Question 173560
5/y-3=y+7/2y-6+1<pre><font size = 4 color = "indigo"><b>
Hmmm! You didn't put parentheses to show where numerators 
and denominators begin "(" and end ")", so I'm guessing you
mean one of two possibilities.  I'll do both:

If you mean this:

{{{matrix(1,5, 5/(y-3),"=",(y+7)/(2y-6),"+",1)}}}

Write the {{{1}}} as {{{1/1}}} so every term will
be a fraction:

{{{matrix(1,5, 5/(y-3),"=",(y+7)/(2y-6),"+",1/1)}}}

Enclose each term in parentheses:

{{{matrix(1,5, (5/(y-3)),"=",((y+7)/(2y-6)),"+",(1/1))}}}

Factor the denominator {{{2y-6}}} as {{{2(y-3)}}}

{{{matrix(1,5, (5/(y-3)),"=",((y+7)/(2(y-3))),"+",(1/1))}}}

Find the LCD, which is {{{2(y-3)}}}, which we will
write as {{{((2(y-3))/1)}}} and multiply each term
by it:

{{{matrix(1,5, ((2(y-3))/1)(5/(y-3)),"=",((2(y-3))/1)((y+7)/(2(y-3))),"+",((2(y-3))/1)(1/1))}}}

Cancel what will cancel:

{{{matrix(1,5, ((2(cross(y-3)))/1)(5/(cross(y-3))),"=",((cross(2)(cross(y-3)))/1)((y+7)/(cross(2)(cross(y-3)))),"+",((2(y-3))/1)(1/1))}}}

All that's left is:

{{{matrix(1,3,2*5,"=", y+7+2(y-3))}}}

Distribute:

{{{matrix(1,3,2*5,"=", y+7+2y-6))}}}

Combine like terms:

{{{matrix(1,3,10,"=", 3y+1))}}}

Add -1 to both sides

{{{matrix(1,3,9,"=",3y)}}}

Divide both sides by 3:

{{{matrix(1,3,3,"=",y)}}}

Now we must check this answer, by
substituting it into the original, to make sure it
is not an extraneous solution:

{{{matrix(1,5, 5/(y-3),"=",(y+7)/(2y-6),"+",1)}}}

{{{matrix(1,5, 5/(3-3),"=",(3+7)/(2(3)-6),"+",1)}}}

{{{matrix(1,5, 5/0,"=",(y+7)/(6-6),"+",1)}}}

{{{matrix(1,5, 5/0,"=",(y+7)/0,"+",1)}}}

But since division by 0 is undefined, there is
no solution. 3 was an extraneous, or "bogus"
solution, not an actual solution.

================

If you meant this:

{{{matrix(1,7, 5/(y-3),"=",y,"+",7/(2y-6),"+",1)}}}

Use 1 denominators so that all terms will be
fractions:

{{{matrix(1,7, 5/(y-3),"=",y/1,"+",7/(2y-6),"+",1/1)}}}

Enclose each term in parentheses:

{{{matrix(1,7, (5/(y-3)),"=",(y/1),"+",(7/(2y-6)),"+",(1/1))}}}

Factor the denominator {{{2y-6}}} as {{{2(y-3)}}}

{{{matrix(1,7, (5/(y-3)),"=",(y/1),"+",(7/(2(y-3))),"+",(1/1))}}}

Find the LCD, which is {{{2(y-3)}}}, which we will
write as {{{((2(y-3))/1)}}} and multiply each term
by it:

{{{matrix(1,7, ((2(y-3))/1)(5/(y-3)),"=",((2(y-3))/1)(y/1),"+",((2(y-3))/1)(7/(2(y-3))),"+",((2(y-3))/1)(1/1))}}}

Cancel what will cancel:

{{{matrix(1,7, ((2(cross(y-3)))/1)(5/(cross(y-3))),"=",((2(y-3))/1)(y/1),"+",((cross(2)(cross(y-3)))/1)(7/(cross(2)(cross(y-3)))),"+",((2(y-3))/1)(1/1))}}}

All that's left is:

{{{matrix(1,7, 2*5,"=",2*(y-3)*y, "+", 7, "+", 2(y-3))}}}

{{{matrix(1,7, 10,"=",2y(y-3), "+", 7, "+", 2(y-3))}}}

Distribute to remove parentheses:

{{{matrix(1,3,10,"=",2y^2-6y+7+2y-6)}}}

Collect like terms:

{{{matrix(1,3,10,"=",2y^2-4y+1)}}}

Get 0 on the left by subtracting 10 from both sides:

{{{matrix(1,3,0,"=",2y^2-4y-9)}}}

or, since I like to have 0 on the right, not left,
I'll switch the sides:


{{{matrix(1,3,2y^2-4y-9,"=",0)}}}

That doesn't factor, so we use the quadratic
formula:

{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{y = (-(-4) +- sqrt( (-4)^2-4*(2)*(-9) ) )/(2*(2)) }}}

{{{y = (4 +- sqrt(16+72) )/4 }}}

{{{y = (4 +- sqrt(88) )/4 }}}

{{{y = (4 +- sqrt(4*22) )/4 }}}

{{{y = (4 +- 2sqrt(22) )/4 }}}

Factor 2 out of the numerator:

{{{y = (2(2 +- sqrt(22)) )/4 }}}

Cancel the 2 on top into the 4 on the bottom,
getting 2 on the bottom:

{{{y = (cross(2)(2 +- sqrt(22)) )/(cross(4))[2] }}}

{{{y = (2 +- sqrt(22) )/2 }}}

This is difficult to check. However if we were
to substitute that in the original, none of the
denominators would be 0, as they were in the
first way I assumed you meant.  So these are both
solutions.

Edwin</pre>