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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 #21 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

Where do the following lines intersect?

y = 7 x - 35 and y = 31 x + 71

Possible Answers:

${x = - \frac{53}{6}}$

${x=- \frac{53}{12}}$

$\uptext{x}=6$

$\uptext{x}=-5$

${x=\frac{53}{12}}$

Correct answer:

${x=- \frac{53}{12}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

7 x - 35 = 31 x + 71

7 x - 35+35=31 x + 71+35

7 x=31 x + 106

7 x-31 x=31 x + 106-31 x

- 24 x=106

$x=- \frac{53}{12}$

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Example Question #22 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

Where do the following lines intersect?

y = 17 x - 31 and y = 42 x + 33

Possible Answers:

$\uptext{x}=6$

$\uptext{x}=-5$

${x = - \frac{128}{25}}$

${x=- \frac{64}{25}}$

${x=\frac{64}{25}}$

Correct answer:

${x=- \frac{64}{25}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

17 x - 31 = 42 x + 33

17 x - 31+31=42 x + 33+31

17 x=42 x + 64

17 x-42 x=42 x + 64-42 x

- 25 x=64

$x=- \frac{64}{25}$

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Example Question #563 : High School: Algebra

Where do the following lines intersect?

y = 44 x + 18 and y = 5 x + 34

Possible Answers:

${x=- \frac{16}{39}}$

$\uptext{x}=-5$

$\uptext{x}=6$

${x = \frac{32}{39}}$

${x=\frac{16}{39}}$

Correct answer:

${x=\frac{16}{39}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

44 x + 18 = 5 x + 34

44 x + 18-18=5 x + 34-18

44 x=5 x + 16

44 x-5 x=5 x + 16-5 x

39 x=16

$x=\frac{16}{39}$

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Example Question #564 : High School: Algebra

Where do the following lines intersect?

y = 45 x - 35 and y = 15 x - 58

Possible Answers:

$\uptext{x}=6$

$\uptext{x}=-5$

${x = - \frac{23}{15}}$

${x=\frac{23}{30}}$

${x=- \frac{23}{30}}$

Correct answer:

${x=- \frac{23}{30}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

45 x - 35 = 15 x - 58

45 x - 35+35=15 x - 58+35

45 x=15 x - 23

45 x-15 x=15 x - 23-15 x

30 x=-23

$x=- \frac{23}{30}$

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Example Question #565 : High School: Algebra

Where do the following lines intersect?

y = 4 x + 93 and y = 34 x - 68

Possible Answers:

$\uptext{x}=-5$

${x=- \frac{161}{30}}$

${x = \frac{161}{15}}$

$\uptext{x}=6$

${x=\frac{161}{30}}$

Correct answer:

${x=\frac{161}{30}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

4 x + 93 = 34 x - 68

4 x + 93-93=34 x - 68-93

4 x=34 x - 161

4 x-34 x=34 x - 161-34 x

- 30 x=-161

$x=\frac{161}{30}$

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Example Question #566 : High School: Algebra

Where do the following lines intersect?

y = 49 x + 81 and y = 13 x + 75

Possible Answers:

${x=\frac{1}{6}}$

${x=- \frac{1}{6}}$

$\uptext{x}=-5$

$\uptext{x}=6$

${x = - \frac{1}{3}}$

Correct answer:

${x=- \frac{1}{6}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

49 x + 81 = 13 x + 75

49 x + 81-81=13 x + 75-81

49 x=13 x - 6

49 x-13 x=13 x - 6-13 x

36 x=-6

$x=- \frac{1}{6}$

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Example Question #104 : Reasoning With Equations & Inequalities

Where do the following lines intersect?

y = 16 x + 57 and y = 23 x - 24

Possible Answers:

$\uptext{x}=-5$

$\uptext{x}=6$

${x = \frac{162}{7}}$

${x=- \frac{81}{7}}$

${x=\frac{81}{7}}$

Correct answer:

${x=\frac{81}{7}}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

16 x + 57 = 23 x - 24

16 x + 57-57=23 x - 24-57

16 x=23 x - 81

16 x-23 x=23 x - 81-23 x

- 7 x=-81

$x=\frac{81}{7}$

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Example Question #31 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

What is the x-coordinate where the following lines intersect?

y = x + 35 and y = 22 x + 25

Possible Answers:

$x = \frac{20}{21}$

x = 6

$x = \frac{10}{21}$

$x = - \frac{10}{21}$

x = -5

Correct answer:

$x = \frac{10}{21}$

Explanation:

In order to figure out this problem, we need to set each of the equations equal to each other, and solve for $\uptext{x}$ .

x + 35 = 22 x + 25

x + 35 - 35 = 22 x + 25 - 35

x = 22 x - 10

x - 22 x = 22 x - 10 - 22 x

- 21 x = -10

$x = \frac{10}{21}$

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Example Question #1 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 51 x^{2} - 46$ and y = 26 x + 27 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 1.48 , 102.48 \right), \left( -0.97 , -22.47 \right)$

$\left( -4.39 , -17.43 \right) , \left( 2.5 , 10.63 \right)$

$\left( 1.12 , -3.84 \right) , \left( 8.17 , -15.05 \right)$

$\left( 8.96 , 5.74 \right) , \left( -8.69 , 7.97 \right)$

$\left( -4.02 , 5.3 \right) , \left( -2.85 , 0.66 \right)$

Correct answer:

$\left( 1.48 , 102.48 \right), \left( -0.97 , -22.47 \right)$

Explanation:

The first step to solving this problem is to set the equations equal to each other.

$51 x^{2} - 46 = 26 x + 27$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$51 x^{2} - 46 - 26 x - 27 = 0$

$51 x^{2} - 26 x - 73 = 0$

Recall the quadratic equation.

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

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

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

In this case, a = 51 , b = -26 , c = -73 .

Now plug these values in.

$x =\frac{ 26 \pm\sqrt{\left( -26 \right)^2- 4 \cdot 51 \cdot -73 }}{ 2 \cdot 51 }$

$x =\frac{ 26 \pm 4 \sqrt{973} }{ 102 }$

Split this up into equations.

$x =\frac{ 26 + 4 \sqrt{973} }{ 102 }$

x = 1.48

$x = - \frac{2 \sqrt{973}}{51} + \frac{13}{51}$

x = -0.97

Since we want points, we need to plug these values into one of the original equations.

$p_{1} = 51 \cdot 1.48 + 27$

$p_{1} = 75.48 + 27$

$p_{1} = 102.48$

So the first intersection point is at $\left( 1.48 , 102.48 \right)$

Now we need to find the last point of intersection.

$p_{2} = 51 \cdot -0.97 + 27$

$p_{2} = -49.47 + 27$

$p_{2} = -22.47$

So the second intersection point is at $\left( -0.97 , -22.47 \right)$

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Example Question #2 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 24 x^{2} - 70$ and y = 48 x + 93 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 7.61 , 14.19 \right) , \left( -6.06 , -16.25 \right)$

$\left( -2.8 , 3.57 \right) , \left( -8.72 , -7.49 \right)$

$\left( 3.79 , 183.96 \right), \left( -1.79 , 50.04 \right)$

$\left( 8.87 , -6.51 \right) , \left( -2.84 , -3.34 \right)$

$\left( 8.38 , 18.56 \right) , \left( -2.93 , 4.26 \right)$

Correct answer:

$\left( 3.79 , 183.96 \right), \left( -1.79 , 50.04 \right)$

Explanation:

The first step to solving this problem is to set the equations equal to each other.

$24 x^{2} - 70 = 48 x + 93$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$24 x^{2} - 70 - 48 x - 93 = 0$

$24 x^{2} - 48 x - 163 = 0$

Recall the quadratic equation.

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

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

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

In this case, a = 24 , b = -48 , c = -163 .

Now plug these values in.

$x =\frac{ 48 \pm\sqrt{\left( -48 \right)^2- 4 \cdot 24 \cdot -163 }}{ 2 \cdot 24 }$

$x =\frac{ 48 \pm 4 \sqrt{1122} }{ 48 }$

Split this up into equations.

$x =\frac{ 48 + 4 \sqrt{1122} }{ 48 }$

x = 3.79

$x = - \frac{\sqrt{1122}}{12} + 1$

x = -1.79

Since we want points, we need to plug these values into one of the original equations.

$p_{1} = 24 \cdot 3.79 + 93$

$p_{1} = 90.96 + 93$

$p_{1} = 183.96$

So the first intersection point is at $\left( 3.79 , 183.96 \right)$

Now we need to find the last point of intersection.

$p_{2} = 24 \cdot -1.79 + 93$

$p_{2} = -42.96 + 93$

$p_{2} = 50.04$

So the second intersection point is at $\left( -1.79 , 50.04 \right)$

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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 #21 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

Example Question #22 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

Example Question #563 : High School: Algebra

Example Question #564 : High School: Algebra

Example Question #565 : High School: Algebra

Example Question #566 : High School: Algebra

Example Question #104 : Reasoning With Equations & Inequalities

Example Question #31 : Solve Systems Of Linear Equations Exactly And Approximately: Ccss.Math.Content.Hsa Rei.C.6

Example Question #1 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Example Question #2 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

All Common Core: High School - Algebra Resources

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