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PSAT Math : Inequalities

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Example Questions

Example Question #155 : Algebra

Solve for :

4x-23<7x+5

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

4x-23<7x+5

becomes

-3x-23<5

Next, move the to the right side:

-3x<28

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

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

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Example Question #156 : Algebra

Solve for :

17x +122>20x-22

Possible Answers:

x < 48

x > 48

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

x < 92

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

Correct answer:

x < 48

Explanation:

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

17x +122>20x-22

becomes

-3x +122>-22

Next, subtract 122 from both sides:

-3x>-144

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

x < 48

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Example Question #2 : How To Find The Solution To An Inequality With Division

Solve the inequality

$\frac{3x}{5} > 6$

Possible Answers:

x > 9

x < 10

x > 3

x > 10

Correct answer:

x > 10

Explanation:

First, multiplying each side of the equality by gives 3x > 30 . Next, dividing each side of the inequality by will solve for ; x > 10 .

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Example Question #5 : Inequalities

What is the solution set of the inequality $\dpi{100} \small 3x+8<35$ ?

Possible Answers:

$\dpi{100} \small x<9$

$\dpi{100} \small x>27$

$\dpi{100} \small x>9$

$\dpi{100} \small x<35$

$\dpi{100} \small x<27$

Correct answer:

$\dpi{100} \small x<9$

Explanation:

We simplify this inequality similarly to how we would simplify an equation

$\dpi{100} \small 3x+8-8<35-8$

$\dpi{100} \small \frac{3x}{3}<\frac{27}{3}$

Thus $\dpi{100} \small x<9$

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Example Question #6 : Inequalities

What is a solution set of the inequality 2x+12>42 ?

Possible Answers:

x>4

x>15

x<9

$x>\frac{3}{2}$

x<15

Correct answer:

x>15

Explanation:

In order to find the solution set, we solve 2x+12>42 as we would an equation:

2x+12>42

2x>30

x>15

Therefore, the solution set is any value of x>15 .

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Example Question #1 : Systems Of Inequalities

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

$\sqrt{x + 7} > x + 1$

Possible Answers:

2.0 < x < -3.0

-3.0 < x < 2.0

$-3.0 \leq x$

2.0 < x

x = 2.0

Correct answer:

-3.0 < x < 2.0

Explanation:

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

$\left( \sqrt{x + 7} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 7 > x^{2} + 2 x + 1$

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

$x + 7 -\left( x + 7 \right)> x^{2} + 2 x + 1 -\left( x + 7 \right)$

$0 > x^{2} + x - 6$

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 a = 1 , b = 1 , and c = -6 .

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

x = 2.0

x = -3.0

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.

-3.0 < x < 2.0

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Example Question #2 : Inequalities And Absolute Value

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

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

Possible Answers:

-3.5 < x < 2.5

2.5 < x

$-3.5 \leq x$

2.5 < x < -3.5

x = 2.5

Correct answer:

-3.5 < x < 2.5

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 a = 1 , b = 1 , and c = -9 .

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

x = 2.5

x = -3.5

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.

-3.5 < x < 2.5

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Example Question #1 : Systems Of Inequalities

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

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

Possible Answers:

0.0 < x

$-1.0 \leq x$

x = 0.0

-1.0 < x < 0.0

0.0 < x < -1.0

Correct answer:

-1.0 < x < 0.0

Explanation:

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

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

Now simplify each side.

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

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

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

$0 > x^{2} + x$

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 a = 1 , b = 1 , and c = 0 .

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

x = 0.0

x = -1.0

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.

-1.0 < x < 0.0

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Example Question #3 : Inequalities And Absolute Value

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

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

Possible Answers:

$-1.0 \leq x$

0.0 < x < -1.0

0.0 < x

x = 0.0

-1.0 < x < 0.0

Correct answer:

-1.0 < x < 0.0

Explanation:

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

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

Now simplify each side.

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

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

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

$0 > x^{2} + x$

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 a = 1 , b = 1 , and c = 0 .

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

x = 0.0

x = -1.0

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.

-1.0 < x < 0.0

Report an Error

Example Question #11 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

$\frac{3x+4}{4}<\frac{x}{2}$

Which of the following provides the complete solution set for given the above inequality?

Possible Answers:

x<-2

x<-4

x>4

$x<\frac{1}{2}$

x<2

Correct answer:

x<-4

Explanation:

To solve this problem, first cross-multiply the inequality to eliminate the denominators. Note that while this is an inequality, you can safely multiply by both denominators since both are positive so there is no need to consider flipping the direction of the inequality. The result of this step is:

6x+8<4x

Then you can combine like terms by subtracting from both sides:

2x+8<0

Then to isolate the variable term, subtract from both sides:

2x<-8

Finally, divide both sides by to get the variable alone:

x<-4

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PSAT Math : Inequalities

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #155 : Algebra

Example Question #156 : Algebra

Example Question #2 : How To Find The Solution To An Inequality With Division

Example Question #5 : Inequalities

Example Question #6 : Inequalities

Example Question #1 : Systems Of Inequalities

Example Question #2 : Inequalities And Absolute Value

Example Question #1 : Systems Of Inequalities

Example Question #3 : Inequalities And Absolute Value

Example Question #11 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

All PSAT Math Resources

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