Asked by Nikki
How do I solve these equations? I was out sick a whole week :( please no direct answers. I want to know the process as well.
-a=5
4/5x=20
-2/3y=-18
-x/3=18
Please note that the / are fractions. Thank you so much in advance!! :)
-a=5
4/5x=20
-2/3y=-18
-x/3=18
Please note that the / are fractions. Thank you so much in advance!! :)
Answers
Answered by
Jai
for the first problem:
-a = 5
to get a, you divide both sides of equation by the numerical coefficient (number before the variable) of a. the numerical coefficient of a is -1, thus:
-a/(-1) = 5/(-1)
a = -5
for problem 2, 3 and 4:
do the same procedure as what we did to problem 1. for problem 2, the numerical coefficient of x is 4/5, thus divide both sides of equation by 4/5:
(4/5)x / (4/5) = 20/(4/5)
x = 20*5/4 = ?
for problem 3, numerical coeff is -2/3. thus:
(-2/3)y / (-2/3) = -18/(-2/3)
y = 18*3/2 = ?
then also do the same for problem 4.
hope this helps~ :)
-a = 5
to get a, you divide both sides of equation by the numerical coefficient (number before the variable) of a. the numerical coefficient of a is -1, thus:
-a/(-1) = 5/(-1)
a = -5
for problem 2, 3 and 4:
do the same procedure as what we did to problem 1. for problem 2, the numerical coefficient of x is 4/5, thus divide both sides of equation by 4/5:
(4/5)x / (4/5) = 20/(4/5)
x = 20*5/4 = ?
for problem 3, numerical coeff is -2/3. thus:
(-2/3)y / (-2/3) = -18/(-2/3)
y = 18*3/2 = ?
then also do the same for problem 4.
hope this helps~ :)
Answered by
Nikki
Thanks! That helped a lot! But now I have more questions...please see my other algebra question, as well as my health question if you can. Thanks!!
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