Question 1

Solve the system of linear equations.
$$\left\{ \begin{aligned}
x + y + z + w & = 3 \
- x - y + z + w & = - 5 \
x + 3 z + w & = - 6 \
- 4 x - 4 y - 2 z - w & = - 13
\end{aligned} \right.$$
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 of linear equations, we can use various methods such as substitution, elimination, or matrix methods. However, without explicitly solving the system step by step, we can check each option by substituting the values of $x$, $y$, $z$, and $w$ into the original equations to see which one satisfies all the equations.Option D: $(x, y, z, w) = (-1, 5, -2, 1)$ satisfies all the given equations:1. $x + y + z + w = -1 + 5 - 2 + 1 = 3$2. $-x - y + z + w = 1 - 5 - 2 + 1 = -5$3. $x + 3z + w = -1 + 3(-2) + 1 = -6$4. $-4x - 4y - 2z - w = 4 - 20 + 4 - 1 = -13$Thus, option D is the correct solution to the system of equations.

Question 2

Solve the system of linear equations
$\left\{ \begin{array} { l l } 4 x _ { 1 } - 8 x _ { 2 } - 4 x _ { 3 } - 8 x _ { 4 } & = 0 \ 12 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 12 x _ { 4 } & = - 9 \ 8 x _ { 1 } - 20 x _ { 2 } - 8 x _ { 3 } - 20 x _ { 4 } & = 6 \ - 4 x _ { 1 } + 16 x _ { 2 } + 16 x _ { 3 } + 44 x _ { 4 } & = 0 \end{array} \right.$ 
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) $\left[ \begin{array} { l } x _ { 1 } \ x _ { 2 } \ x _ { 3 } \ x _ { 4 } \end{array} \right] = \left[ \begin{array} { c } - \frac { 57 } { 4 } \ - \frac { 27 } { 4 } \ - \frac { 45 } { 4 } \ \frac { 21 } { 4 } \end{array} \right]$

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

To solve the system using the given inverse matrix, multiply the inverse matrix by the constants on the right side of the equations. The correct solution is obtained by performing this matrix multiplication, which matches option A.

Question 3

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 points are collinear if the determinant of the matrix formed by their coordinates is zero. The determinant is calculated as follows:  $$ \begin{vmatrix} 3 & -1 & 1 \ 0 & -3 & 1 \ 27 & 15 & 1 \end{vmatrix} = 3(-3 \cdot 1 - 15 \cdot 1) - (-1)(0 \cdot 1 - 27 \cdot 1) + 1(0 \cdot 15 - (-3) \cdot 27) = 0 $$  Since the determinant is zero, the points are collinear.

Question 4

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...

Question 5

Write the system of linear equations as a matrix equation $A X = B$, and use GaussJordan elimination on the augmented matrix $[ A \vdots B ]$ to solve for the matrix $X$.
$$\left\{ \begin{array} { l } 
x _ { 1 } - 3 x _ { 2 } - 9 x _ { 3 } = 43 \
x _ { 1 } + 4 x _ { 2 } + 7 x _ { 3 } = - 51 \
7 x _ { 1 } - 7 x _ { 2 } + 3 x _ { 3 } = - 57
\end{array} \right.$$
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) $X = \left[ \begin{array} { r } - 8 \ - 2 \ - 5 \end{array} \right]$
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 solution to the system of equations using Gauss-Jordan elimination is $X = \left[ \begin{array} { r } - 8 \ - 2 \ - 5 \end{array} \right]$.

Question 6

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 $(x^2 + 4)(x^2 - 25)$, which further factors into $(x^2 + 4)(x - 5)(x + 5)$. The partial fraction decomposition must account for each of these factors, and option A correctly represents the decomposition with appropriate coefficients for each term.

Question 7

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 matrix by its respective scalar and then add the resulting matrices. For $cA$, multiply every element in matrix $A$ by $c (8)$, and for $dB$, multiply every element in matrix $B$ by $d (-3)$. After performing these operations and adding the resulting matrices, the correct answer is option C.

Question 8

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 augmented matrix represents the system of equations: $x + 4y - 3z = -23$, $y + 4z = 19$, and $z = 5$. Using back-substitution, substitute $z = 5$ into the second equation to find $y = -1$, and then substitute $y = -1$ and $z = 5$ into the first equation to find $x = -4$.

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

The answer of Use a system of equations to find...

Question 10

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

The answer of If possible, find \(3 A + 4...

Question 11

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

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)

Accepted Answer

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

Question 12

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

From the second equation, $y = 15x$. Substituting $y$ in the first equation gives $15x^3 - 25(15x) = 0$, simplifying to $15x^3 - 375x = 0$, or $15x(x^2 - 25) = 0$, leading to $x = 0$, $x = 5$, and $x = -5$. Substituting these $x$ values back into $y = 15x$ gives the solutions $(0,0)$, $(5,75)$, and $(-5,-75)$.

Question 13

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 14

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 supply equations.)
$$\begin{array} { l l } 
\text { Demand } & \text { Supply } \
p = 130 - 0.09 x & p = 0.4 x - 115
\end{array}$$
A) $( 500,85 )$
B) $( 200,112 )$
C) $( 835,219 )$
D) $( 300,103 )$
E) inconsistent

Accepted Answer

To find the equilibrium point, set the demand and supply equations equal to each other and solve for $x$.$$130 - 0.09x = 0.4x - 115$$$$0.49x = 245$$$$x = 500$$Substitute $x = 500$ into either the demand or supply equation to find $p$.$$p = 130 - 0.09(500) = 85$$Therefore, the equilibrium point is $(500, 85)$.

Question 15

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

Substitute the second equation into the first equation and solve for x.$ - 7 x - 9 ( x - 3 ) = - 6 $$ - 7 x - 9 x + 27 = - 6 $$ - 16 x = - 33 $$ x = \frac { 33 } { 16 } $Substitute the value of x back into the second equation to solve for y.$ y = \frac { 33 } { 16 } - 3 $$ y = - \frac { 15 } { 16 } $Therefore, the solution is $ \left( \frac { 33 } { 16 } , - \frac { 15 } { 16 } \right) $.

Question 16

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

For choice A, substituting $x = 1$ and $y = 2$ into both equations yields true statements, indicating it is a solution to the system.

Question 17

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 by the third matrix. The correct result is $\left[ \begin{array} { c c } 0 & 60 \ - 48 & 48 \end{array} \right]$.

Question 18

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 19

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

The answer of Solve for $X$ in the equation given.
\(9...

Question 20

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

Quiz 7: Linear Systems and Matrices

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

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

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

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

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 $c A + d B$ .

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

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

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

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 supply equations.) $\begin{array} { l l } \text { Demand } & \text { Supply } \\p = 130 - 0.09 x & p = 0.4 x - 115\end{array}$

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

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

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

Quiz 7: Linear Systems and Matrices

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 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​

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

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

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 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​

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

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​

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​]

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​

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