How do you use cross products to solve #2/t=5/(t-6)#?

2 Answers
May 14, 2018

Answer:

#t=-4#

Explanation:

Cross multiply the denominator with numerator like:

#2/t=5/(t-6)#

#2xx(t-6) = 5 xx t#

#2t-12 = 5t#

Add #12# both sides:
#2t-12+12# = 5t+12#

#2t cancel(-12+12)# = 5t+12#

#2t = 5t+12#
#2t-5t = 12# -----> making #t# the subject by subtracting #-5t# both sides:

#-3t=12#

#t=-12/3#

#t=-4#

May 14, 2018

Answer:

#t=-4#

Explanation:

#"using the method of "color(blue)"cross-products"#

#•color(white)(x)a/b=c/drArrbc=ad#

#rArr5t=2(t-6)#

#rArr5t=2t-12#

#"subtract "2t" from both sides"#

#5t-2t=cancel(2t)cancel(-2t)-12#

#rArr3t=-12#

#"divide both sides by 3"#

#(cancel(3) t)/cancel(3)=(-12)/3#

#rArrt=-4#

#color(blue)"As a check"#

Substitute this value into the equation and if both sides are equal then it is the solution.

#"left "=2/(-4)=-1/2#

#"right "=5/(-4-6)=5/(-10)=-1/2#

#rArrt=-4" is the solution"#