.
x=30-25 =5
\(\huge\color{blue}{ \rm S }\) \(\huge\color{blue}{ \rm S }\) \(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \color{lightcyan}{\Huge\frac{~~\frac{~}{~~~\frac{~\frac{}{}~}{~\frac{~}{~}}~~~~}~~~}{~\frac{\frac{~}{~\frac{~\frac{}{~}}{~}}~~}{~}~~} } }}}\) \(\huge\color{blue}{ \rm S }\) \(\huge\color{blue}{ \rm S }\) \(\huge\color{blue}{ \rm Perimeter~of~a~square = 4\times s }\) ("s" is juts a side)
this is because all sides of a square are equal.
Now, you know that a side of a square is \(\large\color{slate}{ 7x+3 }\) yes yes... \(\large\color{slate}{ 4( 7x+3 )= 25x+30 }\)
that is the set up, b/c `25x+30` is the perimeter.
and your perimeter is equal to the "side times 4" So we are setting that up, P=4 * s, BUT we are now actually plugging the value of your perimeter (P) and the value of the side (s).
im confussed
why?
how di u get 12
oh,Pandeee_ expanded the parenthesis on the left side:
anyway, so \(\large\color{slate}{ 4( 7x+3 )= 25x+30 }\) \(\large\color{slate}{ 28x+12= 25x+30 }\) (as you already said for the left side)
Pandeee_ , do you understand what is being done?
oh ok
so can you find the "x" ?
Shirtcut the math , lol.... \(\large\color{slate}{ 28x+12\color{red}{-25x+12}= 25x+30\color{red}{-25x+12} }\)
I am subtracting the red from both sides, and what will this give me?
yes, but, \(\large\color{slate}{ 28x+12\color{red}{-25x+12}= 25x+30\color{red}{-25x+12} }\) \(\large\color{slate}{ 25x\color{blue}{+3x}+12\color{red}{-25x+12}= 25x+12\color{blue}{+18}\color{red}{-25x+12} }\)
oh I forgot parenthesis
\(\large\color{slate}{ 28x+12=25x+30 }\) you expression we are at. \(\large\color{slate}{ 28x+12\color{red}{-25x-12}=25x+30\color{red}{-25x-12} }\)
i wrote that we are adding 12 to both sides, but we are not. We are subtracting 12 from both sides.
I am subtracting the `smaller term with variable` together with the `smaller constant` from both sides.
30-12 = ?
yes
3x=18 so, x=?
yes
Not a problem:)
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