Question

Solve 3/(X^2-7x+10) + 2 = (x-4)/ (x-5)
and eplain. thanksss!

Answer 1

first Factor:
factors of 10 that add up to -7
\[x^{2}-7x+10 = (x-5)(x-2)\]
add fractions getting common denominator
\[\frac{3}{(x-5)(x-2)} + \frac{2(x-5)(x-2)}{(x-5)(x-2)} = \frac{3+2(x-5)(x-2)}{(x-5)(x-2)}\]
expand numerator by distributing and adding like terms
\[=\frac{2x ^{2}-14x+23}{(x-5)(x-2)}\]
Notice 2 fractions equal to each other, we can cross-multiply
\[(x-5)(2x^{2}-14x+23) = (x-4)(x-5)(x-2)\]
Divide by (x-5) on both sides, they cancel
FOIL right side,
\[2x^{2}-14x+23 = x^{2}-6x+8\]
move everything to left side so it equals 0
\[x^{2} -8x +15 =0\]
Factor and solve
\[(x-5)(x-3) = 0\]
\[x=5, x=3\]