1.#1/2t+36=-5?" "# 2. Solve for #q:(5q-3n)/m=f" "#3.Two numbers add up to #157# and the first is #17# bigger than the second. What is are the two numbers? #" "#4.Solve #3x-2=4(x-1)+2x" "# 5.Solve for r:#" "A=2 pi " radius " h#

1 Answer
Feb 28, 2018
  1. #t=-82#
  2. #q = (fm+3n)/5" " or q= (fm)/5 +(3n)/5#
  3. The numbers are #120 and 137#
  4. #x =2/3#
  5. #r = A/(2pih)#

Explanation:

  1. Solve for #t#

#1/2 t+36 -36 = -5-36" "larr# subtract 36 from both sides
#1/2t = -41#

#(cancel2xx1)/cancel2t = 2xx-41" "larr# multiply both sides by #2#
#t =-82#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2. Make # q # the subject

#(5q-3n)/m=f" "larr# multiply by #m

#5q -3n = fm#

#5q = fm +3n" "larr# add #3n# to both sides

#q = (fm+3n)/5" " or (fm)/5 +(3n)/5#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
3. Let the smaller number be #x#
The larger number is #17# more, so it will be #x+17#

#x+x+17 = 257" "larr# their sum is #157#

#2x = 240" "larr# subtract 17 from both sides

#x = 120#

The numbers are #120 and 137#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
4. #" "3x-2=4(x-1)+2x" "larr# remove the brackets

#3x-2 = 4x-4+2x" "larr# keep the #x# terms on the right

#-2+4 = 6x-3x#

#2 = 3x" "larr# divide by #3#

#2/3 =x#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
5. Make #r# the subject in #A = 2 pi rh#

#2pirh = A" "larr# divide both sides by #2pih#

#r = A/(2pih)#