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Common Core: High School - Algebra : Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

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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 #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 $\left( -2.06 , 7.62 \right)$

Report an Error

Example Question #10 : Solve Simple System Of Two Variable Linear And Quadratic Equations: Ccss.Math.Content.Hsa Rei.C.7

Find the points of intersection of $y = 21 x^{2} - 94$ and y = 18 x + 95 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -9.57 , 0.39 \right) , \left( -2.92 , 18.97 \right)$

$\left( 6.64 , 1.93 \right) , \left( -1.81 , -7.35 \right)$

$\left( 4.81 , 14.19 \right) , \left( -4.75 , -12.3 \right)$

$\left( 3.46 , 167.66 \right), \left( -2.6 , 40.4 \right)$

$\left( -1.12 , 9.28 \right) , \left( -1.22 , 6.44 \right)$

Correct answer:

$\left( 3.46 , 167.66 \right), \left( -2.6 , 40.4 \right)$

Explanation:

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

$21 x^{2} - 94 = 18 x + 95$

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

$21 x^{2} - 94 - 18 x - 95 = 0$

$21 x^{2} - 18 x - 189 = 0$

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 = 21 , b = -18 , c = -189 .

Now plug these values in.

$x =\frac{ 18 \pm\sqrt{\left( -18 \right)^2- 4 \cdot 21 \cdot -189 }}{ 2 \cdot 21 }$

$x =\frac{ 18 \pm 90 \sqrt{2} }{ 42 }$

Split this up into equations.

$x =\frac{ 18 + 90 \sqrt{2} }{ 42 }$

x = 3.46

$x = - \frac{15 \sqrt{2}}{7} + \frac{3}{7}$

x = -2.6

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

$p_{1} = 21 \cdot 3.46 + 95$

$p_{1} = 72.66 + 95$

$p_{1} = 167.66$

So the first intersection point is at $\left( 3.46 , 167.66 \right)$

Now we need to find the last point of intersection.

$p_{2} = 21 \cdot -2.6 + 95$

$p_{2} = -54.6 + 95$

$p_{2} = 40.4$

So the second intersection point is at $\left( -2.6 , 40.4 \right)$

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If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

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A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

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St. Louis, MO 63105

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I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

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Common Core: High School - Algebra : Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

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

All Common Core: High School - Algebra Resources

Example Questions

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

Example Question #10 : 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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