Question

2(a-1)=3(a+1)

3(x-3)=5(1.5+x)

2(1.5m+3)=3.5m-1

5n+3=2(n+2)-3n

5(f+2)=9+5f
please help me and show all work please

Answer 1

all these are single variable equations and i presume you're supposed to solve for the variables...

i'll explain to you how to solve for equations with that structure and solve one for u :) hopefully you can do the rest :D

according to the distributive property of math, a(b+c) = ab + ac and a(b-c) = ab - ac

that's all there is to it, solve for your variable and you're through :)
so with the first one for example...

2(a-1)=3(a+1)
2(a) - 2(1) = 3(a) + 3(1)
2a - 2 = 3a + 3
making "a" the subject...
-2-3 =3a-2a
-5 = a
=> a = -5 :D can u solve the rest?

Answer 2

help me someone

Answer 3

i think i did lol :P did i miss something?

Answer 4

where ???

Answer 5

I explained it to you right up there... :/

Answer 6

ok thanks i need help on the second one and im pretty sure i can do the rest but tht has a decimal and i tried but idk

Answer 7

3(x-3)=5(1.5+x)
3(x-3)=5((3/2)+x) 1.5 = (3/2)
3x-9 = 15/2 + 5x multiply through by 2 to cancel the fraction
6x - 18 = 15 +10x
-18-15 = 10x - 6x
-33 = 4x
-33/4 = x

=> x = -(33/4)

Answer 8

ok what??

Answer 9

did that make any sense to u?

Answer 10

no it didn't

Answer 11

ok umm lemme break it down... umm err... hmmm.... which part didn't make sense...?

Answer 12

all of it

Answer 13

ok umm lemme break it down... umm err... hmmm.... which part didn't make sense...?

Answer 14

k, lemme try again... now watch carefully though it'll take a while while i type so wait patiently :)

Answer 15

normally, when you have an equation with a single variable in it, you will be required to find the value of that variable or letter... so to find those out, we have to understand something called change of subject...

that basically means that if you have an equation with a variable in it that you're looking for, play with the equation until u make that variable stand alone on one side of the equation.

there is also a rule that, if you do something to the left side of an equation, you do the same thing to the right side. you'll know why later in my explanation.

now, on to the distributive property, just accept it as it is for now that a(b+c) = a(b)+a(c)
that is to say that a times b plus a times c. it is called factorization but lets not go there yet.

***taking everything i have said above... if you don't understand anything up to this ***point, don't continue reading, start again from "normally..." . if you still don't get it up ***to this point, let me know, otherwise keep reading :)

let's solve your first equation as an example to explaining how to solve such equations.
2(a-1)=3(a+1)
lets take the left had side first... 2(a-1)
from the distributive property,
2(a-1) = 2(a) - 2(1)
= 2a -2

now lets take the right hand side... 3(a+1)
from the distributive property,
3(a+1) = 3(a) + 3(1)
= 3a + 3

now lets combine both...
2(a-1) = 3(a+1)
2a-2 = 3a +3


I hope that made sense... now lets move on to the next concept... change of subject... making "a" stand alone...
2a-2 = 3a+3
lets make all the "a" go to the right side...
for that to happen, we will subtract 2a from the left side so that the 2a that is already there can vanish since 2a - 2a = 0 .... but remember that whatever you do to the left hand side... you do the same to the right hand side...
2a - 2a - 2 = 3a - 2a + 3
0-2 = a + 3

the "a" is still not alone because + 3 is on the same side of the equation as "a" so we have to subtract 3 on both sides of the equation to get rid of the 3 on the right side...
-2-3 = a + 3 - 3
-5=a + 0
-5=a
this means that a=-5 ...... I hope this made more sense... let me know if it didn't make sense... i'll try again :) if u don't understand any part, point out that part to me :)

Answer 16

good job sasogeek

Answer 17

thanks :)

Answer 18

:D