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

Put the equations 39 x + 34 y - 35 z = 2 , 16 x + 7 y + 31 z = -44 , - 48 x + 29 y + 8 z = 37 into proper matrix form.

Possible Answers:

$\begin{bmatrix} 39 & 34 & -35 \\ 16 & 7 & 31 \\ -48 & 29 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

$\begin{bmatrix} 34 & 39 & -35 \\ 7 & 16 & 31 \\ 44 & 29 & -48 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

$\begin{bmatrix} 39 & 34 & -35 \\ 16 & 7 & 31 \\ -48 & 29 & 44 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

$\begin{bmatrix} 39 & 34 & -35 \\ 16 & 31 & 7 \\ -48 & 44 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

$\begin{bmatrix} 39 & 16 & -48 \\ 34 & 7 & 29 \\ -35 & 31 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 39 & 34 & -35 \\ 16 & 7 & 31 \\ -48 & 29 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ 37 \end{bmatrix}$

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.

$\begin{bmatrix} 39 & 34 & -35 \\ 16 & 7 & 31 \\ -48 & 29 & 44 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 2 \\ -44 \\ -35 \end{bmatrix}$

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Example Question #9 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 31 x - 26 y - 48 z = 19 , 4 x + 42 y - 12 z = 35 , 36 x + 29 y + 9 z = 10 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & 4 & 36 \\ -26 & 42 & 29 \\ -48 & -12 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

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.

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ -48 \end{bmatrix}$

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Example Question #131 : Reasoning With Equations & Inequalities

Put the equations - 31 x - 26 y - 48 z = 19 , 4 x + 42 y - 12 z = 35 , 36 x + 29 y + 9 z = 10 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -26 & -31 & -48 \\ 42 & 4 & -12 \\ 49 & 29 & 36 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & -12 & 42 \\ 36 & 49 & 29 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

$\begin{bmatrix} -31 & 4 & 36 \\ -26 & 42 & 29 \\ -48 & -12 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ 10 \end{bmatrix}$

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.

$\begin{bmatrix} -31 & -26 & -48 \\ 4 & 42 & -12 \\ 36 & 29 & 49 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 19 \\ 35 \\ -48 \end{bmatrix}$

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Example Question #11 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 34 x - 28 y + 3 z = -41 , - 21 x + 20 y - 4 z = 40 , 39 x - 48 y + 10 z = -36 into proper matrix form.

Possible Answers:

$\begin{bmatrix} -34 & -28 & 3 \\ -21 & -4 & 20 \\ 39 & 8 & -48 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

$\begin{bmatrix} -34 & -28 & 3 \\ -21 & 20 & -4 \\ 39 & -48 & 8 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

$\begin{bmatrix} -34 & -21 & 39 \\ -28 & 20 & -48 \\ 3 & -4 & 8 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

$\begin{bmatrix} -34 & -28 & 3 \\ -21 & 20 & -4 \\ 39 & -48 & 8 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

$\begin{bmatrix} -28 & -34 & 3 \\ 20 & -21 & -4 \\ 8 & -48 & 39 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -34 & -28 & 3 \\ -21 & 20 & -4 \\ 39 & -48 & 8 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ -36 \end{bmatrix}$

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.

$\begin{bmatrix} -34 & -28 & 3 \\ -21 & 20 & -4 \\ 39 & -48 & 8 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -41 \\ 40 \\ 3 \end{bmatrix}$

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Example Question #12 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations - 22 x + 21 y = -5 , - 18 x - 6 y - 26 z = 11 , - 9 x - 3 y + 11 z = -37 into proper matrix form.

Possible Answers:

$\begin{bmatrix} 21 & -22 & 0 \\ -6 & -18 & -26 \\ -41 & -3 & -9 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

$\begin{bmatrix} -22 & -18 & -9 \\ 21 & -6 & -3 \\ 0 & -26 & -41 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

$\begin{bmatrix} -22 & 21 & 0 \\ -18 & -26 & -6 \\ -9 & -41 & -3 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

$\begin{bmatrix} -22 & 21 & 0 \\ -18 & -6 & -26 \\ -9 & -3 & -41 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

$\begin{bmatrix} -22 & 21 & 0 \\ -18 & -6 & -26 \\ -9 & -3 & -41 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} -22 & 21 & 0 \\ -18 & -6 & -26 \\ -9 & -3 & -41 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ -37 \end{bmatrix}$

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.

$\begin{bmatrix} -22 & 21 & 0 \\ -18 & -6 & -26 \\ -9 & -3 & -41 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} -5 \\ 11 \\ 0 \end{bmatrix}$

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Example Question #13 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Put the equations 30 x - 48 y - 28 z = 24 , 21 x - 46 y - 5 z = -32 , 27 x + 27 y + 12 z = 43 into proper matrix form.

Possible Answers:

$\begin{bmatrix} 30 & 21 & 27 \\ -48 & -46 & 27 \\ -28 & -5 & -37 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

$\begin{bmatrix} 30 & -48 & -28 \\ 21 & -5 & -46 \\ 27 & -37 & 27 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

$\begin{bmatrix} 30 & -48 & -28 \\ 21 & -46 & -5 \\ 27 & 27 & -37 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

$\begin{bmatrix} -48 & 30 & -28 \\ -46 & 21 & -5 \\ -37 & 27 & 27 \end{bmatrix} \cdot \begin{bmatrix} x \\ z \\ y \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

$\begin{bmatrix} 30 & -48 & -28 \\ 21 & -46 & -5 \\ 27 & 27 & -37 \end{bmatrix} \cdot \begin{bmatrix} y \\ z \\ x \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 30 & -48 & -28 \\ 21 & -46 & -5 \\ 27 & 27 & -37 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ 43 \end{bmatrix}$

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.

$\begin{bmatrix} 30 & -48 & -28 \\ 21 & -46 & -5 \\ 27 & 27 & -37 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 24 \\ -32 \\ -28 \end{bmatrix}$

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Example Question #1 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Does the following matrix have an inverse?

$A=\begin{bmatrix} 84 & -21 \\ 0 & 0 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}$ , $\uptext{b}$ , $\uptext{c}$ , and $\uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 84 \cdot 0 - -21 \cdot 0$

$\det(A)= 0 - 0$

$\det(A)= 0$

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Example Question #2 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Does the following matrix have an inverse?

$A=\begin{bmatrix} 33 & -28 \\ -21 & 84 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c}$ , and $\uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 33 \cdot -21 - -28 \cdot 84$

$\det(A)= -693 - -2352$

$\det(A)= 1659$

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Example Question #3 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Does the following matrix have an inverse?

$A=\begin{bmatrix} 55 & -83 \\ 45 & -51 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c}$ , and $\uptext{d$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 55 \cdot 45 - -83 \cdot -51$

$\det(A)= 2475 - 4233$

$\det(A)= -1758$

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Example Question #4 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Does the following matrix have an inverse?

$A=\begin{bmatrix} 85 & -90 \\ 96 & -67 \end{bmatrix}$

Possible Answers:

Yes

Correct answer:

Yes

Explanation:

In order to determine if a matrix has an inverse is to calculate the determinant.

$\det(A)=a\cdotc-b\cdot d$

Where $\uptext{a}, \uptext{b}, \uptext{c},$ and $\uptext{d}$ correspond to the entries in the following matrix.

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}$

If the determinant is not equal to zero, an inverse exists, and if it's equal to zero, no inverse exists.

Now let's calculate the determinant.

$\det(A)= 85 \cdot 96 - -90 \cdot -67$

$\det(A)= 8160 - 6030$

$\det(A)= 2130$

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

Example Question #9 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #131 : Reasoning With Equations & Inequalities

Example Question #11 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #12 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #13 : Matrix Representation Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.8

Example Question #1 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Example Question #2 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Example Question #3 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

Example Question #4 : Use Matrix Inverse To Solve System Of Linear Equations: Ccss.Math.Content.Hsa Rei.C.9

All Common Core: High School - Algebra Resources

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