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

Card 0 of 8

Question

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

Answer

N_part_1

N_part_2

N_part_3

N_part_4

N_part_5

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Question

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

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

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.

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.

Answer

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Question

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

Answer

is such that:

Add to each side of the inequality:

Multiply each side of the inequality by :

Multiply each side of the inequality by :

Divide each side of the inequality by :

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

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

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Question

Give the solution set of this inequality:

Answer

The absolute value inequality

can be rewritten as the compound inequality

or

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

Subtract 17 from both sides:

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

,

which in interval notation is

The same steps are performed with the other inequality:

which in interval notation is .

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

.

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Question

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

Answer

First step is to rewrite

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

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