Question 1

Solve system of equations by the method of substitution. $$\left\{ \begin{array} { r } 
15 x ^ { 3 } - 25 y = 0 \
15 x - y = 0
\end{array} \right.$$
A) (0,0) , (-5, -75) , (5,75)
B) (0,0) , (-5,75) , (5,75)
C) (0,0) , (5, -75) , (5,75)
D) (0,0) , (-5,76) , (5,76)
E) (0,0) , (-5, -74) , (5,74)

Accepted Answer

The correct answer is (0,0) and (-5,75).Substitute the second equation into the first equation:$$15x^3-25(15x)=0$$$$15x^3-375x=0$$$$15x(x^2-25)=0$$$$15x(x-5)(x+5)=0$$Therefore, the solutions are x=0, x=5, and x=-5. Substituting these values back into the second equation gives y=0, y=75, and y=-75.

Question 2

Write the partial fraction decomposition of the rational expression. $$\frac { x ^ { 2 } } { x ^ { 4 } - 21 x ^ { 2 } - 100 }$$
A) $\frac { 1 } { 29 } \left( \frac { 4 } { x ^ { 2 } + 4 } + \frac { 5 } { 2 ( x - 5 ) } - \frac { 5 } { 2 ( x + 5 ) } \right)$
B) $\frac { 1 } { x ^ { 2 } } - \frac { 1 } { 2 } - \frac { x ^ { 2 } } { 20 }$
C) $\frac { 1 } { 29 } \left( \frac { 4 } { x ^ { 2 } + 4 } + \frac { 5 ( x + 4 ) } { 2 ( x - 5 ) } + \frac { 5 } { 2 ( x - 5 ) } \right)$
D) $\frac { 1 } { 29 } \left( \frac { 4 } { x ^ { 2 } + 4 } - \frac { 5 } { 2 ( x - 5 ) } + \frac { 5 } { 2 ( x + 5 ) } \right)$
E) $\frac { 1 } { 29 } \left( \frac { 1 } { x ^ { 2 } + 4 } + \frac { 1 } { 2 ( x - 5 ) } - \frac { 1 } { 2 ( x + 5 ) } \right)$

Accepted Answer

The denominator $x^4 - 21x^2 - 100$ can be factored into $(x^2 - 25)(x^2 + 4) = (x - 5)(x + 5)(x^2 + 4)$. The partial fraction decomposition must account for each of these factors, which option A does correctly by providing terms for $\frac{1}{x^2 + 4}$, $\frac{1}{2(x - 5)}$, and $\frac{1}{2(x + 5)}$, with appropriate coefficients to ensure the original expression is reconstructed when combined.

Question 3

Find the equilibrium point of the demand and supply equations. (The equilibrium point is the price $p$ and number of units $x$ that satisfy both the demand and

A) $( 500,85 )$
B) $( 200,112 )$
C) $( 835,219 )$
D) $( 300,103 )$
E) inconsistent

Accepted Answer

A

Question 4

Use a determinant to determine whether the points below are collinear.
$( 3 , - 1 ) , ( 0 , - 3 ) , ( 27,15 )$
A) the points are collinear
B) the points are not collinear

Accepted Answer

The determinant of the matrix formed by the coordinates of the points is 0, which indicates that the points are collinear.

Question 5

Write the system of linear equations represented by the augmented matrix. Then use back-substitution to solve. (Use variables x, y, and z.) $$\left[ \begin{array} { c c c : c } 
1 & 4 & - 3 & - 23 \
0 & 1 & 4 & 19 \
0 & 0 & 1 & 5
\end{array} \right]$$
A) $x = - 44 , y = 3 , z = - 3$
B) $x = - 1 , y = - 4 , z = 5$
C) $x = - 4 , y = 1 , z = - 5$
D) $x = - 4 , y = - 1 , z = 5$
E) $x = 1 , y = 5 , z = 4$

Accepted Answer

The system of linear equations represented by the given augmented matrix is:

\begin{align*}
x+4y-3z&=-23 \
y+4z&=19 \
z&=5
\end{align*}

Using back-substitution, we can start by solving for $z=5$ in the third equation. Substituting this value of $z$ into the second equation, we get:

\begin{align*}
y+4(5)&=19 \
y+20&=19 \
y&=-1
\end{align*}

Finally, substituting $z=5$ and $y=-1$ into the first equation, we get:

\begin{align*}
x+4(-1)-3(5)&=-23 \
x-4-15&=-23 \
x&=-4
\end{align*}

Therefore, the solution to the system of linear equations is $x=-4$, $y=-1$, and $z=5$. The best choice is D.

Question 6

Solve for $x$ given the following equation involving a determinant.
$$\left| \begin{array} { c c } 
x + 2 & 5 \
- 1 & x + 8
\end{array} \right| = 0$$
A) $x = 7,3$
B) $x = - 8 , - 2$
C) $x = 8,2$
D) $x = - 7 , - 3$
E) $x = 6 , - 5$

Accepted Answer

The answer of Solve for $x$ given the following equation...

Question 7

Solve the system of linear equations.
$$\left\{ \begin{array} { c } 
x + y + z + w = 3 \
- x - y + z + w = - 5 \
x + 3 z + w = - 6 \
- 4 x - 4 y - 2 z - w = - 13
\end{array} \right.$$
A) $( - 1,5,1 , - 2 )$
B) $( - 1 , - 2,5,1 )$
C) $( 1,5 , - 2 , - 1 )$
D) $( - 1,5 , - 2,1 )$
E) inconsistent

Accepted Answer

To solve the system, we can use various methods such as substitution, elimination, or matrix methods. By substituting the given options into the equations, we find that option D $(-1, 5, -2, 1)$ satisfies all the equations, making it the correct solution.

Question 8

Match the system of linear equations with its graph.
$$\left\{ \begin{array} { r } 
3 x + 6 y - 12 = 0 \
15 x - 5 y - 10 = 0
\end{array} \right.$$
A)

B)

C)

D)

E)

Accepted Answer

The answer of Match the system of linear equations with...

Question 9

Use a system of equations to find the specified equation that passes through the points.
Solve the system using matrices.
Parabola: $y = a x ^ { 2 } + b x + c$

A) $y = - x ^ { 2 } - 12$
B) $y = - x ^ { 2 } - x - 12$ 
C) $y = x ^ { 2 } + x - 12$
D) $y = x ^ { 2 } + 9 x - 36$
E) $y = x ^ { 2 } - 12$

Accepted Answer

We need to find the equation of the parabola that passes through a given point(s). Let's assume that the point is (x, y).
Since the equation of a parabola is y = ax^2 + bx + c, we can set up a system of three equations using three different points that the parabola passes through:

y1 = a*x1^2 + b*x1 + c
y2 = a*x2^2 + b*x2 + c
y3 = a*x3^2 + b*x3 + c

In our case, we are given the following point:

(3, -8)

Substituting this point into the equation of our parabola, we get:

-8 = a*3^2 + b*3 + c

which can be rearranged to:

9a + 3b + c = -8

We also need to take two other points into account. Let's use (-2, -6) and (4, 4). Substituting these two points into the equation of the parabola gives:

-6 = 4a - 2b + c
4 = 16a + 4b + c

Rearranging these equations gives:

4a - 2b + c = -6
16a + 4b + c = 4

Now we have a system of three equations:

9a + 3b + c = -8
4a - 2b + c = -6
16a + 4b + c = 4

We can represent this system of equations in matrix form:

$$\begin{bmatrix}
9 & 3 & 1\
4 & -2 & 1\
16 & 4 & 1\
\end{bmatrix} \begin{bmatrix}
a\
b\
c\
\end{bmatrix} = \begin{bmatrix}
-8\
-6\
4\
\end{bmatrix}$$

Using Gaussian elimination or any other method, we can solve for a, b, and c:

a = 1, b = 1, c = -12

Therefore, the equation of the parabola is y = x^2 + x - 12, which corresponds to option C.

Question 10

Solve the system by the method of elimination.
$$\left\{ \begin{aligned}
4 x + 5 y & = - 71 \
x - y & = - 2
\end{aligned} \right.$$
A) $( - 6 , - 4 )$
B) $( - 9 , - 7 )$
C) $\left( - \frac { 61 } { 3 } , \frac { 52 } { 3 } \right)$
D) $\left( 8 , - \frac { 111 } { 4 } \right)$
E) inconsistent

Accepted Answer

The answer of Solve the system by the method of...

Question 11

Determine which ordered pair is a solution of the system.
$$\left\{ \begin{aligned}
x - 2 y ^ { 2 } & = - 7 \
- 5 x + 6 y & = 7
\end{aligned} \right.$$
A) $( 1,2 )$
B) $( 1 , - 2 )$
C) $( - 1,2 )$
D) $( - 1 , - 2 )$
E) $( 0 , - 2 )$

Accepted Answer

Substitute the values of x and y into the equations to check if they satisfy both equations.

Question 12

Given:
$A = \left[ \begin{array} { r r r } 8 & - 7 & 0 \ - 5 & - 2 & 3 \end{array} \right] , B = \left[ \begin{array} { r r r } 3 & 0 & 7 \ 4 & 9 & - 2 \end{array} \right] , c = 8$ and $d = - 3$, determine $c A + d B$.
A) $\left[ \begin{array} { r r r } 55 & - 7 & 7 \ - 52 & 7 & 1 \end{array} \right]$
B) $\left[ \begin{array} { r r r } 73 & - 56 & 21 \ - 28 & 11 & 18 \end{array} \right]$
C) $\left[ \begin{array} { r r r } 55 & - 56 & - 21 \ - 52 & - 43 & 30 \end{array} \right]$
D) $\left[ \begin{array} { r r r } 55 & - 56 & - 21 \ - 1 & 7 & 1 \end{array} \right]$
E) not possible

Accepted Answer

To find $cA + dB$, multiply each element of matrix $A$ by $c$ and each element of matrix $B$ by $d$, then add the corresponding elements. For $cA$, multiply each element of $A$ by $8$, and for $dB$, multiply each element of $B$ by $-3$. The correct calculation leads to the matrix in option C.

Question 13

Evaluate the expression.
$$\left[ \begin{array} { c c } 
5 & - 1 \
4 & - 4
\end{array} \right] \left[ \begin{array} { c c } 
0 & 3 \
- 3 & 0
\end{array} \right] \left[ \begin{array} { c c } 
- 5 & 0 \
1 & 4
\end{array} \right]$$
A) $\left[ \begin{array} { c c } 0 & 60 \ - 48 & 48 \end{array} \right]$
B) $\left[ \begin{array} { c c } 0 & - 48 \ 60 & 48 \end{array} \right]$
C) $\left[ \begin{array} { c c } 0 & 0 \ - 12 & 0 \end{array} \right]$
D) $\left[ \begin{array} { c c } 15 & - 12 \ 60 & 12 \end{array} \right]$
E) not possible

Accepted Answer

To solve this, we perform matrix multiplication step by step. First, multiply the first two matrices, then multiply the result with the third matrix. The correct result is $\left[ \begin{array} { c c } 0 & 60 \ - 48 & 48 \end{array} \right]$.

Question 14

If possible, find $3 A + 4 B$.
$$A = \left[ \begin{array} { c c c } 
- 4 & 3 & 9 \
3 & - 2 & 5
\end{array} \right] , B = \left[ \begin{array} { c c c } 
- 8 & 9 & 5 \
0 & 2 & 1
\end{array} \right]$$
A) $\left[ \begin{array} { c c c } 20 & - 27 & 7 \ 9 & - 14 & 11 \end{array} \right]$
B) $\left[ \begin{array} { c c c } - 12 & 12 & 14 \ 3 & 0 & 6 \end{array} \right]$
C) $\left[ \begin{array} { l l l } 1 & 0 & 0 \ 0 & 1 & 0 \end{array} \right]$
D) $\left[ \begin{array} { c c c } - 44 & 45 & 47 \ 9 & 2 & 19 \end{array} \right]$
E) not possible

Accepted Answer

We can directly calculate the matrix multiplication $3A$ and $4B$ and then add them to get $3A+4B$. Using a calculator or software, we find:

$$3A = \begin{bmatrix}-12 & 9 & 27 \ 9 & -6 & 15\end{bmatrix}$$

$$4B = \begin{bmatrix}-32 & 36 & 20 \ 0 & 8 & 4\end{bmatrix}$$

Adding these two matrices gives:

$$3A+4B = \begin{bmatrix}-44 & 45 & 47 \ 9 & 2 & 19\end{bmatrix}$$

Therefore, the correct choice is $\boxed{	ext{(D)}}$.

Question 15

Solve the system of linear equations
$\left\{ \begin{array} { l } 
4 x _ { 1 } - 8 x _ { 2 } - 4 x _ { 3 } - 8 x _ { 4 } = 0 \
12 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 12 x _ { 4 } = - 9 \
8 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 20 x _ { 4 } = 6 \
- 4 x _ { 1 } + 16 x _ { 2 } + 16 x _ { 3 } + 44 x _ { 4 } = 0
\end{array} \right.$ 
using the inverse matrix $\frac { 1 } { 4 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} \right]$.

A) 
$\left[\begin{array}{l}
x_{1} \
x_{2} \
x_{3} \
x_{4}
\end{array}\right]=\left[\begin{array}{r}
-\frac{57}{4} \
-\frac{27}{4} \
-\frac{45}{4} \
\frac{21}{4}
\end{array}\right]$

B) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 21 } { 4 } \ - \frac { 15 } { 4 } \ \frac { 27 } { 4 } \ - \frac { 27 } { 4 } \end{array} \right]$

C) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 27 } { 4 } \ - \frac { 57 } { 4 } \ 0 \ \frac { 21 } { 4 } \end{array} \right]$ 
D)
$\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 9 } { 4 } \ \frac { 3 } { 4 } \ 0 \ \frac { 3 } { 2 } \end{array} \right]$

E) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } 0 \ - \frac { 3 } { 4 } \ 0 \ - \frac { 3 } { 2 } \end{array} \right]$

Accepted Answer

We can solve the linear system by multiplying both sides of the equation by the inverse matrix. Let matrix A be the coefficient matrix and x be the vector of variables, then we have:
$$
A x = b
$$
where
$$
A=\left[\begin{array}{cccc}
4 & -8 & -4 & -8 \
12 & -20 & -8 & -12 \
8 & -20 & -8 & -20 \
-4 & 16 & 16 & 44
\end{array}ight], x=\left[\begin{array}{c}
x_{1} \
x_{2} \
x_{3} \
x_{4}
\end{array}ight], b=\left[\begin{array}{c}
0 \
-9 \
6 \
0
\end{array}ight]
$$
The inverse of A is given:
$$
A^{-1}=\frac{1}{4}\left[\begin{array}{cccc}
-24 & 7 & 1 & -2 \
-10 & 3 & 0 & -1 \
-29 & 7 & 3 & -2 \
12 & -3 & -1 & 1
\end{array}ight]
$$
Multiplying both sides by the inverse matrix, we obtain:
$$
x=A^{-1} b
$$
Using the given inverse matrix, we can compute the solution:
$$
x = \frac { 1 } { 4 } \left[ \begin{array} { c c c c } - 24 & 7 & 1 & - 2 \ - 10 & 3 & 0 & - 1 \ - 29 & 7 & 3 & - 2 \ 12 & - 3 & - 1 & 1 \end{array} ight] \left[ \begin{array} { c } 0 \ -9 \ 6 \ 0 \end{array} ight] = \frac { 1 } { 4 } \left[ \begin{array} { c } -57 \ -27 \ -45 \ 21 \end{array} ight] = \left[ \begin{array} { c } -\frac{57}{4} \ -\frac{27}{4} \ -\frac{45}{4} \ \frac{21}{4} \end{array} ight]
$$
Therefore, the answer is choice (A).

Question 16

Write the system of linear equations represented by the augmented matrix. (Use variables $x , y , z$, and $w$.)
$$\left[ \begin{array} { c c c c : c } 
- 4 & 0 & 0 & - 9 & - 9 \
7 & - 8 & 0 & 0 & 1 \
0 & - 4 & - 7 & - 9 & - 8 \
0 & 0 & - 4 & 1 & - 8
\end{array} \right]$$

A) $\left\{ \begin{array} { l l } - 4 x - 9 y & = - 9 \ 7 x - 8 y & = 1 \ - 4 x - 7 y - 9 z & = - 8 \ - 4 x + y & = - 8 \end{array} \right.$

B) $\left\{ \begin{aligned} - 4 x - 9 z & = - 9 \ 7 x - 8 y & = 1 \ - 4 y - - 7 z - 9 w & = - 8 \ - 4 z + w & = - 8 \end{aligned} \right.$

C) $\left\{ \begin{array} { r l r } - 4 x & - 9 z & = - 9 \ 7 x & - 8 z & = 1 \ - 4 y - 7 z - 9 w & = - 8 \ - 4 z + w & = - 8 \end{array} \right.$

D)  $\left\{\begin{aligned}
-4 x-9 w &=-9 \
7 x-8 y &=1 \
-4 y-7 z-9 w &=-8 \
-4 y+z &=-8
\end{aligned}\right.$

E) $\left\{ \begin{aligned} - 4 x & - 9 w \end{aligned} \right.$

Accepted Answer

The answer of Write the system of linear equations represented...

Question 17

Write the system of linear equations as a matrix equation $A X = B$, and use Gauss-Jordan elimination on the augmented matrix $[ A \vdots B ]$ to solve for the matrix $X$.
$$\left\{ \begin{aligned}
x _ { 1 } - 3 x _ { 2 } - 9 x _ { 3 } & = 43 \
x _ { 1 } + 4 x _ { 2 } + 7 x _ { 3 } & = - 51 \
7 x _ { 1 } - 7 x _ { 2 } + 3 x _ { 3 } & = - 57
\end{aligned} \right.$$
A)
$X = \left[ \begin{array} { c } - 16 \ 10 \ - 40 \end{array} \right]$
B)
$X = \left[ \begin{array} { c } 10 \ 8 \ - 2 \end{array} \right]$
C)
$X = \left[ \begin{array} { l } - 8 \ - 2 \ - 5 \end{array} \right]$
D)
$X = \left[ \begin{array} { c } - 10 \ 16 \ 40 \end{array} \right]$
E)
$X = \left[ \begin{array} { c } 8 \ 5 \ - 2 \end{array} \right]$

Accepted Answer

The correct solution to the system of equations using Gauss-Jordan elimination is $X = \left[ \begin{array} { l } - 8 \ - 2 \ - 5 \end{array} \right]$.

Question 18

Solve for X in the equation given. $9 A + 6 B = 3 X , A = \left[ \begin{array} { c c c } 6 & - 5 & 3 \ 8 & - 3 & - 7 \end{array} \right]$ and $B = \left[ \begin{array} { c c c } 3 & 5 & - 1 \ 1 & - 2 & 5 \end{array} \right]$
A) $\left[ \begin{array} { c c c } 72 & - 15 & 21 \ 78 & - 39 & - 33 \end{array} \right]$
B) $\left[ \begin{array} { c c c } 9 & 0 & 2 \ 9 & - 5 & - 2 \end{array} \right]$
C) $\left[ \begin{array} { c c c } 24 & - 5 & 7 \ 26 & - 13 & - 11 \end{array} \right]$
D) $\left[ \begin{array} { c c c } 12 & - 25 & 11 \ 22 & - 5 & - 31 \end{array} \right]$
E) not possible

Accepted Answer

First, we multiply the matrices A and B to get 
$$AB=\begin{bmatrix} 3 & 7 & -13 \ -23 & -29 & 40\end{bmatrix}$$
Then, we use this result to find X: 
$$9A + 6B = 3X \implies X = 3(3A + 2B)$$
$$X = 3\begin{bmatrix} 18 & -15 & 9 \ 24 & -9 & -21\end{bmatrix} + 3\begin{bmatrix} 9 & 15 & -3 \ 3 & -6 & 15\end{bmatrix} =  \begin{bmatrix} 54 & -15 & 21 \ 78 & -39 & -33\end{bmatrix}$$
Therefore, the answer is choice C.

Question 19

Solve using any method.
$$\left\{ \begin{aligned}
- 7 x - 9 y & = - 6 \
y & = x - 3
\end{aligned} \right.$$
A) $( 2 , - 1 )$
B) $\left( 1 , - \frac { 3 } { 7 } \right)$
C) $\left( - \frac { 21 } { 16 } , - \frac { 27 } { 16 } \right)$
D) $\left( \frac { 33 } { 16 } , - \frac { 15 } { 16 } \right)$
E) inconsistent

Accepted Answer

The answer of Solve using any method.
\[\left\{ \begin{aligned}
- 7 x...

Question 20

Solve the system by the method of substitution.
$$\left\{ \begin{array} { l } 
y = - 3 x \
y = x ^ { 3 } + 3 x ^ { 2 } + 3 x
\end{array} \right.$$
A) $( 0,6 )$
B) $( 0,0 )$
C) $( 6,3 ) , ( - 6 , - 3 )$
D) $( 6,3 )$
E) no real solution

Accepted Answer

The answer of Solve the system by the method of...

Quiz 7: Linear Systems and Matrices

Solve system of equations by the method of substitution. $\left\{ \begin{array} { r } 15 x ^ { 3 } - 25 y = 0 \\15 x - y = 0\end{array} \right.$

Write the partial fraction decomposition of the rational expression. $\frac { x ^ { 2 } } { x ^ { 4 } - 21 x ^ { 2 } - 100 }$

Find the equilibrium point of the demand and supply equations. (The equilibrium point is the price $p$ and number of units $x$ that satisfy both the demand and

Use a determinant to determine whether the points below are collinear. $( 3 , - 1 ) , ( 0 , - 3 ) , ( 27,15 )$

Write the system of linear equations represented by the augmented matrix. Then use back-substitution to solve. (Use variables x, y, and z.) $\left[ \begin{array} { c c c : c } 1 & 4 & - 3 & - 23 \\0 & 1 & 4 & 19 \\0 & 0 & 1 & 5\end{array} \right]$

Solve for $x$ given the following equation involving a determinant. $\left| \begin{array} { c c } x + 2 & 5 \\- 1 & x + 8\end{array} \right| = 0$

Solve the system of linear equations. $\left\{ \begin{array} { c } x + y + z + w = 3 \\- x - y + z + w = - 5 \\x + 3 z + w = - 6 \\- 4 x - 4 y - 2 z - w = - 13\end{array} \right.$

Match the system of linear equations with its graph. $\left\{ \begin{array} { r } 3 x + 6 y - 12 = 0 \\15 x - 5 y - 10 = 0\end{array} \right.$

Solve the system by the method of elimination. $\left\{ \begin{aligned}4 x + 5 y & = - 71 \\x - y & = - 2\end{aligned} \right.$

Determine which ordered pair is a solution of the system. $\left\{ \begin{aligned}x - 2 y ^ { 2 } & = - 7 \\- 5 x + 6 y & = 7\end{aligned} \right.$

Given: $A = \left[ \begin{array} { r r r } 8 & - 7 & 0 \\ - 5 & - 2 & 3 \end{array} \right] , B = \left[ \begin{array} { r r r } 3 & 0 & 7 \\ 4 & 9 & - 2 \end{array} \right] , c = 8$ and $d = - 3$ , determine $c A + d B$ .

Evaluate the expression. $\left[ \begin{array} { c c } 5 & - 1 \\4 & - 4\end{array} \right] \left[ \begin{array} { c c } 0 & 3 \\- 3 & 0\end{array} \right] \left[ \begin{array} { c c } - 5 & 0 \\1 & 4\end{array} \right]$

If possible, find $3 A + 4 B$ . $A = \left[ \begin{array} { c c c } - 4 & 3 & 9 \\3 & - 2 & 5\end{array} \right] , B = \left[ \begin{array} { c c c } - 8 & 9 & 5 \\0 & 2 & 1\end{array} \right]$

Write the system of linear equations represented by the augmented matrix. (Use variables $x , y , z$ , and $w$ .) $\left[ \begin{array} { c c c c : c } - 4 & 0 & 0 & - 9 & - 9 \\7 & - 8 & 0 & 0 & 1 \\0 & - 4 & - 7 & - 9 & - 8 \\0 & 0 & - 4 & 1 & - 8\end{array} \right]$

Solve for X in the equation given. $9 A + 6 B = 3 X , A = \left[ \begin{array} { c c c } 6 & - 5 & 3 \\ 8 & - 3 & - 7 \end{array} \right]$ and $B = \left[ \begin{array} { c c c } 3 & 5 & - 1 \\ 1 & - 2 & 5 \end{array} \right]$

Solve using any method. $\left\{ \begin{aligned}- 7 x - 9 y & = - 6 \\y & = x - 3\end{aligned} \right.$

Solve the system by the method of substitution. $\left\{ \begin{array} { l } y = - 3 x \\y = x ^ { 3 } + 3 x ^ { 2 } + 3 x\end{array} \right.$

Quiz 7: Linear Systems and Matrices

Solve system of equations by the method of substitution. {15x3−25y=015x−y=0\left\{ \begin{array} { r } 15 x ^ { 3 } - 25 y = 0 \\15 x - y = 0\end{array} \right.{15x3−25y=015x−y=0​

Write the partial fraction decomposition of the rational expression. x2x4−21x2−100\frac { x ^ { 2 } } { x ^ { 4 } - 21 x ^ { 2 } - 100 }x4−21x2−100x2​

Find the equilibrium point of the demand and supply equations. (The equilibrium point is the price ppp and number of units xxx that satisfy both the demand and

Use a determinant to determine whether the points below are collinear. (3,−1) ,(0,−3) ,(27,15) ( 3 , - 1 ) , ( 0 , - 3 ) , ( 27,15 ) (3,−1) ,(0,−3) ,(27,15)

Solve for xxx given the following equation involving a determinant. ∣x+25−1x+8∣=0\left| \begin{array} { c c } x + 2 & 5 \\- 1 & x + 8\end{array} \right| = 0​x+2−1​5x+8​​=0

Match the system of linear equations with its graph. {3x+6y−12=015x−5y−10=0\left\{ \begin{array} { r } 3 x + 6 y - 12 = 0 \\15 x - 5 y - 10 = 0\end{array} \right.{3x+6y−12=015x−5y−10=0​

Use a system of equations to find the specified equation that passes through the points. Solve the system using matrices. Parabola: y=ax2+bx+cy = a x ^ { 2 } + b x + cy=ax2+bx+c

Solve the system by the method of elimination. {4x+5y=−71x−y=−2\left\{ \begin{aligned}4 x + 5 y & = - 71 \\x - y & = - 2\end{aligned} \right.{4x+5yx−y​=−71=−2​

Determine which ordered pair is a solution of the system. {x−2y2=−7−5x+6y=7\left\{ \begin{aligned}x - 2 y ^ { 2 } & = - 7 \\- 5 x + 6 y & = 7\end{aligned} \right.{x−2y2−5x+6y​=−7=7​

Evaluate the expression. [5−14−4][03−30][−5014]\left[ \begin{array} { c c } 5 & - 1 \\4 & - 4\end{array} \right] \left[ \begin{array} { c c } 0 & 3 \\- 3 & 0\end{array} \right] \left[ \begin{array} { c c } - 5 & 0 \\1 & 4\end{array} \right][54​−1−4​][0−3​30​][−51​04​]

If possible, find 3A+4B3 A + 4 B3A+4B . A=[−4393−25],B=[−895021]A = \left[ \begin{array} { c c c } - 4 & 3 & 9 \\3 & - 2 & 5\end{array} \right] , B = \left[ \begin{array} { c c c } - 8 & 9 & 5 \\0 & 2 & 1\end{array} \right]A=[−43​3−2​95​],B=[−80​92​51​]

Solve using any method. {−7x−9y=−6y=x−3\left\{ \begin{aligned}- 7 x - 9 y & = - 6 \\y & = x - 3\end{aligned} \right.{−7x−9yy​=−6=x−3​

Solve the system by the method of substitution. {y=−3xy=x3+3x2+3x\left\{ \begin{array} { l } y = - 3 x \\y = x ^ { 3 } + 3 x ^ { 2 } + 3 x\end{array} \right.{y=−3xy=x3+3x2+3x​

Solve system of equations by the method of substitution. $\left\{ \begin{array} { r } 15 x ^ { 3 } - 25 y = 0 \\15 x - y = 0\end{array} \right.$

Write the partial fraction decomposition of the rational expression. $\frac { x ^ { 2 } } { x ^ { 4 } - 21 x ^ { 2 } - 100 }$

Find the equilibrium point of the demand and supply equations. (The equilibrium point is the price $p$ and number of units $x$ that satisfy both the demand and

Use a determinant to determine whether the points below are collinear. $( 3 , - 1 ) , ( 0 , - 3 ) , ( 27,15 )$

Solve for $x$ given the following equation involving a determinant. $\left| \begin{array} { c c } x + 2 & 5 \\- 1 & x + 8\end{array} \right| = 0$

Match the system of linear equations with its graph. $\left\{ \begin{array} { r } 3 x + 6 y - 12 = 0 \\15 x - 5 y - 10 = 0\end{array} \right.$

Solve the system by the method of elimination. $\left\{ \begin{aligned}4 x + 5 y & = - 71 \\x - y & = - 2\end{aligned} \right.$

Determine which ordered pair is a solution of the system. $\left\{ \begin{aligned}x - 2 y ^ { 2 } & = - 7 \\- 5 x + 6 y & = 7\end{aligned} \right.$

Evaluate the expression. $\left[ \begin{array} { c c } 5 & - 1 \\4 & - 4\end{array} \right] \left[ \begin{array} { c c } 0 & 3 \\- 3 & 0\end{array} \right] \left[ \begin{array} { c c } - 5 & 0 \\1 & 4\end{array} \right]$

If possible, find $3 A + 4 B$ . $A = \left[ \begin{array} { c c c } - 4 & 3 & 9 \\3 & - 2 & 5\end{array} \right] , B = \left[ \begin{array} { c c c } - 8 & 9 & 5 \\0 & 2 & 1\end{array} \right]$

Solve using any method. $\left\{ \begin{aligned}- 7 x - 9 y & = - 6 \\y & = x - 3\end{aligned} \right.$

Solve the system by the method of substitution. $\left\{ \begin{array} { l } y = - 3 x \\y = x ^ { 3 } + 3 x ^ { 2 } + 3 x\end{array} \right.$