Solved Examples and Worksheet for Solving Equations for a Specified variable

Q1Solve i2d + 6g = 7 for i.
A. i7 + 6g
B. g + 72d
C. (2d + 6g)
D. 2d(7 - 6g)

Solutions
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Step: 1
i2d = 7 - 6g
  [Subtracting 6g from the two sides of the equation.]
Step: 2
i = 2d(7 - 6g)
  [Multiply throughout by 2d.]
Correct Answer is :   2d(7 - 6g)
Q2The kinetic energy, E, of a body moving with a velocity v is given by the equation E = 12mv2, where m is the mass of the body. Find the mass of a body moving with a velocity 30 m/s and having an energy of 2400 N.
A. 5.33 kg
B. 5.23 kg
C. 8.33 kg
D. 5.63 kg

Solutions
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Step: 1
E = 12mv2
Step: 2
2E = mv2
Step: 3
2Ev2 = m
  [Solve the equation for m.]
Step: 4
m = 2 × 240030 × 30
  [Substitute: E = 2400, v = 30.]
Step: 5
m = 2×8030 = 5.33
Step: 6
The mass of the body is 5.33 kg.
Correct Answer is :   5.33 kg
Q3The kinetic energy, E, of a body moving with a velocity v is given by the equation E = 12mv2, where m is the mass of the body. Find the mass of a body moving with a velocity 50 m/s and having an energy of 4850 N.

A. 0.14 kg
B. 0.08 kg
C. 0.08 kg
D. 3.88 kg

Solutions
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Step: 1
E = 12mv2
Step: 2
2E = mv2
Step: 3
2Ev2 = m
  [Solve the equation for m.]
Step: 4
m = 2 × 485050 × 50
  [Substitute: E = 4850, v = 50.]
Step: 5
m = 2×9750 = 3.88
Step: 6
The mass of the body is 3.88 kg.
Correct Answer is :   3.88 kg
Q4Solve the formula for P, l = p2 - w .
A. p = 2(l + w)
B. p = 2wl
C. p = 2lw
D. p = 2(l - w)

Solutions
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Step: 1
l = p2 - w
  [Original equation.]
Step: 2
l + w = p2
  [add w on both sides.]
Step: 3
2(I + w) = p
  [Multiply by 2 on both sides.]
Step: 4
p = 2(I + w)
  [Symmetry property.]
Correct Answer is :   p = 2(l + w)
Q5Solve 1f = 1u + 1v for v.

A. v = fuf + u
B. v = fuf - u
C. v = u - fu + f
D. v = u - fuf

Solutions
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Step: 1
1f = 1u + 1v for v.
  [Original equation.]
Step: 2
1f - 1u = 1v
  [Subtract 1v from both sides.]
Step: 3
f - ufu = 1v
  [Simplify.]
Step: 4
v = fuf - u
Correct Answer is :   v = fuf - u
Q6Solve I = PTR100 for T.

A. T = 100 RPI
B. T = PR100 I
C. T = 100 PIR
D. T = 100 IPR

Solutions
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Step: 1
I = PTR100
  [Original equation.]
Step: 2
100 I = PTR
  [Multiply by 100 on both sides.]
Step: 3
100 IPR = T
  [divide by PR on both sides.]
Step: 4
T = 100  IPR
  [Symmetry property.]
Correct Answer is :   T = 100 IPR
Q7Solve V = π(R - s)h for h.

A. h = Vπ(R - s)
B. h = π RVs
C. h =π(R - s)V
D. h = VRπs

Solutions
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Step: 1
v = π(R - s)h
  [Original equation.]
Step: 2
vπ(R - s) = h
  [Divide both sides by π(R - s).]
Correct Answer is :   h = Vπ(R - s)
Q8Pythagorean formula, In a right triangle PQR, PR2 = PQ2 + QR2.Find the length of one side PQ, if the hypotenuse PR and the other side QR are known.

A. PQ = PR 2 +QR 2
B. PQ = (PR) - (QR)
C. PQ = PR2 -QR 2
D. PQ = PR2 -QR 2

Solutions
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Step: 1
In a right triangle PQR, PR2 = PQ2 + QR2
  [Given.]
Step: 2
PR2 - QR2 = PQ2
  [Subtract QR2 from both sides.]
Step: 3
PR 2 -QR 2 = PQ
  [Take square root on both sides.]
Step: 4
therefore, the length of the side PQ = PR 2 -QR 2
Correct Answer is :   PQ = PR2 -QR 2
Q9In the formula, s = ut + 12 at2, s stands for displacement of an object, u for its initial speed, t for time taken. Select a formula that would help you to find the initial speed (u) of the object.
A. u = st - at 22
B. u = st2 - at2
C. u = at - st2
D. u = st - at2

Solutions
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Step: 1
s = ut + 12 at2
  [Original Formula.]
Step: 2
s - 12at2 = ut + 12 at2 - 12at2
  [Subtract 12at2 from both the sides.]
Step: 3
s - 12at2 = ut
Step: 4
st - 12at = u
  [Divide by t on both sides.]
Step: 5
u = st - at2
  [Symmetry property,]
Correct Answer is :   u = st - at2

HighSchool Math