Question

Solve: $ 8x+\dfrac{21}{4}=3x+7 $
A. $ \dfrac{2}{9} $
B. $ \dfrac{7}{20} $
C. $ \dfrac{6}{17} $
D. $ \dfrac{4}{21} $

Answer 1

Hint: If two quantities are equal, then they will still be equal if we add / subtract or multiply / divide both of them by the same quantity.
In order to solve for x, we must keep only the variable on one side of the equality sign and only numbers on the other side.
A quick calculation reveals that the denominator of the answer must be a multiple of 4.

Complete step by step answer:
Given that $ 8x+\dfrac{21}{4}=3x+7 $ .
On multiplying both sides by 4, we get:
⇒ $ 32x+21=12x+28 $
Subtracting $ 12x $ from both sides, we get:
⇒ $ 20x+21=28 $
Subtracting 21 from both sides, give us:
⇒ $ 20x=7 $
And dividing both sides by 20, will give us:
⇒ $ x=\dfrac{7}{20} $
The correct answer is B. $ \dfrac{7}{20} $ .

Note: Some Rules of Equality:
If $ a=b $ , then $ a\pm x>b\pm y $ , $ a{{(x)}^{\pm 1}}=b{{(x)}^{\pm 1}} $ and $ \dfrac{x}{a}=\dfrac{x}{b} $ for all x.
If $ a=b $ and $ x>y $ , then $ a+x>b+y $ , but we cannot say anything definite about $ a-x,b-x $ or $ ax,bx $ .

Some Rules of Inequalities:
If $ a>b $ , then $ a\pm x>b\pm x $ .
If $ a>b $ , then $ a{{(x)}^{\pm 1}}>b{{(x)}^{\pm 1}} $ if x>0 and $a{(x)^{\pm 1}}$ < $b{(x)^{\pm}}$ if $ x<0 $ .
If $ a>b $ , then $ \dfrac{x}{a}$ > $\dfrac{x}{b}$ if a,b>0 OR a,b<0.
If a>b and x>y, then a+x>b+y, but we cannot say anything definite about a-x and b-x or ax and bx.