How do you solve #7(4x +9)-13=-87#?
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Start with the easiest step, which is adding #13# to both left hand side and right hand side.
#7(4x+9) cancel(-13 + 13) = -87 + 13#
#7(4x+9) = -74#
Next, divide both LHS and RHS by #7#
#(cancel7(4x+9))/cancel7 = -74/7#
#4x+9 = -74/7#
Subtracting LHS and RHS by #9#,
#4x cancel(+9-9) = (-74/7)-9#
#4x = -137/7#
Dividing LHS and RHS by #4#,
#cancel4x/cancel4 = -137/28#
#x = -137/28 or -4.89#
First of all you will have to expand your bracket and this is how you will get a normal solving equation and be able to find what #x# is.
To expand the bracket you have to know that you need to multiply everything inside the bracket by the outside:
#7(4x+9)-13=-87#
#28x+63-13=-87# Expand the bracket
#28x+63=-74# Get rid of -13 by ading it to both sides because you have to always do the same to both sides of the equal sign. Therefore if you have -13 you +13.
#28x=-137# You then subtract 63 because you need #x# on its own. To do this you look at the sign before the number 63, in this case you have +63 which means you have to subtract 63 and you get that #28x=-137# and you divide by how many #x# you have to find what 1 #x# is
#x=-4.892857143#
or round it #-4.9#
I hope this helps you!
#color(blue)(x=4.893# to the nearest 3 decimal places
#7(4x+9)-13=-87#
#:.28x+63-13=-87#
#:.28x=-87-63+13#
#:.28x=-150+13#
#:.28x=-137#
#:.x=-137/28#
#:.color(blue)(x=4.893# to the nearest 3 decimal places
