Inequalities

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PSAT Math › Inequalities

Questions 1 - 10

Solve for :

$x<-\frac{1}{4}$

$x>-\frac{25}{4}$

$x>-\frac{35}{8}$

$x<-\frac{25}{4}$

$x>-\frac{12}{25}$

Explanation

Begin by moving all of the values to the left side of the inequality:

becomes

Next, move the to the right side:

Finally, divide both sides by :

$x>-\frac{25}{4}$

Solve for :

$x<-\frac{1}{4}$

$x>-\frac{25}{4}$

$x>-\frac{35}{8}$

$x<-\frac{25}{4}$

$x>-\frac{12}{25}$

Explanation

Begin by moving all of the values to the left side of the inequality:

becomes

Next, move the to the right side:

Finally, divide both sides by :

$x>-\frac{25}{4}$

Solve for :

$x>-\frac{28}{3}$

$x<-\frac{28}{3}$

$x>\frac{18}{3}$

$x<\frac{18}{3}$

$x<\frac{31}{4}$

Explanation

First, move the values to the left side of the inequality:

becomes

Next, move the to the right side:

Finally, divide by . Remember:you must flip the inequality sign when you multiply or divide by a negative number.

$x>-\frac{28}{3}$

Solve for :

$x>-\frac{28}{3}$

$x<-\frac{28}{3}$

$x>\frac{18}{3}$

$x<\frac{18}{3}$

$x<\frac{31}{4}$

Explanation

First, move the values to the left side of the inequality:

becomes

Next, move the to the right side:

Finally, divide by . Remember:you must flip the inequality sign when you multiply or divide by a negative number.

$x>-\frac{28}{3}$

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

$\frac{1}{2}$

Explanation

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.

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

$\frac{1}{2}$

Explanation

Solve for :

$x < \frac{137}{3}$

$x > \frac{130}{4}$

Explanation

First, get the factors on the left side of the inequality:

becomes

Next, subtract from both sides:

Now, divide by . Remember: Dividing or multiplying by a negative number requires you to flip the inequality sign:

Solve for :

$x < \frac{137}{3}$

$x > \frac{130}{4}$

Explanation

First, get the factors on the left side of the inequality:

becomes

Next, subtract from both sides:

Now, divide by . Remember: Dividing or multiplying by a negative number requires you to flip the inequality sign:

Solve for $x$ .

$-2x+5\leq 10$

$x\geq \frac{5}{2}$

$x\leq \frac{5}{2}$

$x\geq -\frac{5}{2}$

$x\leq 5$

$None\ of\ the\ above$

Explanation

Move +5 using subtraction rule which will give you $-2x\leq 5$ .

Divide both sides by 2 (using division rule) and you will get $-x\leq \frac{5}{2}$ which is the same as $x\geq \frac{5}{2}$

Solve the following inequality for $\uptext{x}$ , round your answer to the nearest tenth.

$\sqrt{x + 10} > x + 1$

$-3.5 \leq x$

Explanation

The first step is to square each side of the inequality.

$\left( \sqrt{x + 10} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 10 > x^{2} + 2 x + 1$

Now subtract the left side of the inequality to make it zero, so that we can use the quadratic formula.

$x + 10 -\left( x + 10 \right)> x^{2} + 2 x + 1 -\left( x + 10 \right)$

$0 > x^{2} + x - 9$

Now we can use the quadratic formula.

Recall the quadratic formula.

$\frac{ -b \pm \sqrt{ b ^2 -4 \cdot a \cdot c }}{ 2 \cdot a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to coefficients in the quadratic equation.

$a x^{2} + b x + c = 0$

In this case , , and .

Now plug these values into the quadratic equation, and we get.

Now since we are dealing with an inequality, we put the least value on the left side, and the greatest value on the right. It will look like the following.

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