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Common Core: High School - Algebra : Reasoning with Equations & Inequalities

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

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Example Questions

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Example Question #7 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

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

$\sqrt{x + 35} > x + 1$

Possible Answers:

$-6.4 \leq x$

x = 5.4

5.4 < x < -6.4

5.4 < x

-6.4 < x < 5.4

Correct answer:

-6.4 < x < 5.4

Explanation:

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

$\left( \sqrt{x + 35} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 35 > x^{2} + 2 x + 1$

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

$x + 35 -\left( x + 35 \right)> x^{2} + 2 x + 1 -\left( x + 35 \right)$

$0 > x^{2} + x - 34$

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 = -34 .

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

x = 5.4

x = -6.4

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.

-6.4 < x < 5.4

Report an Error

Example Question #681 : Algebra

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

$\sqrt{x + 17} > x + 1$

Possible Answers:

x = 3.5

3.5 < x < -4.5

-4.5 < x < 3.5

3.5 < x

$-4.5 \leq x$

Correct answer:

-4.5 < x < 3.5

Explanation:

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

$\left( \sqrt{x + 17} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 17 > x^{2} + 2 x + 1$

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

$x + 17 -\left( x + 17 \right)> x^{2} + 2 x + 1 -\left( x + 17 \right)$

$0 > x^{2} + x - 16$

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 = -16 .

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

x = 3.5

x = -4.5

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.

-4.5 < x < 3.5

Report an Error

Example Question #2 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

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

$\sqrt{x + 8} > x + 1$

Possible Answers:

$-3.2 \leq x$

x = 2.2

2.2 < x

2.2 < x < -3.2

-3.2 < x < 2.2

Correct answer:

-3.2 < x < 2.2

Explanation:

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

$\left( \sqrt{x + 8} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 8 > x^{2} + 2 x + 1$

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

$x + 8 -\left( x + 8 \right)> x^{2} + 2 x + 1 -\left( x + 8 \right)$

$0 > x^{2} + x - 7$

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 = -7 .

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

x = 2.2

x = -3.2

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.

-3.2 < x < 2.2

Report an Error

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

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

$\sqrt{x + 15} > x + 1$

Possible Answers:

x = 3.3

-4.3 < x < 3.3

3.3 < x

$-4.3 \leq x$

3.3 < x < -4.3

Correct answer:

-4.3 < x < 3.3

Explanation:

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

$\left( \sqrt{x + 15} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 15 > x^{2} + 2 x + 1$

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

$x + 15 -\left( x + 15 \right)> x^{2} + 2 x + 1 -\left( x + 15 \right)$

$0 > x^{2} + x - 14$

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 = -14 .

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

x = 3.3

x = -4.3

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.

-4.3 < x < 3.3

Report an Error

Example Question #11 : Systems Of Inequalities

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

$\sqrt{x + 16} > x + 1$

Possible Answers:

3.4 < x < -4.4

3.4 < x

x = 3.4

-4.4 < x < 3.4

$-4.4 \leq x$

Correct answer:

-4.4 < x < 3.4

Explanation:

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

$\left( \sqrt{x + 16} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 16 > x^{2} + 2 x + 1$

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

$x + 16 -\left( x + 16 \right)> x^{2} + 2 x + 1 -\left( x + 16 \right)$

$0 > x^{2} + x - 15$

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 = -15 .

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

x = 3.4

x = -4.4

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.

-4.4 < x < 3.4

Report an Error

Example Question #12 : Systems Of Inequalities

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

$\sqrt{x + 34} > x + 1$

Possible Answers:

x = 5.3

5.3 < x < -6.3

5.3 < x

-6.3 < x < 5.3

$-6.3 \leq x$

Correct answer:

-6.3 < x < 5.3

Explanation:

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

$\left( \sqrt{x + 34} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 34 > x^{2} + 2 x + 1$

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

$x + 34 -\left( x + 34 \right)> x^{2} + 2 x + 1 -\left( x + 34 \right)$

$0 > x^{2} + x - 33$

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 = -33 .

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

x = 5.3

x = -6.3

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.

-6.3 < x < 5.3

Report an Error

Example Question #13 : Systems Of Inequalities

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

$\sqrt{x + 19} > x + 1$

Possible Answers:

-4.8 < x < 3.8

$-4.8 \leq x$

3.8 < x < -4.8

x = 3.8

3.8 < x

Correct answer:

-4.8 < x < 3.8

Explanation:

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

$\left( \sqrt{x + 19} \right)^2> \left( x + 1 \right)^2$

Now simplify each side.

$x + 19 > x^{2} + 2 x + 1$

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

$x + 19 -\left( x + 19 \right)> x^{2} + 2 x + 1 -\left( x + 19 \right)$

$0 > x^{2} + x - 18$

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 = -18 .

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

x = 3.8

x = -4.8

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.

-4.8 < x < 3.8

Report an Error

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

x = 2.0

-3.0 < x < 2.0

$-3.0 \leq x$

2.0 < x < -3.0

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

Report an Error

Example Question #511 : New Sat

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

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

Possible Answers:

x>4

x<-4

x<2

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

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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Example Question #1 : Use Completing The Square To Derive The Quadratic Formula: Ccss.Math.Content.Hsa Rei.B.4a

Solve $x^{2} + 17 x - 33 = 0$ by completing the square. Round your answer to the nearest hundredth.

Possible Answers:

x = -0.52, x = -18.64

x = -6.76, x = 4.76

x = 1.26, x = -18.26

x = 1.76 , x = -18.76

x = -0.74, x = -16.26

Correct answer:

x = 1.76 , x = -18.76

Explanation:

The first step is to add to both sides.

$x^{2} + 17 x - 33 + 33 = 0 + 33$

$x^{2} + 17 x = 33$

Now we take the coefficient in front of the $\uptext{x}$ term, divide it by , square it and add it to each side.

$x^{2} + 17 x + \frac{17}{2} ^2 = 33 + \frac{17}{2} ^2$

$x^{2} + 17 x + \frac{289}{4} = 33 + \frac{289}{4}$

Now we factor the left hand side, and add up the right hand side.

$\left( x + \frac{17}{2} \right)^2= \frac{421}{4}$

Now we take the square root of each side.

$\sqrt{\left( x + \frac{17}{2} \right)^2}= \sqrt{ \frac{421}{4} }$

$x + \frac{17}{2} = \sqrt{ \frac{421}{4} }$

Now we subtract $\frac{17}{2}$ from each side.

$x + \frac{17}{2} - \frac{17}{2} = \sqrt{ \frac{421}{4} }- \frac{17}{2}$

$x = - \frac{17}{2} + \frac{\sqrt{421}}{2}$

Since we are taking the square root, we need to set up equations to solve for $\uptext{x}$ .

$x = - \frac{17}{2} + \frac{\sqrt{421}}{2}$

x = 1.76

$x = - \frac{\sqrt{421}}{2} - \frac{17}{2}$

x = -18.76

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : Reasoning with Equations & Inequalities

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #7 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #681 : Algebra

Example Question #2 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #1 : Solve One Variable Linear Equations And Inequalities: Ccss.Math.Content.Hsa Rei.B.3

Example Question #11 : Systems Of Inequalities

Example Question #12 : Systems Of Inequalities

Example Question #13 : Systems Of Inequalities

Example Question #1 : Systems Of Inequalities

Example Question #511 : New Sat

Example Question #1 : Use Completing The Square To Derive The Quadratic Formula: Ccss.Math.Content.Hsa Rei.B.4a

All Common Core: High School - Algebra Resources

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