How to find the solution to an inequality with multiplication

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SAT Math › How to find the solution to an inequality with multiplication

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1

We have , find the solution set for this inequality.

CORRECT

0

0

0

0

Explanation

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

2

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

n2 < n

0

|n2 - 1| > 1

CORRECT

(n-1)2 > n

0

16n2 - 1 = 0

0

n2 < 2n

0

Explanation

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

3

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

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

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

CORRECT

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

0

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

0

None of the other answers are correct.

0

The rational expression is undefined.

0

Explanation

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

Let us simplify the second expression. We know that:

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

So we can cancel out as follows:

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

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

4

Give the solution set of this inequality:

$|-4x+ 17| \ge 67$

$(-\infty, -12.5 ] \cup [21, \infty )$

CORRECT

0

0

$(-\infty, -21 ] \cup [12.5, \infty )$

0

The set of all real numbers

0

Explanation

The absolute value inequality

$|-4x+ 17| \ge 67$

can be rewritten as the compound inequality

$-4x+ 17 \le - 67$ or $-4x+ 17 \ge 67$

Solve each inequality separately, using the properties of inequality to isolate the variable on the left side:

$-4x+ 17 \le - 67$

Subtract 17 from both sides:

$-4x+ 17 - 17 \le - 67 - 17$

$-4x \le - 84$

Divide both sides by , switching the inequality symbol since you are dividing by a negative number:

$-4x \div (-4) \ge - 84 \div (-4)$

$x \ge 21$ ,

which in interval notation is $[21, \infty )$

The same steps are performed with the other inequality:

$-4x+ 17 \ge 67$

$-4x+ 17- 17 \ge 67 - 17$

$-4x \ge 50$

$-4x \div (-4) \le 50 \div (-4)$

$x \le -12.5$

which in interval notation is $(-\infty, -12.5 ]$ .

The correct response is the union of these two sets, which is

$(-\infty, -12.5 ] \cup [21, \infty )$ .

5

Find the maximum value of , from the system of inequalities.

$y\geq 2$

$4x+5\geq y+3$

$1\leq x\leq5$

CORRECT

0

0

0

0

Explanation

First step is to rewrite $4x+5\geq y+3$

$4x+5\geq y+3$

$4x+2\geq y$

Next step is to find the vertices of the bounded region. We do this by plugging in the x bounds into the equation. Don't forgot to set up the other x and y bounds, which are given pretty much.

The vertices are

Now we plug each coordinate into , and what the maximum value is.

So the maximum value is

6

What value must take in order for the following expression to be greater than zero?

$\frac{3}{k}-\frac{5}{7}$

CORRECT

0

0

0

0

Explanation

is such that:

$\frac{3}{k}-\frac{5}{7}>0$

Add $\frac{5}{7}$ to each side of the inequality:

$\frac{3}{k}>\frac{5}{7}$

Multiply each side of the inequality by :

$\frac{21}{k}>5$

Multiply each side of the inequality by :

Divide each side of the inequality by :

$k< \frac{21}{5}$

You can now change the fraction on the right side of the inequality to decimal form.

$k<\frac{21}{5}=\frac{20}{5}+\frac{1}{5}= 4.2$

The correct answer is , since k has to be less than for the expression to be greater than zero.

7

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

All of the answers choices are valid.

0

-4

0

-3

0

-2

0

-1

CORRECT

Explanation

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

8

Solve for x

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

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

CORRECT

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

0

$\small x \leq \frac{3}{5}$

0

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

0

Explanation

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

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

$\small 5x \geq -3$

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

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