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

Card 0 of 70

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

Compare your answer with the correct one above

Question

Solve for x

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

Answer

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

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

$\small 5x \geq -3$

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

Compare your answer with the correct one above

Question

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

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

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:

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

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

Compare your answer with the correct one above

Question

We have , find the solution set for this inequality.

Answer

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

Compare your answer with the correct one above

Question

Give the solution set of the inequality:

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

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,

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

Compare your answer with the correct one above

Question

Give the solution set of the following inequality:

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

Answer

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

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

or, in interval notation, .

Compare your answer with the correct one above

Question

What is the greatest value of that makes

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

a true statement?

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.

Compare your answer with the correct one above

Question

What is the least value of that makes

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

a true statement?

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.

Compare your answer with the correct one above

Question

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

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.

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

Compare your answer with the correct one above

Question

Solve for x

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

Answer

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

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

$\small 5x \geq -3$

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

Compare your answer with the correct one above

Question

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

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

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:

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

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

Compare your answer with the correct one above

Question

We have , find the solution set for this inequality.

Answer

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

Compare your answer with the correct one above

Question

Give the solution set of the inequality:

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

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,

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

Compare your answer with the correct one above

Question

Give the solution set of the following inequality:

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

Answer

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

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

or, in interval notation, .

Compare your answer with the correct one above

Question

What is the greatest value of that makes

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

a true statement?

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.

Compare your answer with the correct one above

Question

What is the least value of that makes

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

a true statement?

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.

Compare your answer with the correct one above

Question

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

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.

Compare your answer with the correct one above

Tap the card to reveal the answer