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Question 1 of 5
Write dd as the subject of the equation below
A=12h(c+d)A=12h(c+d)
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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 22
| AA |
== |
12h(c+d)12h(c+d) |
|
| AA×2×2 |
== |
12h(c+d)12h(c+d)×2×2 |
|
| 2A2A |
== |
h(c+d)h(c+d) |
12×212×2 cancels out |
| 2A2A |
== |
h(c+d)h(c+d) |
| 2A2A÷h÷h |
== |
h(c+d)h(c+d)÷h÷h |
|
| 2Ah2Ah |
== |
c+dc+d |
h÷hh÷h cancels out |
Finally, subtract cc from both sides
| 2Ah2Ah |
== |
c+dc+d |
|
| 2Ah2Ah -c−c |
== |
c+dc+d -c−c |
|
| 2Ah-c2Ah−c |
== |
dd |
c-cc−c cancels out |
|
| dd |
== |
2Ah-c2Ah−c |
Interchange the sides |
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Question 2 of 5
Write xx as the subject of the equation below
G(x+y)=H(x-y)G(x+y)=H(x−y)
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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
| G(x+y)G(x+y) |
== |
H(x-y)H(x−y) |
| Gx+GyGx+Gy |
== |
Hx-HyHx−Hy |
Leave only xx terms on the left
| Gx+GyGx+Gy |
== |
Hx-HyHx−Hy |
| Gx+GyGx+Gy -Hx−Hx |
== |
Hx-HyHx−Hy -Hx−Hx |
Subtract HxHx from both sides |
| Gx+Gy-HxGx+Gy−Hx |
== |
-Hy−Hy |
Hx-HxHx−Hx cancels out |
| Gx+Gy-HxGx+Gy−Hx -Gy−Gy |
== |
-Hy−Hy -Gy−Gy |
Subtract GyGy from both sides |
| Gx-HxGx−Hx |
== |
-Gy-Hy−Gy−Hy |
Gy-GyGy−Gy cancels out |
| Gx-HxGx−Hx |
== |
-Gy-Hy−Gy−Hy |
| x(G-H)x(G−H) |
== |
y(-G-H)y(−G−H) |
| x(G-H)x(G−H) |
== |
-y(G+H)−y(G+H) |
Finally, divide both sides by G-HG−H
| x(G-H)x(G−H) |
== |
-y(G+H)−y(G+H) |
| x(G-H)x(G−H)÷(G-H)÷(G−H) |
== |
-y(G+H)−y(G+H)÷(G-H)÷(G−H) |
|
| xx |
== |
-y(G+H)G-H−y(G+H)G−H |
(G-H)÷(G-H)(G−H)÷(G−H) cancels out |
|
| x |
= |
y(G+H)H-G |
Multiply -1 to the numerator and denominator |
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Question 3 of 5
Write x as the subject of the equation below
1z=1y+1x
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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
Subtract 1y from both sides
| 1z |
= |
1y+1x |
|
| 1z -1y |
= |
1y+1x -1y |
|
| 1z-1y |
= |
1x |
1y-1y cancels out |
|
| y-zzy |
= |
1x |
Finally, divide both sides by y-z
| x(y-z) |
= |
zy |
| x(y-z)÷(y-z) |
= |
zy÷(y-z) |
|
| x |
= |
zyy-z |
(y-z)÷(y-z) cancels out |
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Question 4 of 5
Solve for t given that v=20, u=8 and a=2
v=u+at
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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
Substitute the given values to solve for t
| v |
= |
u+at |
| 20 |
= |
8+2t |
Substitute known values |
| 20 -8 |
= |
8+2t -8 |
Subtract 8 from both sides |
| 12 |
= |
2t |
8-8 cancels out |
| 12÷2 |
= |
2t÷2 |
Divide both sides by 2 |
| 6 |
= |
t |
2÷2 cancels out |
| t |
= |
6 |
Interchange the sides |
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Question 5 of 5
Solve for r given that a=2 and S=25
S=a1-r
Write fractions in the format “a/b”
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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 1-r
| S |
= |
a1-r |
| S×(1-r) |
= |
a1-r×(1-r) |
|
| S(1-r) |
= |
a |
11-r×(1-r) cancels out |
Divide both sides of the equation to S
| S(1-r) |
= |
a |
| S(1-r)÷S |
= |
a÷S |
|
| 1-r |
= |
aS |
Leave only r terms on the left
| 1-r |
= |
aS |
|
| 1-r -1 |
= |
aS -1 |
Subtract 1 from both sides |
|
| -r |
= |
aS-1 |
1-1 cancels out |
Next, multiply both sides by -1
| -r |
= |
aS-1 |
|
| -r×(-1) |
= |
(aS-1)×(-1) |
|
| r |
= |
1-aS |
Finally, substitute the given values to solve for r
| r |
= |
1-aS |
|
| r |
= |
1-225 |
Substitute known values |
|
| r |
= |
2525-225 |
Convert the whole number to a fraction |
|
| r |
= |
2325 |
