+0

# Hello May I have assitence

+4
373
7

I'm given the equation $$-\frac{1}{2}$$ x + 3 = –x + 7

and my options are

1/2

-8

I keep getting -2.67 (Note that number is rounded to the nearest hundreth)

The tatic I have been using is multiplying each term by -2 or the denominator of -1/2

THANKS SO MUCH FOR THE TIME YOU SPEND IN EFFORT OF TEACH THIS CONCEPT TO ME.

Jun 28, 2019

#1
+2

Right now, you are given -1/2x+3=-x+7

1/2x+3=7

Subtract  3 on both sides

1/2x=4

mutiply by 2

x=8

Jun 28, 2019
#2
0

Thanks alot for your effort : ) but the one part I'm having trouble grasping is when you add x to both side's do you mean are you carrying x from the rigth side of the equation over to the left? and if so what were the effects exactly because x seemed to have vanished formt he problem entrily leaving only the x that is being multiplied by the 1/2 in it's original state...

Wait sorry XD for that incovience I see now 1/2 times x becomes positive when you carry over the x formt he left side THANKS ALOT and have a great day!

edited by HiylinLink  Jun 28, 2019
#3
+1

If you want to simply “move” the x over, you need to reverse the sign

for example

x+3=2x

x-2x+3=0

-x+3=0

and to move 3 over,

-x=-3

then mutiply by -1

x=3

Guest Jun 28, 2019
#4
+3

Also, x didn’t vanish like you mentioned, it was originally -1/2x it is now positive 1/2x

Guest Jun 28, 2019
#5
0

Yes, I understand how to do a varibles inverse I didn't realize for fractions it's considered adding on varibles in the this situation. thanks!

#6
+5

Using the tactic that you had been using.....

-$$\frac12$$x + 3  =  -x + 7

Multiply both sides of the equation by  -2

-2( -$$\frac12$$x + 3 )  =  -2( -x + 7 )

Distribute the  -2  to each term

x - 6  =  2x - 14

Add  14  to both sides of the equation.

x - 6 + 14  =  2x - 14 + 14

x + 8  =  2x

Subtract  x  from both sides.

x + 8 - x  =  2x - x

On the left,  we have  x - x = 0

8  =  2x - x

On the right,  we have  2x - 1x = 1x  because  2 things - 1 thing = 1 thing

8  =  1x

8  =  x

Jun 28, 2019
#7
0

THANKS ALOT!  it's definetly helpful to see this done both ways....