How do you evaluate #\frac { 2x ^ { 2} + 2x } { 2x ^ { 2} } \cdot \frac { x ^ { 2} - 3x } { x ^ { 2} - 2x - 3}#?

2 Answers
Dr. S
Jan 14, 2018

1

Explanation:

#(2x^2+2x)/(2x^2)*(x^2-3x)/(x^2-2x-3)#

#=(2x(x+1))/(2x^2)*(x(x-3))/((x+1)(x-3))#

#=((cancel(2x)cancel((x+1)))/cancel(2x^2)*(cancel(x)cancel((x-3)))/(cancel((x+1))cancel((x-3))))#

#=1#

Jim G.
Jan 14, 2018

#1#

Explanation:

#"factorise and cancel common factors where possible"#

#2x^2+2x=2x(x+1)larrcolor(blue)"common factor of 2x"#

#x^2-3x=x(x-3)larrcolor(blue)"common factor of x"#

#x2-2x-3=(x-3)(x+1)larrcolor(blue)"quadratic factorising"#

#rArr(2x^2+2x)/(2x^2)xx(x^2-3x)/(x^2-2x-3)#

#=(cancel(2x)cancel((x+1)))/(cancel(2x)cancel(x))xx(cancel(x)cancel((x-3)))/(cancel((x-3))cancel((x+1)))#

#=1xx1=1#

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