How do you solve #\frac { x + 5} { 4} = \frac { x ^ { 2} - x } { 8}#?

First eliminate the fractions by multiplying both sides by the least common multiply 8

# 8 xx ( x+5)/4 = 8 xx ( x^2-x)/8# This gives

# 2 xx ( x + 5) = x^2 -x # removing the parenthesis gives

# 2x + 10 = x^2 -x# subtract -2x and -10 from both sides

# 2x-2x + 10-10 = x^2 -x -2x -10# combining like terms gives.

# 0 = x^2 -3x -10# Since the C term (-10) is negative one binomial must be negative and one binomial must be positive. Since the B term -3x is negative the negative binomial must be greater than the positive binomial when they are algebraically added.

so # 0 = ( x -5) xx ( x +2) # solve for x in each binomial

# 0 = x -5 # add five to both sides

# 0 + 5 = x -5 +5# which gives.

# 5 = x#

# 0 = x +2 # subtract 2 from both sides

# 0 -2 = x + 2 -2# which gives

# -2 = x #

#8(x+5)=4(x^2-x)#

#2(x+5)=(x^2-x)#

#2x+10=x^2-x#

#x^2-x-2x-10=0#

#x^2-3x-10=0#

#(x-5)(x+2)=0#

#x=5,-2#

#(x+5)/4=(x^2-x)/8#

multily both sides by #32#

#:.32 xx (x+5)/4=32 xx (x^2-x)/8#

#:.(cancel32^8(x+5))/cancel4^1=(cancel32^4(x^2-8))/cancel8^1#

#:.8(x+5)=4(x^2-8)#

#:.8x+40=4x^2-32#

#:.4x^2-32=8x+40#

#:.4x^2-12x-40#

#:.(2x+4)(2x-10)#

#:.2x=-4 or 2x=10#

#:.x=-4/2 or x=10/2#

#:.color(blue)(x=-2 or x= 5#

substitute #color(blue)(x=-2#

#:.((color(blue)(-2))+5)/4=((color(blue)(-2))^2-(color(blue)(-2)))/8#

#:.3/4=(4+2)/8#

#:.3/4=cancel6^color(blue)3/cancel8^color(blue)4#

#:.3/4=3/4#

substitute #color(blue)(x=5#

#:.((color(blue)5)+5)/4=((color(blue)5)^2-(color(blue)5))/8#

#:.cancel10^color(blue)5/cancel4^color(blue)2=(25-5)/8#

#:.5/2=cancel20^color(blue)5/cancel8^color(blue)2#

#:.color(blue)(5/2=5/2#