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Given that 1 + 3x > 4 and 2x - 3 < 5, all values of x

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Given that 1 + 3x > 4 and 2x - 3 < 5, all values of x

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Given that 1 + 3x > 4 and 2x - 3 < 5, all values of x must be between which of the following pairs of integers?

A. 3 and 8
B. 1 and 4
C. 3 and 12
D. 4/3 and 5/2
E. -5 and 1

The OA is B.

In this PS question, I have to clear x of each inequality, right? Then I can get the values of x that meet the inequalities and those will be the solution.

Experts, am I correct? Can you assist me with the solution, please? Thanks.

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AAPL wrote:
Given that 1 + 3x > 4 and 2x - 3 < 5, all values of x must be between which of the following pairs of integers?

A. 3 and 8
B. 1 and 4
C. 3 and 12
D. 4/3 and 5/2
E. -5 and 1

The OA is B.

In this PS question, I have to clear x of each inequality, right? Then I can get the values of x that meet the inequalities and those will be the solution.

Experts, am I correct? Can you assist me with the solution, please? Thanks.
You can simply isolate x in each inequality.

1 + 3x > 4
3x > 3
x > 1

And
2x -3 < 5
2x < 8
x < 4

Together 1 < x < 4. The answer is B

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Quote:
Given that 1 + 3x > 4 and 2x - 3 < 5, all values of x must be between which of the following pairs of integers?

A. 3 and 8
B. 1 and 4
C. 3 and 12
D. 4/3 and 5/2
E. -5 and 1

The OA is B.

In this PS question, I have to clear x of each inequality, right? Then I can get the values of x that meet the inequalities and those will be the solution.

Experts, am I correct? Can you assist me with the solution, please? Thanks.
Hi AAPL,
Let's take a look at your question.

$$1+3x>4$$
$$3x>4-1$$
$$3x>3$$
$$x>\frac{3}{3}$$
$$x>1 ... (i)$$

The next inequality is:
$$2x-3<5$$
$$2x<5+3$$
$$2x<8$$
$$x<\frac{8}{2}$$
$$x<4... (ii)$$

Combining (i) and (ii), we get:
1 < x <4
Which shows that the values of x must be between 1 and 4.

Therefore, Option B is correct.

Hope it helps.
I am available if you'd like any follow up.

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