Question 8308
1)Solve:

{{{x/2 + x/5 + x/3 = 31}}} Add the fractions on the left side. Common denominator is 30.

{{{(15x + 6x + 10x)/30 = 31}}} 

{{{31x/30 = 31}}} Multiply both sides by 30.

{{{31x = 31(30)}}} Divide both sides by 31.

{{{x = 30}}}

2)Solve {{{R(r1+r2) = r1r2}}} for R Divide both sides by (r1+r2)

{{{R = r1r2/(r1+r2)}}}

3) Solve {{{b(5px-3c) = a(qx-4)}}} for x. Divide both sides by b.

{{{5px-3c = a(qx-4)/b}}} Separate the right side into 2 terms.

{{{5px-3c = aqx/b - 4a/b}}} Add 3c to both sides.

{{{5px = 3c + aqx/b - 4a/b}}} Subtract {{{aqx/b}}} from both sides.

{{{5px - aqx/b = 3c - 4a/b}}} Factor the x from the left side.

{{{x(5p - aq/b) = 3c - 4a/b}}} Simplify both sides then divide both sides by {{{(5bp - aq)/b}}}

{{{x = ((3bc-4a)/b)/((5bp-aq)/b)}}} = {{{((3bc-4a)/b)*(b/(5bp-aq))}}} Cancel the b's

{{{x = (3bc-4a)/(5bp-aq)}}}  Restriction: aq may not = 5bp

4) Solve {{{3ax/5 - 4c = ax/5}}} Add 4c to both sides.

{{{3ax/5 = ax/5 + 4c}}} Subtract ax/5 from both sides.

{{{2ax/5 = 4c}}} Multiply both sides by 5.

{{{2ax = 20c}}} Divide both sides by 2a and simplify.

{{{x = 10c/a}}} Restriction: a cannot = 0