Answer :

VectorFundament120

Answer:

x ≤ 2

Step-by-step explanation:

We are given the inequality:

[tex]\displaystyle{2x-3 \leq \dfrac{x}{2}}[/tex]

First, get rid of the denominator by multiplying both sides by 2:

[tex]\displaystyle{2x\cdot 2-3\cdot 2 \leq \dfrac{x}{2}\cdot 2}\\\\\displaystyle{4x-6 \leq x}[/tex]

Add both sides by 6 then subtract both sides by x:

[tex]\displaystyle{4x-6+6 \leq x+6}\\\\\displaystyle{4x \leq x+6}\\\\\displaystyle{4x-x \leq x+6-x}\\\\\displaystyle{4x-x \leq 6}\\\\\displaystyle{3x \leq 6}[/tex]

Then divide both sides by 3:

[tex]\displaystyle{\dfrac{3x}{3} \leq \dfrac{6}{3}}\\\\\displaystyle{x \leq 2}[/tex]

Therefore, the answer is x ≤ 2

1055666

Answer: [tex]x \leq 2[/tex]

Step-by-step explanation: Given [tex]2x - 3 \leq \frac{x}{2}[/tex], we multiply 2 by both sides to cancel out the 2 in the denominator (multiplying by a number in a fraction turns it into 1, and since the denominator is one, it is the same as saying the number [or variable] on the numerator by itself.)

We then get [tex]4x - 6 \leq x[/tex].

Adding 6 to both sides, we get [tex]4x \leq x + 6[/tex].

Subtracting x from both sides, we get [tex]3x \leq 6[/tex]

Dividing by 3 from both sides, we get [tex]x \leq 2[/tex]

Hope this helped!

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