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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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Example Questions

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

Report an Error

Example Question #121 : Reasoning With Equations & Inequalities

Find the points of intersection of $y = 39 x^{2} - 97$ and y = 14 x + 66 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -4.2 , 18.15 \right) , \left( 2.68 , -19.5 \right)$

$\left( 2.23 , 152.97 \right), \left( -1.87 , -6.93 \right)$

$\left( -0.73 , -16.91 \right) , \left( -0.96 , 16.36 \right)$

$\left( 7.46 , -19.52 \right) , \left( -1.09 , -16.79 \right)$

$\left( 8.95 , -17.41 \right) , \left( 4.01 , 10.85 \right)$

Correct answer:

$\left( 2.23 , 152.97 \right), \left( -1.87 , -6.93 \right)$

Explanation:

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, a = 39 , b = -14 , c = -163 .

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 = 2.23

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

x = -1.87

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 $\left( -1.87 , -6.93 \right)$

Report an Error

Example Question #122 : Reasoning With Equations & Inequalities

Find the points of intersection of $y = 49 x^{2} - 76$ and y = 25 x + 33 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 7.51 , -18.71 \right) , \left( -6.11 , -0.05 \right)$

$\left( 1.77 , 119.73 \right), \left( -1.26 , -28.74 \right)$

$\left( -0.75 , 0.71 \right) , \left( -5.82 , -6.31 \right)$

$\left( 8.38 , -2.92 \right) , \left( -9.33 , -2.63 \right)$

$\left( -0.9 , 3.49 \right) , \left( 4.07 , -10.53 \right)$

Correct answer:

$\left( 1.77 , 119.73 \right), \left( -1.26 , -28.74 \right)$

Explanation:

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, a = 49 , b = -25 , c = -109 .

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 = 1.77

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

x = -1.26

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 $\left( -1.26 , -28.74 \right)$

Report an Error

Example Question #1 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations 44 x - 5 y + 33 z = -1 , 40 x + 18 y - 28 z = -8 , 18 x - 18 y + 3 z = -17 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -5 & 44 & 33 \\ 18 & 40 & -28 \\ -1 & -18 & 18 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

$\begin{bmatrix} 44 & 40 & 18 \\ -5 & 18 & -18 \\ 33 & -28 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

$\begin{bmatrix} 44 & -5 & 33 \\ 40 & 18 & -28 \\ 18 & -18 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

$\begin{bmatrix} 44 & -5 & 33 \\ 40 & 18 & -28 \\ 18 & -18 & -1 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

$\begin{bmatrix} 44 & -5 & 33 \\ 40 & -28 & 18 \\ 18 & -1 & -18 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 44 & -5 & 33 \\ 40 & 18 & -28 \\ 18 & -18 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ -17 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} 44 & -5 & 33 \\ 40 & 18 & -28 \\ 18 & -18 & -1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -1 \\ -8 \\ 33 \end{bmatrix}$

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Example Question #2 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 29 x - 30 y + 36 z = -37 , - 14 x + 48 y + 35 z = -37 , - 23 x + 19 y + z = -1 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -29 & -30 & 36 \\ -14 & 35 & 48 \\ -23 & -43 & 19 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

$\begin{bmatrix} -30 & -29 & 36 \\ 48 & -14 & 35 \\ -43 & 19 & -23 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

$\begin{bmatrix} -29 & -14 & -23 \\ -30 & 48 & 19 \\ 36 & 35 & -43 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

$\begin{bmatrix} -29 & -30 & 36 \\ -14 & 48 & 35 \\ -23 & 19 & -43 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

$\begin{bmatrix} -29 & -30 & 36 \\ -14 & 48 & 35 \\ -23 & 19 & -43 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -29 & -30 & 36 \\ -14 & 48 & 35 \\ -23 & 19 & -43 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ -1 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -29 & -30 & 36 \\ -14 & 48 & 35 \\ -23 & 19 & -43 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -37 \\ -37 \\ 36 \end{bmatrix}$

Report an Error

Example Question #3 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 5 x + 12 y + 35 z = 0 , 3 x - 13 y - 6 z = 9 , 48 x + 15 y + 2 z = -19 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -5 & 3 & 48 \\ 12 & -13 & 15 \\ 35 & -6 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

$\begin{bmatrix} 12 & -5 & 35 \\ -13 & 3 & -6 \\ 44 & 15 & 48 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

$\begin{bmatrix} -5 & 12 & 35 \\ 3 & -13 & -6 \\ 48 & 15 & 44 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

$\begin{bmatrix} -5 & 12 & 35 \\ 3 & -13 & -6 \\ 48 & 15 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

$\begin{bmatrix} -5 & 12 & 35 \\ 3 & -6 & -13 \\ 48 & 44 & 15 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -5 & 12 & 35 \\ 3 & -13 & -6 \\ 48 & 15 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ -19 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -5 & 12 & 35 \\ 3 & -13 & -6 \\ 48 & 15 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 9 \\ 35 \end{bmatrix}$

Report an Error

Example Question #4 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 37 x - 43 y + 33 z = 13 , 34 x - 29 y + 2 z = -6 , - 14 x + 27 y + 4 z = 35 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -37 & -43 & 33 \\ 34 & -29 & 2 \\ -14 & 27 & 35 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

$\begin{bmatrix} -37 & 34 & -14 \\ -43 & -29 & 27 \\ 33 & 2 & 35 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

$\begin{bmatrix} -37 & -43 & 33 \\ 34 & -29 & 2 \\ -14 & 27 & 35 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

$\begin{bmatrix} -43 & -37 & 33 \\ -29 & 34 & 2 \\ 35 & 27 & -14 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

$\begin{bmatrix} -37 & -43 & 33 \\ 34 & 2 & -29 \\ -14 & 35 & 27 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -37 & -43 & 33 \\ 34 & -29 & 2 \\ -14 & 27 & 35 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 35 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -37 & -43 & 33 \\ 34 & -29 & 2 \\ -14 & 27 & 35 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 13 \\ -6 \\ 33 \end{bmatrix}$

Report an Error

Example Question #5 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 34 x + 29 y - 19 z = 40 , - 48 x + 9 y - 13 z = 30 , - 47 x + 37 y + 5 z = 23 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -34 & -48 & -47 \\ 29 & 9 & 37 \\ -19 & -13 & -3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

$\begin{bmatrix} -34 & 29 & -19 \\ -48 & 9 & -13 \\ -47 & 37 & -3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

$\begin{bmatrix} -34 & 29 & -19 \\ -48 & -13 & 9 \\ -47 & -3 & 37 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

$\begin{bmatrix} 29 & -34 & -19 \\ 9 & -48 & -13 \\ -3 & 37 & -47 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

$\begin{bmatrix} -34 & 29 & -19 \\ -48 & 9 & -13 \\ -47 & 37 & -3 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -34 & 29 & -19 \\ -48 & 9 & -13 \\ -47 & 37 & -3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ 23 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -34 & 29 & -19 \\ -48 & 9 & -13 \\ -47 & 37 & -3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 40 \\ 30 \\ -19 \end{bmatrix}$

Report an Error

Example Question #6 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 47 x + 3 y + 37 z = 2 , 18 x + 37 y - 27 z = 42 , 31 x - 20 y + 6 z = -28 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -47 & 3 & 37 \\ 18 & 37 & -27 \\ 31 & -20 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

$\begin{bmatrix} -47 & 3 & 37 \\ 18 & -27 & 37 \\ 31 & 29 & -20 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

$\begin{bmatrix} -47 & 18 & 31 \\ 3 & 37 & -20 \\ 37 & -27 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

$\begin{bmatrix} 3 & -47 & 37 \\ 37 & 18 & -27 \\ 29 & -20 & 31 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

$\begin{bmatrix} -47 & 3 & 37 \\ 18 & 37 & -27 \\ 31 & -20 & 29 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -47 & 3 & 37 \\ 18 & 37 & -27 \\ 31 & -20 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ -28 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -47 & 3 & 37 \\ 18 & 37 & -27 \\ 31 & -20 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ 42 \\ 37 \end{bmatrix}$

Report an Error

Example Question #7 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 46 x + 10 y + 19 z = -3 , - 11 x - 16 y + 49 z = -31 , 13 x - 35 y + 7 z = -2 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -46 & 10 & 19 \\ -11 & -16 & 49 \\ 13 & -35 & -46 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

$\begin{bmatrix} -46 & 10 & 19 \\ -11 & 49 & -16 \\ 13 & -46 & -35 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

$\begin{bmatrix} 10 & -46 & 19 \\ -16 & -11 & 49 \\ -46 & -35 & 13 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

$\begin{bmatrix} -46 & -11 & 13 \\ 10 & -16 & -35 \\ 19 & 49 & -46 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

$\begin{bmatrix} -46 & 10 & 19 \\ -11 & -16 & 49 \\ 13 & -35 & -46 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -46 & 10 & 19 \\ -11 & -16 & 49 \\ 13 & -35 & -46 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ -2 \end{bmatrix}$

Explanation:

In order to put these equations into proper matrix form, let's look at the general form.

$\begin{bmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a \\ b \\ c\end{bmatrix}$

The variables in the first matrix correspond to the coefficients in the equations and the variables in the third matrix correspond to the answers of the equations.

Now, let's substitute for each variable in the matrices.

$\begin{bmatrix} -46 & 10 & 19 \\ -11 & -16 & 49 \\ 13 & -35 & -46 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -3 \\ -31 \\ 19 \end{bmatrix}$

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

Example Question #121 : Reasoning With Equations & Inequalities

Example Question #122 : Reasoning With Equations & Inequalities

Example Question #1 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #2 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #3 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #4 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #5 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #6 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #7 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

All Common Core: High School - Algebra Resources

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