Question 972802
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n, n+1, n+2= three consecutive numbers
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{{{(1/4)n+(1/5)(n+1)=(1/7)(n+2)-5}}}
{{{(35/140)n+(28/140)n+(28/140)=(20/140)n+(40/140)-(700/140)}}}
{{{(63/140)n+(28/140)=(20/140)n-(660/140)}}} Multiply by 140.
{{{63n+28=20n-660}}} Subtract (20n+28) from each side.
{{{43n=-668}}} Divide each side by 43.
{{{n=-16}}} The first number is -16.
{{{n+1=-16+1=-15}}} The second number is -15.
{{{n+2=-16+2=-14}}} The third number is -14.
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CHECK:
{{{(1/4)-16+(1/5)-15=(1/7)-14-5}}}
{{{-4-3=-2-5}}}
{{{-7=-7}}}