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

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Question

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

Answer

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, , , .

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.

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

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

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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Question

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

Answer

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, , , .

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.

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

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

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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Question

Find the points of intersection of $y = 6 x^{2} - 57$ and . Round your answers to the nearest hundredth.

Answer

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, , , .

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.

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

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

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

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Question

Find the points of intersection of $y = 9 x^{2} - 89$ and . Round your answers to the nearest hundredth.

Answer

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, , , .

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.

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

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

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

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Question

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

Answer

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, .

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.

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

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

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

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Question

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

Answer

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, .

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.

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

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

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

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Question

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

Answer

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, .

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.

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

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

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

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Question

Find the points of intersection of $y = 37 x^{2} - 89$ and . Round your answers to the nearest hundredth.

Answer

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, .

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.

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

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

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

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Question

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

Answer

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, .

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.

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

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

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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Question

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

Answer

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, .

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.

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

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

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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Question

Find the points of intersection of $y = 39 x^{2} - 97$ and . Round your answers to the nearest hundredth.

Answer

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

$39 x^{2} - 97 = 14 x + 66$

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

$39 x^{2} - 97 - 14 x - 66 = 0$

$39 x^{2} - 14 x - 163 = 0$

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, .

Now plug these values in.

$x =\frac{ 14 \pm\sqrt{\left( -14 \right)^2- 4 \cdot 39 \cdot -163 }}{ 2 \cdot 39 }$

$x =\frac{ 14 \pm 2 \sqrt{6406} }{ 78 }$

Split this up into equations.

$x =\frac{ 14 + 2 \sqrt{6406} }{ 78 }$

$x = - \frac{\sqrt{6406}}{39} + \frac{7}{39}$

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

$p_{1} = 39 \cdot 2.23 + 66$

$p_{1} = 86.97 + 66$

$p_{1} = 152.97$

So the first intersection point is at $\left( 2.23 , 152.97 \right)$

Now we need to find the last point of intersection.

$p_{2} = 39 \cdot -1.87 + 66$

$p_{2} = -72.93 + 66$

$p_{2} = -6.93$

So the second intersection point is at $\left( -1.87 , -6.93 \right)$

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Question

Find the points of intersection of $y = 49 x^{2} - 76$ and . Round your answers to the nearest hundredth.

Answer

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

$49 x^{2} - 76 = 25 x + 33$

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

$49 x^{2} - 76 - 25 x - 33 = 0$

$49 x^{2} - 25 x - 109 = 0$

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, .

Now plug these values in.

$x =\frac{ 25 \pm\sqrt{\left( -25 \right)^2- 4 \cdot 49 \cdot -109 }}{ 2 \cdot 49 }$

$x =\frac{ 25 \pm \sqrt{21989} }{ 98 }$

Split this up into equations.

$x =\frac{ 25 + \sqrt{21989} }{ 98 }$

$x = - \frac{\sqrt{21989}}{98} + \frac{25}{98}$

x =

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

$p_{1} = 49 \cdot 1.77 + 33$

$p_{1} = 86.73 + 33$

$p_{1} = 119.73$

So the first intersection point is at $\left( 1.77 , 119.73 \right)$

Now we need to find the last point of intersection.

$p_{2} = 49 \cdot -1.26 + 33$

$p_{2} = -61.74 + 33$

$p_{2} = -28.74$

So the second intersection point is at $\left( -1.26 , -28.74 \right)$

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