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 $r$ as the subject of the equation below

$V=pir^2h$
- 1. $r=sqrt((pih)/V)$
- 2. $r=sqrt(V/(pih))$
- 3. $r=sqrt(V/(pi^2h))$
- 4. $r=(V^2)/(pih)$
Hint

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Correct

Excellent

Incorrect

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 $pih$

$V$ $=$ $pir^2h$

$V$ $-:pih$ $=$ $pir^2h$ $-:pih$

$V/(pih)$ $=$ $r^2$ $pih-:pih$ cancels out

Get the square root of both sides

$V/(pih)$ $=$ $r^2$

$sqrt(V/(pih))$ $=$ $sqrt(r^2)$

$sqrt(V/(pih))$ $=$ $r$

$r$ $=$ $sqrt(V/(pih))$ Interchange the sides

$r=sqrt(V/(pih))$

Question 2 of 5

2. Question

Write

$a$ as the subject of the equation below

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

1. $a=(4b+5)/(b-3)$
2. $a=(b+3)/(b-1)$
3. $a=(b-3)/(b-4)$
4. $a=(4b-3)/(b-1)$

Hint

Help Video

Correct

Well Done

Incorrect

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 $a-4$

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

$b(a-4)$	$=$	$a-3$	$1/(a-4)xx(a-4)$ cancels out

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

Leave only

$a$ terms on the left

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

Factor out

$a$

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

Finally, divide both sides by $b-1$

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

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

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

Question 3 of 5

3. Question

Write

$x$ as the subject of the equation below

$7-3x=mx+t$

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

Hint

Help Video

Correct

Fantastic

Incorrect

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

Keep only

$x$ terms on the right side

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

Factor out

$x$

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

Finally, divide both sides by $m+3$

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

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

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

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

Question 4 of 5

4. Question

Write

$x$ as the subject of the equation below

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

1. $x=(1-y)/(y+1)$
2. $x=(y-1)/2$
3. $x=2-3y$
4. $x=(1-y)/2$

Hint

Help Video

Correct

Great Work

Incorrect

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 $x+y$

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

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

$1-x$	$=$	$x+y$	$1/(x+y)xx(x+y)$ cancels out

Leave only the

$x$ terms on the right side

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

Finally, divide both sides by $2$

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

$(1-y)/2$	$=$	$x$	$2-:2$ cancels out

$x$	$=$	$(1-y)/2$	Interchange the sides

$x=(1-y)/2$

Question 5 of 5

5. Question

Write

$b$ as the subject of the equation below

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

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

Hint

Help Video

Correct

Good Job

Incorrect

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 $3b+y$

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

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

$x(3b+y)$	$=$	$2b$	$1/(3b+y)xx(3b+y)$ cancels out
$3bx+xy$	$=$	$2b$	Distribute $x$

Leave only

$b$ terms on the left side

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

Factor out

$b$

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

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

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

$b$	$=$	$(-xy)/(3x-2)$	$(3x-2)-:(3x-2)$ cancels out

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

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$V$	$=$	$pir^2h$
$V$ $-:pih$	$=$	$pir^2h$ $-:pih$

$V/(pih)$	$=$	$r^2$	$pih-:pih$ cancels out

$V/(pih)$	$=$	$r^2$

$sqrt(V/(pih))$	$=$	$sqrt(r^2)$

$sqrt(V/(pih))$	$=$	$r$

$r$	$=$	$sqrt(V/(pih))$	Interchange the sides

Change the Subject of an Equation 1

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