How to find the solution to an inequality with multiplication - PSAT Math

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Question

We have , find the solution set for this inequality.

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Answer

$x^2-4<0 \Rightarrow x^2<4 \Rightarrow -2<x<2$

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Question

If –1 < n < 1, all of the following could be true EXCEPT:

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Answer

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Question

(√(8) / -x ) < 2. Which of the following values could be x?

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Answer

The equation simplifies to x > -1.41. -1 is the answer.

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Question

Solve for x

$\small 3x+7 \geq -2x+4$

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Answer

$\small 3x+7 \geq -2x+4$

$\small 3x \geq -2x-3$

$\small 5x \geq -3$

$\small x\geq -\frac{3}{5}$

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Question

Fill in the circle with either $<$ , $>$ , or $=$ symbols:

$(x-3)\circ\frac{x^2-9}{x+3}$ for $x\geq 3$ .

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Answer

$(x-3)\circ\frac{x^2-9}{x+3}$

Let us simplify the second expression. We know that:

$(x^2-9)=(x+3)(x-3)$

So we can cancel out as follows:

$\frac{x^2-9}{x+3}=\frac{(x+3)(x-3)}{(x+3)}=x-3$

$(x-3)=\frac{x^2-9}{x+3}$

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Question

Give the solution set of the inequality:

$-7 \leq - \frac{3}{4}x < -4$

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Answer

Divide each of the three expressions by $-\frac{3}{4}$ , or, equivalently, multiply each by its reciprocal, $-\frac{4}{3}$ :

$-7 \leq - \frac{3}{4}x < -4$

$-7 \cdot \left ( - \frac{4}{3} \right ) \geq - \frac{3}{4}x \cdot \left ( - \frac{4}{3} \right ) > -4 \cdot \left ( - \frac{4}{3} \right )$

$\frac{28}{3} \geq x > \frac{16}{3}$

$5 \frac{1}{3} < x \leq 9 \frac{1}{3}$

or, in interval form,

$\left ( 5\frac{1}{3}, 9 \frac{1}{3} \right ]$ .

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Question

Give the solution set of the following inequality:

$32 < \frac{x - 4}{5} < 42$

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Answer

$32 < \frac{x - 4}{5} < 42$

$32\cdot 5 < \frac{x - 4}{5} \cdot 5< 42 \cdot 5$

or, in interval notation, .

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Question

What is the greatest value of that makes

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

a true statement?

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Answer

Find the solution set of the three-part inequality as follows:

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

$-45 \cdot (-6) \geq \frac{x - 7}{- 6} \cdot (-6) \geq - 22 \cdot (-6)$

$270 \geq x-7 \geq 132$

$270 + 7 \geq x-7 + 7 \geq 132 + 7$

$27 7 \geq x \geq 139$

The greatest possible value of is the upper bound of the solution set, which is 277.

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Question

What is the least value of that makes

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

a true statement?

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Answer

Find the solution set of the three-part inequality as follows:

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

$-45 \cdot (-6) \geq \frac{x - 7}{- 6} \cdot (-6) \geq - 22 \cdot (-6)$

$270 \geq x-7 \geq 132$

$270 + 7 \geq x-7 + 7 \geq 132 + 7$

$27 7 \geq x \geq 139$

The least possible value of is the lower bound of the solution set, which is 139.

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Question

Which of the following numbers could be a solution to the inequality ?

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Answer

In order for a negative multiple to be greater than a number and a positive multiple to be less than that number, that number must be negative itself. -4 is the only negative number available, and thus the correct answer.

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Question

If –1 < n < 1, all of the following could be true EXCEPT:

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Answer

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Question

(√(8) / -x ) < 2. Which of the following values could be x?

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Answer

The equation simplifies to x > -1.41. -1 is the answer.

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Question

Solve for x

$\small 3x+7 \geq -2x+4$

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Answer

$\small 3x+7 \geq -2x+4$

$\small 3x \geq -2x-3$

$\small 5x \geq -3$

$\small x\geq -\frac{3}{5}$

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Question

Fill in the circle with either $<$ , $>$ , or $=$ symbols:

$(x-3)\circ\frac{x^2-9}{x+3}$ for $x\geq 3$ .

Tap to reveal answer

Answer

$(x-3)\circ\frac{x^2-9}{x+3}$

Let us simplify the second expression. We know that:

$(x^2-9)=(x+3)(x-3)$

So we can cancel out as follows:

$\frac{x^2-9}{x+3}=\frac{(x+3)(x-3)}{(x+3)}=x-3$

$(x-3)=\frac{x^2-9}{x+3}$

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Question

We have , find the solution set for this inequality.

Tap to reveal answer

Answer

$x^2-4<0 \Rightarrow x^2<4 \Rightarrow -2<x<2$

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Question

Give the solution set of the inequality:

$-7 \leq - \frac{3}{4}x < -4$

Tap to reveal answer

Answer

Divide each of the three expressions by $-\frac{3}{4}$ , or, equivalently, multiply each by its reciprocal, $-\frac{4}{3}$ :

$-7 \leq - \frac{3}{4}x < -4$

$-7 \cdot \left ( - \frac{4}{3} \right ) \geq - \frac{3}{4}x \cdot \left ( - \frac{4}{3} \right ) > -4 \cdot \left ( - \frac{4}{3} \right )$

$\frac{28}{3} \geq x > \frac{16}{3}$

$5 \frac{1}{3} < x \leq 9 \frac{1}{3}$

or, in interval form,

$\left ( 5\frac{1}{3}, 9 \frac{1}{3} \right ]$ .

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Question

Give the solution set of the following inequality:

$32 < \frac{x - 4}{5} < 42$

Tap to reveal answer

Answer

$32 < \frac{x - 4}{5} < 42$

$32\cdot 5 < \frac{x - 4}{5} \cdot 5< 42 \cdot 5$

or, in interval notation, .

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Question

What is the greatest value of that makes

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

a true statement?

Tap to reveal answer

Answer

Find the solution set of the three-part inequality as follows:

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

$-45 \cdot (-6) \geq \frac{x - 7}{- 6} \cdot (-6) \geq - 22 \cdot (-6)$

$270 \geq x-7 \geq 132$

$270 + 7 \geq x-7 + 7 \geq 132 + 7$

$27 7 \geq x \geq 139$

The greatest possible value of is the upper bound of the solution set, which is 277.

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Question

What is the least value of that makes

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

a true statement?

Tap to reveal answer

Answer

Find the solution set of the three-part inequality as follows:

$-45 \leq \frac{x - 7}{- 6} \leq - 22$

$-45 \cdot (-6) \geq \frac{x - 7}{- 6} \cdot (-6) \geq - 22 \cdot (-6)$

$270 \geq x-7 \geq 132$

$270 + 7 \geq x-7 + 7 \geq 132 + 7$

$27 7 \geq x \geq 139$

The least possible value of is the lower bound of the solution set, which is 139.

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Question

Which of the following numbers could be a solution to the inequality ?

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Answer

In order for a negative multiple to be greater than a number and a positive multiple to be less than that number, that number must be negative itself. -4 is the only negative number available, and thus the correct answer.

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