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

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

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

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

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Solve for x

3x+7 geq -2x+4

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3x+7 geq -2x+4

3x geq -2x-3

5x geq -3

xgeq -$\frac{3}{5}$

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

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

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$(x-3)circ$\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}$

We have , find the solution set for this inequality.

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Give the solution set of the inequality:

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

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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 ]$ .

Give the solution set of the following inequality:

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

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

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

or, in interval notation, .

What is the greatest value of that makes

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

a true statement?

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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.

What is the least value of that makes

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

a true statement?

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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.

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

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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.

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

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

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

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Solve for x

3x+7 geq -2x+4

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3x+7 geq -2x+4

3x geq -2x-3

5x geq -3

xgeq -$\frac{3}{5}$

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

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

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$(x-3)circ$\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}$

We have , find the solution set for this inequality.

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Give the solution set of the inequality:

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

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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 ]$ .

Give the solution set of the following inequality:

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

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

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

or, in interval notation, .

What is the greatest value of that makes

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

a true statement?

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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.

What is the least value of that makes

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

a true statement?

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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.

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

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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.