(VI) Solving Algebraic Equations

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Given an algebraic equation, it is important to be able to solve the equation for any variable or constant.

(Example) D = M/V

(a) Solve this equation for M:

(i) We must isolate the M in the equation. We have to get the V to the other side of the equation.

(ii) Since V is in the denominator we can accomplish this by multiplying both sides of the equation by V:

(V)(D) = (M/V)(V)

(iii) Notice that the V terms on the right side cancel. If we write the M on the left and the MV on the right we have our desired equation:

M = DV

(b) Solve this equation for V:

(i) We must isolate the V. We must get the M to the other side of the equation.

(ii) Since M is in the numerator, we must multiply both sides of the equation by1/M.

(1/M)(D) = (M/V)(1/M)

(iii) Notice now that the M terms on the right side cancel leaving:

D/M = 1/V

(iv) Now we simply take the reciprocal of both sides of the equation and we have the desired equation:

M/D = V/1

or, V = M/D

(Example) E = hv

(a) Solve this equation for v:

(i) We must isolate v. We must get the h to the other side of the equation.

(ii)Since h is in the numerator, we multiply both sides of the equation by 1/h:

(1/h)(E) = (hv)(1/h)

(iii) Note that the h values on the right side cancel leaving us with:

E/h = v

or, v = E/h

(b) Solve this equation for h:

(i) We must isolate the h term.

(ii) Multiply both sides by 1/v:

(1/v)(E) = (hv)(1/v)

(iii) Notice now that the v terms on the right side cancel:

E/v = h

or, h = E/v

(Example) y = k/x

(a) Solve for x:

(i) We must isolate the x term and Get the k on the other side of the equation.

(ii)Since k is in the numerator, we multiply both sides by 1/k:

(1/k)(y) = (k/x)(1/k)

(iii) The k terms cancel on the right side leaving:

y/k = 1/x

(iv) Now take the reciprocal of both sides of the equation:

k/y = x/1, or k/y = x

(v) Now simply write the equation with x on the left side:

x = k/y

(Example) F = k(M1)(M2)/d2

(a) Solve for M1:

(i) Note that k and M2 are in the numerator while d2 is in the denominator. In order to isolate M1, we multiply both sides by d2/(k)(M2):

[(d2)/(k)(M2)][F] =
[(k)(M1)(M2)/d2][(d2)/(k)(M2)]

(ii) The d2, M1, and k have been removed from the right side:

[(F)(d2)/(k)(M2)] = M1

(iii) Now we simply rewrite the equation with the M1 on the left:

M1 = (F)(d2)/(k)(M2)

(b) Solve for d:

(i) We must transpose k, M1, and M2 to the left side of the equation. Since these three terms are all in the numerator, we multiply both sides of the equation by a fraction which has these three terms in the denominator:

[1/(k)(M1)(M2)](F) =
[(k)(M1)(M2)/d2][1/(k)(M1)(M2]

(ii) Now the k, M1, and M2 terms are gone from the right side of the equation:

F/(k)(M1)(M2) = 1/d2

(iii) Next we take the reciprocal of both sides:

(k)(M1)(M2)/F = d2/1

(iv) We write the d2 term on the left side with the remainder of the equation on the right side:

d2 = (k)(M1)(M2)/F

(v) In order to have an equation for the first power of d we take the square root of both sides of the equation which gives:

Problems:
Solve the equations for the indicated variable.

22) a = c/b
Solve for b.

23) y = q/x
Solve for q.

24) y = (k)(m2)/x
k is a constant.

(a) solve for m.
(b) solve for x.

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