$ $mx+t$ $+ 3 x$ $+3x$	Add $3 x$ $3x$ to both sides
$7$ $7$	$=$ $=$	$m x + 3 x + t$ $mx+3x+t$	$- 3 x + 3 x$ $-3x+3x$ cancels out
$7$ $7$ $- t$ $-t$	$=$ $=$	$m x + 3 x + t$ $mx+3x+t$ $- t$ $-t$	Subtract $t$ $t$ from both sides
$7 - t$ $7-t$	$=$ $=$	$m x + 3 x$ $mx+3x$	$t - t$ $t-t$ cancels out

Factor out

x

$x$

$7 - t$ $7-t$	$=$ $=$	$m x + 3 x$ $mx+3x$
$7 - t$ $7-t$	$=$ $=$	$x (m + 3)$ $x(m+3)$

Finally, divide both sides by $m + 3$ $m+3$

$7 - t$ $7-t$	$=$ $=$	$x (m + 3)$ $x(m+3)$
$(7 - t)$ $(7-t)$ $\div (m + 3)$ $-:(m+3)$	$=$ $=$	$x (m + 3)$ $x(m+3)$ $\div (m + 3)$ $-:(m+3)$

$\frac{7 - t}{m + 3}$ $(7-t)/(m+3)$	$=$ $=$	$x$ $x$

$x$ $x$	$=$ $=$	$\frac{7 - t}{m + 3}$ $(7-t)/(m+3)$	Interchange the sides

x = \frac{7 - t}{m + 3}

$x=(7-t)/(m+3)$

Question 4 of 5

4. Question

Write

x

$x$ as the subject of the equation below

\frac{1 - x}{x + y} = 1

$(1-x)/(x+y)=1$

1. $x = 2 - 3 y$ $x=2-3y$
2. $x = \frac{1 - y}{2}$ $x=(1-y)/2$
3. $x = \frac{1 - y}{y + 1}$ $x=(1-y)/(y+1)$
4. $x = \frac{y - 1}{2}$ $x=(y-1)/2$

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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation

Multiply both sides to $x + y$ $x+y$

$\frac{1 - x}{x + y}$ $(1-x)/(x+y)$	$=$ $=$	$1$ $1$

$\frac{1 - x}{x + y}$ $(1-x)/(x+y)$ $\times (x + y)$ $xx(x+y)$	$=$ $=$	$1$ $1$ $\times (x + y)$ $xx(x+y)$

$1 - x$ $1-x$	$=$ $=$	$x + y$ $x+y$	$\frac{1}{x + y} \times (x + y)$ $1/(x+y)xx(x+y)$ cancels out

Leave only the

x

$x$ terms on the right side

$1 - x$ $1-x$	$=$ $=$	$x + y$ $x+y$
$1 - x$ $1-x$ $+ x$ $+x$	$=$ $=$	$x + y$ $x+y$ $+ x$ $+x$	Add $x$ $x$ to both sides
$1$ $1$	$=$ $=$	$2 x + y$ $2x+y$	$- x + x$ $-x+x$ cancels out
$1$ $1$ $- y$ $-y$	$=$ $=$	$2 x + y$ $2x+y$ $- y$ $-y$	Subtract $y$ $y$ from both sides
$1 - y$ $1-y$	$=$ $=$	$2 x$ $2x$	$y - y$ $y-y$ cancels out

Finally, divide both sides by $2$ $2$

$1 - y$ $1-y$	$=$ $=$	$2 x$ $2x$
$(1 - y)$ $(1-y)$ $\div 2$ $-:2$	$=$ $=$	$2 x$ $2x$ $\div 2$ $-:2$

$\frac{1 - y}{2}$ $(1-y)/2$	$=$ $=$	$x$ $x$	$2 \div 2$ $2-:2$ cancels out

$x$ $x$	$=$ $=$	$\frac{1 - y}{2}$ $(1-y)/2$	Interchange the sides

x = \frac{1 - y}{2}

$x=(1-y)/2$

Question 5 of 5

5. Question

Write

b

$b$ as the subject of the equation below

x = \frac{2 b}{3 b + y}

$x=(2b)/(3b+y)$

1. $b = \frac{2 x}{3 x + y}$ $b=(2x)/(3x+y)$
2. $b = \frac{- 3 x y}{2 x + 1}$ $b=(-3xy)/(2x+1)$
3. $b = \frac{- x y}{3 x - 2}$ $b=(-xy)/(3x-2)$
4. $b = \frac{2 y}{3 y - x}$ $b=(2y)/(3y-x)$

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Changing the subject means operating on an equation so that a chosen variable remains alone on the left side of the equation

Multiply both sides to $3 b + y$ $3b+y$

$x$ $x$	$=$ $=$	$\frac{2 b}{3 b + y}$ $(2b)/(3b+y)$

$x$ $x$ $\times (3 b + y)$ $xx(3b+y)$	$=$ $=$	$\frac{2 b}{3 b + y}$ $(2b)/(3b+y)$ $\times (3 b + y)$ $xx(3b+y)$

$x (3 b + y)$ $x(3b+y)$	$=$ $=$	$2 b$ $2b$	$\frac{1}{3 b + y} \times (3 b + y)$ $1/(3b+y)xx(3b+y)$ cancels out
$3 b x + x y$ $3bx+xy$	$=$ $=$	$2 b$ $2b$	Distribute $x$ $x$

Leave only

b

$b$ terms on the left side

$3 b x + x y$ $3bx+xy$	$=$ $=$	$2 b$ $2b$
$3 b x + x y$ $3bx+xy$ $- x y$ $-xy$	$=$ $=$	$2 b$ $2b$ $- x y$ $-xy$	Subtract $x y$ $xy$ from both sides
$3 b x$ $3bx$	$=$ $=$	$2 b - x y$ $2b-xy$	$x y - x y$ $xy-xy$ cancels out
$3 b x$ $3bx$ $- 2 b$ $-2b$	$=$ $=$	$2 b - x y$ $2b-xy$ $- 2 b$ $-2b$	Subtract $2 b$ $2b$ from both sides
$3 b x - 2 b$ $3bx-2b$	$=$ $=$	$- x y$ $-xy$	$2 b - 2 b$ $2b-2b$ cancels out

Factor out

b

$b$

$3 b x - 2 b$ $3bx-2b$	$=$ $=$	$- x y$ $-xy$
$b (3 x - 2)$ $b(3x-2)$	$=$ $=$	$- x y$ $-xy$

Finally, divide both sides by $3 x - 2$ $3x-2$

$b (3 x - 2)$ $b(3x-2)$	$=$ $=$	$- x y$ $-xy$
$b (3 x - 2)$ $b(3x-2)$ $\div (3 x - 2)$ $-:(3x-2)$	$=$ $=$	$- x y$ $-xy$ $\div (3 x - 2)$ $-:(3x-2)$

$b$ $b$	$=$ $=$	$\frac{- x y}{3 x - 2}$ $(-xy)/(3x-2)$	$(3 x - 2) \div (3 x - 2)$ $(3x-2)-:(3x-2)$ cancels out

b = \frac{- x y}{3 x - 2}

$b=(-xy)/(3x-2)$

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$V$ $V$	$=$ $=$	$π r^{2} h$ $pir^2h$
$V$ $V$ $\div π h$ $-:pih$	$=$ $=$	$π r^{2} h$ $pir^2h$ $\div π h$ $-:pih$

$\frac{V}{π h}$ $V/(pih)$	$=$ $=$	$r^{2}$ $r^2$	$π h \div π h$ $pih-:pih$ cancels out

Change the Subject of an Equation 1

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