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Common Core: High School - Algebra : High School: Algebra

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

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All Common Core: High School - Algebra Resources

8 Diagnostic Tests 97 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #152 : Reasoning With Equations & Inequalities

Does the point $\left( -25 , 76 \right)$ exist on the line y = 24 x + 37?

Possible Answers:

The point $\left( -25 , 76 \right)$ exists on the line.

The point $\left( -25 , 76 \right)$ does not exist on the line.

Correct answer:

The point $\left( -25 , 76 \right)$ does not exist on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = 24 \cdot -25 + 37$

y = -600 + 37

y = -563

Since the values are not equal, then the point $\left( -25 , 76 \right)$ does not exist on the line.

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Example Question #1 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Does the point $\left( -1 , 47 \right)$ exist on the line y = - 7 x + 40?

Possible Answers:

The point $\left( -1 , 47 \right)$ does not exist on the line.

The point $\left( -1 , 47 \right)$ exists on the line.

Correct answer:

The point $\left( -1 , 47 \right)$ exists on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = -7 \cdot -1 + 40$

y = 7 + 40

y = 47

Since the values are equal, then the point $\left( -1 , 47 \right)$ exist on the line.

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Example Question #2 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Does the point $\left( 3 , -16 \right)$ exist on the line y = 10 x + 49?

Possible Answers:

The point $\left( 3 , -16 \right)$ exists on the line.

The point $\left( 3 , -16 \right)$ does not exist on the line.

Correct answer:

The point $\left( 3 , -16 \right)$ does not exist on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = 10 \cdot 3 + 49$

y = 30 + 49

y = 79

Since the values are not equal, then the point $\left( 3 , -16 \right)$ does not exist on the line.

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Example Question #153 : Reasoning With Equations & Inequalities

Does the point $\left( 0 , 32 \right)$ exist on the line y = 17 x + 32?

Possible Answers:

The point $\left( 0 , 32 \right)$ does not exist on the line.

The point $\left( 0 , 32 \right)$ exists on the line.

Correct answer:

The point $\left( 0 , 32 \right)$ exists on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = 17 \cdot 0 + 32$

y = 0 + 32

y = 32

Since the values are equal, then the point $\left( 0 , 32 \right)$ exist on the line.

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Example Question #157 : Reasoning With Equations & Inequalities

Does the point $\left( -5 , 2 \right)$ exist on the line y = 44 x + 40?

Possible Answers:

The point $\left( -5 , 2 \right)$ does not exist on the line.

The point $\left( -5 , 2 \right)$ exists on the line.

Correct answer:

The point $\left( -5 , 2 \right)$ does not exist on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = 44 \cdot -5 + 40$

y = -220 + 40

y = -180

Since the values are not equal, then the point $\left( -5 , 2 \right)$ does not exist on the line.

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Example Question #151 : Reasoning With Equations & Inequalities

Does the point $\left( -3 , 11 \right)$ exist on the line y = 11 x + 44?

Possible Answers:

The point $\left( -3 , 11 \right)$ exists on the line.

The point $\left( -3 , 11 \right)$ does not exist on the line.

Correct answer:

The point $\left( -3 , 11 \right)$ exists on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = 11 \cdot -3 + 44$

y = -33 + 44

y = 11

Since the values are equal, then the point $\left( -3 , 11 \right)$ exist on the line.

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Example Question #153 : Reasoning With Equations & Inequalities

Does the point $\left( 49 , 92 \right)$ exist on the line y = - 11 x + 37?

Possible Answers:

The point $\left( 49 , 92 \right)$ exists on the line.

The point $\left( 49 , 92 \right)$ does not exist on the line.

Correct answer:

The point $\left( 49 , 92 \right)$ does not exist on the line.

Explanation:

To see if this point exists on the line, we need to plug in the $\uptext{x}$ value into the equation and see if it equals the $\uptext{y}$ value.

$y = -11 \cdot 49 + 37$

y = -539 + 37

y = -502

Since the values are not equal, then the point $\left( 49 , 92 \right)$ does not exist on the line.

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Example Question #1 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

Where do the following equations intersect?

y = - 45 x + 19

y = 43 x - 4

Possible Answers:

$\left( \frac{64}{47} , \frac{1695}{88} \right)$

$\left( \frac{23}{88} , \frac{637}{88} \right)$

$\left( \frac{64}{47} , 13 \right)$

$\left( \frac{64}{47} , \frac{637}{88} \right)$

$\left( - \frac{45}{43} , \frac{637}{88} \right)$

Correct answer:

$\left( \frac{23}{88} , \frac{637}{88} \right)$

Explanation:

To figure out where these equations intersect, we need to set them equal to each other and solve for $\uptext{x}$ .

- 45 x + 19 = 43 x - 4

- 45 x = 43 x - 23

- 88 x = -23

$x = \frac{23}{88}$

Now we simply plug this $\uptext{x}$ value into one of the equations to get the $\uptext{y}$ value.

$y = -45 \cdot \frac{23}{88} + 19$

$y = - \frac{1035}{88} + 19$

$y = \frac{637}{88}$

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Example Question #2 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

Where do the following equations intersect?

y = 22 x - 47

y = - 44 x + 4

Possible Answers:

$\left( \frac{23}{16} , -65 \right)$

$\left( \frac{17}{22} , -30 \right)$

$\left( \frac{23}{16} , - \frac{1017}{22} \right)$

$\left( - \frac{1}{2} , -30 \right)$

$\left( \frac{23}{16} , -30 \right)$

Correct answer:

$\left( \frac{17}{22} , -30 \right)$

Explanation:

To figure out where these equations intersect, we need to set them equal to each other and solve for $\uptext{x}.$

22 x - 47 = - 44 x + 4

22 x = - 44 x + 51

66 x = 51

$x = \frac{17}{22}$

Now we simply plug this $\uptext{x}$ value into one of the equations to get the $\uptext{y}$ value.

$y = 22 \cdot \frac{17}{22} -47$

y = 17 -47

y = -30

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Example Question #3 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

Where do the following equations intersect?

y = 49 x + 17

y = - 40 x - 23

Possible Answers:

$\left( \frac{32}{17} , 3 \right)$

$\left( \frac{32}{17} , - \frac{447}{89} \right)$

$\left( - \frac{40}{89} , - \frac{447}{89} \right)$

$\left( \frac{32}{17} , \frac{1473}{89} \right)$

$\left( - \frac{49}{40} , - \frac{447}{89} \right)$

Correct answer:

$\left( - \frac{40}{89} , - \frac{447}{89} \right)$

Explanation:

To figure out where these equations intersect, we need to set them equal to each other and solve for $\uptext{x}.$

Subtract from both sides:

49 x + 17 = - 40 x - 23

Add 40x to both sides:

49 x = - 40 x - 40

Divide both sides by $89\textup:$

89 x = -40

$x = - \frac{40}{89}$

Now plug this $\uptext{x}$ value into one of the equations to solve for the $\uptext{y}$ value.

y = 49 x + 17

$y = 49 \cdot - \frac{40}{89} + 17$

Multiply $\frac{49}{1}$ by $-\frac{40}{89}\textup:$

$y = - \frac{1960}{89} + 17$

Make fractions with common denominators:

$\frac{17}{1}=\frac{1513}{89}$

Add fractions:

$y = - \frac{1960}{89} + \frac{1513}{89}$

$y = - \frac{447}{89}$

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All Common Core: High School - Algebra Resources

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Common Core: High School - Algebra : High School: Algebra

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

All Common Core: High School - Algebra Resources

Example Questions

Example Question #152 : Reasoning With Equations & Inequalities

Example Question #1 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Example Question #2 : Plotted Solutions And The Graph Of Two Variable Equations: Ccss.Math.Content.Hsa Rei.D.10

Example Question #153 : Reasoning With Equations & Inequalities

Example Question #157 : Reasoning With Equations & Inequalities

Example Question #151 : Reasoning With Equations & Inequalities

Example Question #153 : Reasoning With Equations & Inequalities

Example Question #1 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

Example Question #2 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

Example Question #3 : Use Graph Points To Approximate Solutions: Ccss.Math.Content.Hsa Rei.D.11

All Common Core: High School - Algebra Resources

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