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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

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

Example Questions

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

Report an Error

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

Report an Error

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 $y = 6 x^{2} - 57$ and y = 43 x + 43 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -4.65 , -19.4 \right) , \left( 1.59 , -0.66 \right)$

$\left( -3.74 , -12.52 \right) , \left( -3.09 , 17.18 \right)$

$\left( -4.5 , -7.13 \right) , \left( -6.93 , 9.84 \right)$

$\left( 7.96 , -16.22 \right) , \left( -3.12 , 6.9 \right)$

$\left( 9.02 , 97.12 \right), \left( -1.85 , 31.9 \right)$

Correct answer:

$\left( 9.02 , 97.12 \right), \left( -1.85 , 31.9 \right)$

Explanation:

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

$6 x^{2} - 57 = 43 x + 43$

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

$6 x^{2} - 57 - 43 x - 43 = 0$

$6 x^{2} - 43 x - 100 = 0$

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 = 6 , b = -43 , c = -100 .

Now plug these values in.

$x =\frac{ 43 \pm\sqrt{\left( -43 \right)^2- 4 \cdot 6 \cdot -100 }}{ 2 \cdot 6 }$

$x =\frac{ 43 \pm \sqrt{4249} }{ 12 }$

Split this up into equations.

$x =\frac{ 43 + \sqrt{4249} }{ 12 }$

x = 9.02

$x = - \frac{\sqrt{4249}}{12} + \frac{43}{12}$

x = -1.85

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

$p_{1} = 6 \cdot 9.02 + 43$

$p_{1} = 54.12 + 43$

$p_{1} = 97.12$

So the first intersection point is at $\left( 9.02 , 97.12 \right)$

Now we need to find the last point of intersection.

$p_{2} = 6 \cdot -1.85 + 43$

$p_{2} = -11.1 + 43$

$p_{2} = 31.9$

So the second intersection point is at $\left( -1.85 , 31.9 \right)$

Report an Error

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 $y = 9 x^{2} - 89$ and y = 7 x + 24 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -9.78 , 5.51 \right) , \left( 3.7 , 6.8 \right)$

$\left( 9.44 , -16.45 \right) , \left( -5.44 , -5.89 \right)$

$\left( 3.95 , 59.55 \right), \left( -3.18 , -4.62 \right)$

$\left( -4.62 , 6.67 \right) , \left( -3.98 , 2.45 \right)$

$\left( -6.84 , 2.4 \right) , \left( 0.21 , -4.34 \right)$

Correct answer:

$\left( 3.95 , 59.55 \right), \left( -3.18 , -4.62 \right)$

Explanation:

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

$9 x^{2} - 89 = 7 x + 24$

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

$9 x^{2} - 89 - 7 x - 24 = 0$

$9 x^{2} - 7 x - 113 = 0$

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 = 9 , b = -7 , c = -113 .

Now plug these values in.

$x =\frac{ 7 \pm\sqrt{\left( -7 \right)^2- 4 \cdot 9 \cdot -113 }}{ 2 \cdot 9 }$

$x =\frac{ 7 \pm \sqrt{4117} }{ 18 }$

Split this up into equations.

$x =\frac{ 7 + \sqrt{4117} }{ 18 }$

x = 3.95

$x = - \frac{\sqrt{4117}}{18} + \frac{7}{18}$

x = -3.18

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

$p_{1} = 9 \cdot 3.95 + 24$

$p_{1} = 35.55 + 24$

$p_{1} = 59.55$

So the first intersection point is at $\left( 3.95 , 59.55 \right)$

Now we need to find the last point of intersection.

$p_{2} = 9 \cdot -3.18 + 24$

$p_{2} = -28.62 + 24$

$p_{2} = -4.62$

So the second intersection point is at $\left( -3.18 , -4.62 \right)$

Report an Error

Example Question #5 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 18 x^{2} - 48$ and y = 10 x + 41 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -5.84 , -9.58 \right) , \left( -4.97 , 3.97 \right)$

$\left( -0.82 , 18.3 \right) , \left( -5.32 , 1.14 \right)$

$\left( 2.52 , 86.36 \right), \left( -1.96 , 5.72 \right)$

$\left( 5.33 , 17.8 \right) , \left( 0.16 , 17.67 \right)$

$\left( -0.65 , -17.01 \right) , \left( 5.26 , -3.04 \right)$

Correct answer:

$\left( 2.52 , 86.36 \right), \left( -1.96 , 5.72 \right)$

Explanation:

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

$18 x^{2} - 48 = 10 x + 41$

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

$18 x^{2} - 48 - 10 x - 41 = 0$

$18 x^{2} - 10 x - 89 = 0$

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 = 18 , b = -10 , c = -89 .

Now plug these values in.

$x =\frac{ 10 \pm\sqrt{\left( -10 \right)^2- 4 \cdot 18 \cdot -89 }}{ 2 \cdot 18 }$

$x =\frac{ 10 \pm 2 \sqrt{1627} }{ 36 }$

Split this up into equations.

$x =\frac{ 10 + 2 \sqrt{1627} }{ 36 }$

x = 2.52

$x = - \frac{\sqrt{1627}}{18} + \frac{5}{18}$

x = -1.96

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

$p_{1} = 18 \cdot 2.52 + 41$

$p_{1} = 45.36 + 41$

$p_{1} = 86.36$

So the first intersection point is at $\left( 2.52 , 86.36 \right)$

Now we need to find the last point of intersection.

$p_{2} = 18 \cdot -1.96 + 41$

$p_{2} = -35.28 + 41$

$p_{2} = 5.72$

So the second intersection point is at $\left( -1.96 , 5.72 \right)$

Report an Error

Example Question #6 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 31 x^{2} - 3$ and y = 40 x + 9 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 2.22 , -8.06 \right) , \left( 6.61 , 12.62 \right)$

$\left( -0.73 , -3.88 \right) , \left( 2.19 , 12.02 \right)$

$\left( -3.61 , 4.51 \right) , \left( -1.0 , -8.72 \right)$

$\left( -4.68 , -15.0 \right) , \left( 2.6 , -13.69 \right)$

$\left( 1.54 , 56.74 \right), \left( -0.25 , 1.25 \right)$

Correct answer:

$\left( 1.54 , 56.74 \right), \left( -0.25 , 1.25 \right)$

Explanation:

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

$31 x^{2} - 3 = 40 x + 9$

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

$31 x^{2} - 3 - 40 x - 9 = 0$

$31 x^{2} - 40 x - 12 = 0$

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 = 31 , b = -40 , c = -12 .

Now plug these values in.

$x =\frac{ 40 \pm\sqrt{\left( -40 \right)^2- 4 \cdot 31 \cdot -12 }}{ 2 \cdot 31 }$

$x =\frac{ 40 \pm 4 \sqrt{193} }{ 62 }$

Split this up into equations.

$x =\frac{ 40 + 4 \sqrt{193} }{ 62 }$

x = 1.54

$x = - \frac{2 \sqrt{193}}{31} + \frac{20}{31}$

x = -0.25

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

$p_{1} = 31 \cdot 1.54 + 9$

$p_{1} = 47.74 + 9$

$p_{1} = 56.74$

So the first intersection point is at $\left( 1.54 , 56.74 \right)$

Now we need to find the last point of intersection.

$p_{2} = 31 \cdot -0.25 + 9$

$p_{2} = -7.75 + 9$

$p_{2} = 1.25$

So the second intersection point is at $\left( -0.25 , 1.25 \right)$

Report an Error

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

Find the points of intersection of $y = 4 x^{2} - 66$ and y = 30 x + 59 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 10.48 , 100.92 \right), \left( -2.98 , 47.08 \right)$

$\left( 1.28 , -17.89 \right) , \left( 3.09 , 3.49 \right)$

$\left( 6.75 , 18.79 \right) , \left( -0.89 , 0.74 \right)$

$\left( 6.14 , 12.61 \right) , \left( -9.16 , 14.79 \right)$

$\left( 0.15 , -10.7 \right) , \left( -9.16 , 13.54 \right)$

Correct answer:

$\left( 10.48 , 100.92 \right), \left( -2.98 , 47.08 \right)$

Explanation:

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

4 $x^{2} - 66 = 30 x + 59$

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

$4 x^{2} - 66 - 30 x - 59 = 0$

$4 x^{2} - 30 x - 125 = 0$

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 = 4 , b = -30 , c = -125 .

Now plug these values in.

$x =\frac{ 30 \pm\sqrt{\left( -30 \right)^2- 4 \cdot 4 \cdot -125 }}{ 2 \cdot 4 }$

$x =\frac{ 30 \pm 10 \sqrt{29} }{ 8 }$

Split this up into equations.

$x =\frac{ 30 + 10 \sqrt{29} }{ 8 }$

x = 10.48

$x = - \frac{5 \sqrt{29}}{4} + \frac{15}{4}$ b

x = -2.98

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

$p_{1} = 4 \cdot 10.48 + 59$

$p_{1} = 41.92 + 59$

$p_{1} = 100.92$

So the first intersection point is at $\left( 10.48 , 100.92 \right)$

Now we need to find the last point of intersection.

$p_{2} = 4 \cdot -2.98 + 59$

$p_{2} = -11.92 + 59$

$p_{2} = 47.08$

So the second intersection point is at $\left( -2.98 , 47.08 \right)$

Report an Error

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 $y = 37 x^{2} - 89$ and y = 18 x + 75 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 5.63 , 8.21 \right) , \left( 7.99 , 0.8 \right)$

$\left( -3.99 , -5.48 \right) , \left( -2.42 , -14.14 \right)$

$\left( -9.37 , 7.6 \right) , \left( -9.12 , -12.89 \right)$

$\left( 9.57 , -2.42 \right) , \left( 3.53 , 2.74 \right)$

$\left( 2.36 , 162.32 \right), \left( -1.88 , 5.44 \right)$

Correct answer:

$\left( 2.36 , 162.32 \right), \left( -1.88 , 5.44 \right)$

Explanation:

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

$37 x^{2} - 89 = 18 x + 75$

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

$37 x^{2} - 89 - 18 x - 75 = 0$

$37 x^{2} - 18 x - 164 = 0$

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 = 37 , b = -18 , c = -164 .

Now plug these values in.

$x =\frac{ 18 \pm\sqrt{\left( -18 \right)^2- 4 \cdot 37 \cdot -164 }}{ 2 \cdot 37 }$

$x =\frac{ 18 \pm 2 \sqrt{6149} }{ 74 }$

Split this up into equations.

$x =\frac{ 18 + 2 \sqrt{6149} }{ 74 }$

x = 2.36

$x = - \frac{\sqrt{6149}}{37} + \frac{9}{37}$

x = -1.88

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

$p_{1} = 37 \cdot 2.36 + 75$

$p_{1} = 87.32 + 75$

$p_{1} = 162.32$

So the first intersection point is at $\left( 2.36 , 162.32 \right)$

Now we need to find the last point of intersection.

$p_{2} = 37 \cdot -1.88 + 75$

$p_{2} = -69.56 + 75$

$p_{2} = 5.44$

So the second intersection point is at $\left( -1.88 , 5.44 \right)$

Report an Error

Example Question #9 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 23 x^{2} - 65$ and y = 11 x + 55 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 1.09 , 13.36 \right) , \left( 3.0 , -18.57 \right)$

$\left( 2.54 , 113.42 \right), \left( -2.06 , 7.62 \right)$

$\left( 3.81 , 7.05 \right) , \left( 5.33 , 16.74 \right)$

$\left( -4.6 , -12.3 \right) , \left( 3.99 , -19.69 \right)$

$\left( -2.78 , 5.1 \right) , \left( 9.57 , -5.89 \right)$

Correct answer:

$\left( 2.54 , 113.42 \right), \left( -2.06 , 7.62 \right)$

Explanation:

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

$23 x^{2} - 65 = 11 x + 55$

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

$23 x^{2} - 65 - 11 x - 55 = 0$

$23 x^{2} - 11 x - 120 = 0$

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 = 23 , b = -11 , c = -120 .

Now plug these values in.

$x =\frac{ 11 \pm\sqrt{\left( -11 \right)^2- 4 \cdot 23 \cdot -120 }}{ 2 \cdot 23 }$

$x =\frac{ 11 \pm \sqrt{11161} }{ 46 }$

Split this up into equations.

$x =\frac{ 11 + \sqrt{11161} }{ 46 }$

x = 2.54

$x = - \frac{\sqrt{11161}}{46} + \frac{11}{46}$

x = -2.06

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

$p_{1} = 23 \cdot 2.54 + 55$

$p_{1} = 58.42 + 55$

$p_{1} = 113.42$

So the first intersection point is at $\left( 2.54 , 113.42 \right)$

Now we need to find the last point of intersection.

$p_{2} = 23 \cdot -2.06 + 55$

$p_{2} = -47.38 + 55$

$p_{2} = 7.62$

So the second intersection point is at $\left( -2.06 , 7.62 \right)$

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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 #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

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

Example Question #5 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Example Question #6 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

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

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

Example Question #9 : 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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