Question 282055
{{{(10a-8)/3=(6a+24)/5}}}
i would multiply both sides by 15 to get rid of the fraction
i chose 15 cause {{{3*5=15}}}
so
{{{(3*5)((10a-8)/3)=(3*5)((6a+24)/5)}}} cancel
{{{(cross(3)*5)((10a-8)/cross(3))=(3*cross(5))((6a+24)/cross(5))}}}
{{{(5)(10a-8)=(3)(6a+24)}}} distribute
{{{(50a-40)=(18a+72)}}} get {{{a}}} by itself, subtract {{{18a}}}
{{{32a-40=72}}} add {{{40}}}
{{{32a=112}}}divide by 32
{{{a=3.5}}}
________
Check
{{{(10(3.5)-8)/3=(6(3.5)+24)/5}}}
{{{((35)-8)/3=((21)+24)/5}}}
{{{27/3=45/5}}}
{{{9=9}}} correct