(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]
= (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)
= (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: 22) a = c/b 23) y = q/x 24) y = (k)(m2)/x (a) solve for m. |