Question

Solve $\dfrac{{3x - 4}}{2} \geqslant \dfrac{{x + 1}}{4} - 1$
$A)x \leqslant 1$
$B)x > 1$
$C)x \geqslant 1$
$D)x < 2$

Answer 1

Hint: The addition is the sum of given two or more than two numbers, or variables and in addition, if we sum the two or more numbers then we obtain a new frame of the number will be found, also in subtraction which is the minus of given two or more than two numbers, but here comes with the condition that in subtraction the greater number sign represented in the number will stay constant example $2 - 3 = - 1$
We also going to make use of the simple concept on the inequalities, which is $ \geqslant $ a symbol of greater or equals to sign, $ > $ a symbol of strictly greater than sign, $ < $ a symbol of strictly less than sign, and $ \leqslant $ a symbol of less than equals to sign, which means it is either less than the given values or equals to the given values.

Complete step by step answer:
Since given that the equation $\dfrac{{3x - 4}}{2} \geqslant \dfrac{{x + 1}}{4} - 1$ and then we need to find the value of the unknown variable $x$, so we will make use of the basic mathematical operations to simplify further.
Starting with the cross-multiplication method on the right-hand side, we have $\dfrac{{3x - 4}}{2} \geqslant \dfrac{{x + 1}}{4} - 1 \Rightarrow \dfrac{{3x - 4}}{2} \geqslant \dfrac{{x + 1 - 4}}{4}$
Now canceling the common terms, we get $\dfrac{{3x - 4}}{2} \geqslant \dfrac{{x + 1 - 4}}{4} \Rightarrow 3x - 4 \geqslant \dfrac{{x + 1 - 4}}{2}$
Again, by the cross multiplication, we have $3x - 4 \geqslant \dfrac{{x + 1 - 4}}{2} \Rightarrow 2(3x - 4) \geqslant x + 1 - 4$
Now by the multiplication operation, we get $6x - 8 \geqslant x + 1 - 4$ and by subtraction $6x - 8 \geqslant x - 3$
now Turing the variables on the left-hand side and also the numbers on the right-hand side we get $6x - 8 \geqslant x - 3 \Rightarrow 6x - x \leqslant - 3 + 8$ while changing the values on the equals to, the sign of the values or the numbers will change.
Again, by subtraction we get $6x - x \leqslant - 3 + 8 \Rightarrow 5x \leqslant 5$
Finally, by the division operation, we have $5x \leqslant 5 \Rightarrow x \leqslant \dfrac{5}{5}$ and hence we get $x \leqslant 1$

So, the correct answer is “Option A”.

Note:
The other two operations are multiplication and division operations.
Since multiplicand refers to the number multiplied. Also, a multiplier refers to multiplying the first number. Have a look at an example; while multiplying $5 \times 7$the number $5$ is called the multiplicand and the number $7$ is called the multiplier. Like $2 \times 3 = 6$ or which can be also expressed in the form of $2 + 2 + 2(3times)$
The process of the inverse of the multiplication method is called division. Like $x \times y = z$ is multiplication thus the division sees as $x = \dfrac{z}{y}$. Like $5x \leqslant 5 \Rightarrow x \leqslant \dfrac{5}{5} = 1$
Hence using simple operations, we solved the given problem.