Question 383392
Use a common denominator, {{{3x(2x+5)}}}
{{{(7x+1)/(2x+5)=((7x+1)(3x))/(3x(2x+5))}}}
{{{1=((3x)(2x+5))/(3x(2x+5))}}}
{{{(10x-3)/(3x)=((10x-3)(2x+5))/(3x(2x+5))}}}
So then,
{{{((7x+1)/(2x+5)) +1 = ((10x-3)/3x)}}}
{{{((7x+1)(3x))/(3x(2x+5))+((3x)(2x+5))/(3x(2x+5))=((10x-3)(2x+5))/(3x(2x+5))}}}
{{{((7x+1)(3x))/(3x(2x+5))+((3x)(2x+5))/(3x(2x+5))-((10x-3)(2x+5))/(3x(2x+5))=0}}}
{{{((7x+1)(3x)+(3x)(2x+5)-(10x-3)(2x+5))/(3x(2x+5))=0}}}
{{{((21x^2+3x)+(6x^2+15x)-(20x^2+50x-6x-15))/(3x(2x+5))=0}}}
{{{((21x^2+6x^2-20x^2)+(3x+15x-44x)+15)/(3x(2x+5))=0}}}
{{{(7x^2-26x+15)/(3x(2x+5))=0}}}
{{{((7x-5)(x-3))/(3x(2x+5))=0}}}
The left hand side equals zero where the numerator equals zero.

{{{(7x-5)(x-3)=0}}}
Two solutions:
{{{7x-5=0}}}
{{{7x=5}}}
{{{highlight(x=5/7)}}}
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{{{x-3=0}}}
{{{highlight(x=3)}}}