Algebra Problesm and Questions with Solutions - Grade 12

Grade 12 Algebra questions and problems, complete with step-by-step solutions, are provided here. Some of these problems may be challenging, requiring time and effort to solve. However, tackling difficult questions helps you develop critical thinking and problem-solving skills.

Working in groups on these problems is a great way to collaborate, share ideas, and learn from each other. If you discover alternative solutions to any of the questions, feel free to share them!

Problem 1

Order the following from greatest to least: \[ \begin{array}{ll} \text{a)} & 25^{100} \\ \text{b)} & 2^{300} \\ \text{c)} & 3^{400} \\ \text{d)} & 4^{200} \\ \text{e)} & 2^{600} \end{array} \]
Solution:
We rewrite the given expressions to the same power and compare them: \[ 25^{100} \] \[ 2^{300} = (2^3)^{100} = 8^{100} \] \[ 3^{400} = (3^4)^{100} = 81^{100} \] \[ 4^{200} = (4^2)^{100} = 16^{100} \] \[ 2^{600} = (2^6)^{100} = 64^{100} \] Now, we order them from greatest to least: \[ 3^{400} , \quad 2^{600}, \quad 25^{100} , \quad 4^{200} , \quad 2^{300} \]

Problem 2

Find all rational zeros of \( P(x) = x^3 - 7 x + 6 \) and factor \(P(x)\)

Solution:
According to the Rational Root Theorem, the possible rational zeros of a polynomial \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 \] are given by: \[ \dfrac{\text{factors of the constant term } a_0}{\text{factors of the leading coefficient } a_n}. \]

The leading coefficient is \( 1 \), and its factors are:

\[ \pm 1 \]

The constant term is \( 6 \), and its factors are:

\[ \pm 1, \pm 2, \pm 3, \pm 6 \] - The possible rational zeros are: \[ \pm 1, \pm 2, \pm 3, \pm 6 \]

Testing for Rational Zeros, using direct substitution and calculations:

\[ P(1) = 0, \quad P(2) = 0, \quad P(-3) = 0 \]

Since these values satisfy \( P(x) = 0 \), they are the zeros of \( P(x) \). The polynomial \( P(x) \) has the following rational zeros: \[ x = 1, \quad x = 2, \quad x = -3 \]

Thus, the polynomial can be factored further using these roots.

\[ P(x) = (x-1)(x-2)(x+3) \)

Problem 3

Round all real zeros in the graph to the nearest integer and find a polynomial function \(P\) of lowest degree, with the absolute value of the leading coefficient equal to 1, that has the indicated graph.

problem 4.

Solution:

From the graph, \(x = -3 \) is a zero of multiplicity \(2\), \(x = 0\) is a zero of multiplicity 1 and \(x = 2\) is a zero of multiplicity 2.

The polynomial has degree 5 and its graph is falling on the right and rising on the left therefore the leading coefficient is negative and equal to \(-1\), hence \( P(x) \) is given by

\[ P(x) = - x (x + 3)^2 (x - 2)^2 \]

Problem 4

Given that \( 2 - i \), where \( i \) is the imaginary unit, is a zero of \[ P(x) = x^4 - 4x^3 + 3x^2 + 8x - 10, \] find all zeros of \( P(x) \).

Solution:
Since the coefficients of \( P(x) \) are real, the complex roots must occur in conjugate pairs. Thus, \( 2 + i \) is also a zero of \( P(x) \). Next, we find the quadratic factor corresponding to these two roots: \[ (x - (2 - i))(x - (2 + i)) \] Expanding using the difference of squares: \[ (x - 2 + i)(x - 2 - i) = (x - 2)^2 - i^2 = (x - 2)^2 - (-1) = (x - 2)^2 + 1 \] \[ = x^2 - 4x + 4 + 1 = x^2 - 4x + 5. \] Since \( P(x) \) is divisible by \( x^2 - 4x + 5 \), we perform polynomial division: \[ (x^4 - 4x^3 + 3x^2 + 8x - 10) \div (x^2 - 4x + 5). \] Dividing, we obtain the quotient \( x^2 - 2 \), which factors further as: \[ x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}). \] Thus, the complete set of zeros of \( P(x) \) is: \[ \mathbf{2 - i, \quad 2 + i, \quad -\sqrt{2}, \quad \sqrt{2}.} \]

Problem 5

Find \( a, b, \) and \( c \) so that the graph of the quadratic function \[ f(x) = ax^2 + bx + c \] has a vertex at \( (-2,1) \) and passes through the point \( (0,-3) \).

Solution:
We use the vertex form of a quadratic function: \[ f(x) = a (x + 2)^2 + 1 \] where the vertex is given as \( (-2,1) \).

Use the Given Point \( (0, -3) \): Substituting \( x = 0 \) and \( f(0) = -3 \): \[ -3 = a(0 + 2)^2 + 1 \] \[ -3 = 4a + 1 \] Solving for \( a \): \[ 4a = -4 \] \[ a = -1. \] Substituting \( a = -1 \) into the vertex form: \[ f(x) = -(x + 2)^2 + 1 \] Expanding: \[ f(x) = -x^2 - 4x - 3. \] Comparing with the standard quadratic form \( f(x) = ax^2 + bx + c \), we find: \[ a = -1, \quad b = -4, \quad c = -3. \]

Problem 6

\( f(x) \) is a quadratic function such that \( f(1) = 3 \) and \( f(5) = 3 \). Find the \( x \) coordinate of the vertex of the graph of \( f \).

Solution:
\[ f(x) = ax^2 + bx + c \] Given \( f(1) = 3 \), we have: \[ 3 = a + b + c \] Given \( f(5) = 3 \), we have: \[ 3 = 25a + 5b + c \] Subtracting equation B from equation C: \[ (25a + 5b + c) - (a + b + c) = 3 - 3 \] \[ 24a + 4b = 0 \] The x-coordinate of the vertex is given by: \[ \dfrac{-b}{2a} = 3 \] Note that this problem might be solved as follows:

Since \( f(x) \) is a quadratic function such that \( f(1) = 3 \) and \( f(5) = 3 \), the two points \( (1, 3) \) and \( (5, 3) \) lie on the parabola, and since the function is quadratic, the x coordinate of the vertex is equal to x coordinte of the midpoint of these two points since they have the same y coordinte \( 3 \).

The \( x \)-coordinate of the midpoint is: \[ x_{\text{vertex}} = \dfrac{1 + 5}{2} = \dfrac{6}{2} = 3. \] The \( x \)-coordinate of the vertex of the graph of \( f(x) \) is \( \boxed{3} \).

Problem 7

Find \(a \) and \( b \) so that the rational function given by: \[ f(x) = \dfrac{a x^4 + b x^3 + 3}{x^3 - 2} \] has an oblique asymptote given by: \( \; y = 2x - 3 \)

Solution:

The oblique asymptote is the quotient resulting from the long division of \( a x^4 + b x^3 + 3 \) by \( x^3 - 2 \).

Using long division, we obtain

\[ \dfrac{ax^4 + bx^3 + 3}{x^3 - 2} = ax + b + \dfrac{2ax + (3 + 2b)}{x^3 - 2} \] The quotient \( ax + b \) is the oblique asymptote and it must be equal to \( 2x - 3 \). \[ ax + b = 2x - 3 \] Since two polynomials are equal if and only if their corresponding coefficients are equal, we compare and obtain: \[ a = 2, \quad b = -3 \]

Problem 8

Solve for \( x \) in the equation: \[ \log_9 (x^3) = \log_2 (8) \]
Solution:

Given \[ \log_9 (x^3) = \log_2(8) \] Simplify right-hand side of the given equation \[ \log_2(8) = \log_2 (2^3) = 3 \] Rewrite the above equation \[ \log_9 (x^3) = 3 \] Rewrite 3 using a logarithm with base 9 \[ \log_9 (x^3) = \log_9(9^3) \] Write the algebraic equation from the previous equation \[ x^3 = 9^3 \] Solve for \(x\) \[ x = 9 \quad \]

Problem 9

Find the value of \( \log_y (x^4) \) if \( \log_x (y^3) = 2 \).

Solution:
Given \[ \log_x(y^3) = 2 \] Rewrite in exponential form \[ x^2 = y^3 \] Square both sides \[ (x^2)^2 = (y^3)^2 \] Simplify \[ x^4 = y^6 \] Take the log of base \(y \) of both sides and simplify \[ \log_y(x^4) = \log_y(y^6) = 6 \]

Problem 10

Solve for \(x\) the equation \[ \log_x (8e^3) = 3 \]

Solution:
Given \[ \log_x (8e^3) = 3 \] To solve for \( x \), rewrite the equation in exponential form: \[ x^3 = 8e^3 \] Taking the cube root on both sides: \[ x = \sqrt[3]{8e^3} \] Since \( 8 = 2^3 \), we simplify: \[ x = \sqrt[3]{2^3 e^3} = \sqrt[3]{(2 e)^3} = 2e \]

Problem 11

Find \( 2^{2x} \) If \[ 16^x + 16^{x-1} = 10 \]
Solution:
Given \[ 16^x + 16^{x - 1} = 10 \] Use the exponential rules to write \[ 16^x = (4^2)^x = 4^{2 x} \quad \text{and } 16^{x - 1} = 16^x 16^{-1} = \dfrac{4^{2x}}{16} \] and substitute in the give equation \[ 4^{2x} + \dfrac{4^{2x}}{16} = 10 \] Solve for \( 4^{2x}: \) \[ 4^{2x} = \dfrac{160}{17} \] Take the square root of both sides of the above equation \[ \sqrt {4^{2x}} = \sqrt {(4^x)^2} = 4^x = \dfrac{4\sqrt{10}}{\sqrt{17}} \] Hence \[ 4^x = (2^2)^x = 2^{2x} = \dfrac{4\sqrt{10}}{\sqrt{17}} \] This question could also be solved starting with \[ 16^x = (2^4)^x = 2^{4 x} \] and follow similar steps as in the given solution above.

Problem 12

\[ \text{If } a^2 - b^2 = 8 \text{ and } ab = 2, \text{ find } a^4 + b^4. \]

Solution:

Given \[ a^2 - b^2 = 8 \] Square both sides and expand \[ a^4 + b^4 - 2a^2b^2 = 8^2 \quad (I) \] Given \[ a b = 2 \] Square both sides of the above \[ (a b)^2 = a^2 b^2 = 2^2 =4 \] Substitute in (I) \[ a^4 + b^4 - 2(4) = 8^2 \] Hence \[ a^4 + b^4 = 72 \]

Problem 13

What are the maximum value and minimum values of the function \[ f(x) = \left|2 \sin \left(2 x - \dfrac{\pi}{3}\right) - 5\right| + 3 \]

Solution:

The range of a sine function is

\[ -1 \leq \sin\left( 2x - \dfrac{\pi}{3} \right) \leq 1 \]

multiply all terms of the double inequality by 2

\[ -2 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) \leq 2 \]

Add -5 to all terms of the inequality \[ -2 - 5 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq 2 - 5 \]

Simplify \[ -7 \leq 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \leq -3 \]

Change the above inequality using absolute value \[ 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| \leq 7 \]

Add 3 to all terms of the double inequality \[ 3 + 3 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 7 + 3 \]

Simplify \[ 6 \leq \left| 2\sin\left( 2x - \dfrac{\pi}{3} \right) - 5 \right| + 3 \leq 10 \] The maximum value of \( f(x) \) is equal to \( 10 \) and the minimum value of \( f(x) \) is equal to \( 6 \).

Problem 14

If \(x \lt -7\), simplify \[ \left| 4 - |3 + x| \right| \].

Solution:

If \( x \lt -7 \) then \[ x \lt - 3 \]

Add \( 3 \) to the above inequatlity and simplify \[ x + 3 \lt 0 \]

According to the definition of the absolute value \[ |3 + x| = -(3 + x) \]

The given expression simplifies to \[ |4 - |3 + x|| = |4 - (- (3 + x))| = |x + 7| \]

Since \( x + 7 \lt 0 \) \[ |x + 7| = - (x + 7) = - x - 7 \]

and hence \[ |4 - |3 + x|| = -x - 7 \]

Problem 15

A car travels from \( A \) to \( B \) at an average speed of 50 km/hour. At what average speed would it have to travel from \( B \) to \( A \) to average 60 km/hour for the whole trip?

Solution:

Let \( d \) be the distance between \( A \) and \( B \) is \( d \). The time taken to travel from A to B is given by: \[ t_1 = \dfrac{d}{50} \] where \( 50 \) km/hour is the average speed from \( A \) to \( B \).

Let the average speed for the return trip (from \( B \) to \( A \)) be \( v \). The time taken for the return trip is: \[ t_2 = \dfrac{d}{v} \]

The total time for the entire trip is: \[ t_{\text{total}} = t_1 + t_2 = \dfrac{d}{50} + \dfrac{d}{v} \]

The total distance (away and return) is: \[ d + d = 2d \]

The average speed for the whole trip is given by: \[ \text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} = \dfrac{2d}{t_{\text{total}}} \] We are told the average speed for the whole trip should be 60 km/hour, so: \[ \dfrac{2d}{\dfrac{d}{50} + \dfrac{d}{v}} = 60 \]

Cancel \( d \) from both the numerator and denominator: \[ \dfrac{2}{\dfrac{1}{50} + \dfrac{1}{v}} = 60 \]

Multiply both sides by \( \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \): \[ 2 = 60 \left( \dfrac{1}{50} + \dfrac{1}{v} \right) \]

Simplify: \[ 2 = \dfrac{60}{50} + \dfrac{60}{v} \] \[ 2 = 1.2 + \dfrac{60}{v} \]

Subtract 1.2 from both sides: \[ 0.8 = \dfrac{60}{v} \]

Solve for \( v \): \[ v = \dfrac{60}{0.8} = 75 \]

Thus, the average speed for the return trip from \( B \) to \( A \) needs to be \( 75 \) km/hour to average \( 60 \) km/hour for the whole trip.

Problem 16

Find \(x\) and \( y \) if \( x^2 - y^2 = -12 \) and \( x + y = 6 \).

Solution:

Given \[ x^2 - y^2 = (x - y)(x + y) = -12 \]

Substitute \( x + y \) by 6 in the above \[ 6(x - y) = -12 \]

Solve for \( x - y \) \[ (x - y) = -2 \] We now have a system of equations to solve \[ \begin{cases} x - y = -2 \\ x + y = 6 \end{cases} \]

Solve the system above \[ x = 2, \quad y = 4 \]

Problem 17

\( f(x) \) is a function such that \( f(x) + 3 f(8 - x) = x \) for all real numbers \( x \). Find the value of \( f(2) \).

Solution:

Given \[ f(x) + 3f(8 - x) = x \]

Substitute \( x = 2\) in the given equation \[ f(2) + 3 f(6) = 2 \tag{A} \]

Substitute \( x = 6\) in the given equation \[ f(6) + 3 f(2) = 6 \tag{B}\]

Solve equation (B) for \( f(6) \): \[ f(6) = 6 - 3f(2) \tag{C} \]

Substitute into equation (A): \[ f(2) + 3(6 - 3f(2)) = 2 \]

Expand and group \[ - 8 f(2) + 18 = 2 \]

Solve for } f(2): \[ f(2) = 2 \]

Problem 18

Let \( f(x) \) be a function such that \[ f(2x + 1) = 2f(x) + 1 \quad \text{for all real numbers } x \] Find the value of \( f(3) \) given that \( f(0) = 2 \).

Solution:
Given \[ f(2x + 1) = 2 f(x) + 1 \quad \tag{A} \] let \( x = 1 \) in \( A \) \[ f(3) = 2 f(1) + 1 \quad \tag{B} \] Let \( x = 0 \) in \( A \) \[ f(1) = 2 f(0) + 1 = 2 \times 2 + 1 = 5 \] Substitute \( f(1) = 5 \) in \( B \) to obtain \[ f(3) = 11 \]

Problem 19

Find \( b \) so that the line with equation \( y = 2 x + b \) is tangent to the circle with equation \[ x^2 + y^2 = 4\]

Solution:
Let us find the points of intersections of the circle and the line by solving the system of the two given equations. Substitute \( y \) by \( 2 x + b \) in the eqaution of the circle \[ x^2 + (2 x + b)^2 = 4 \] Expand and simplify \[ 5 x^2 + 4 b x + b^2 - 4 = 0 \] The number of points of intersection is given by the number of solutions of the above equation. The line and circle are tangent if the above quadratic equation has only one solution, which means that the discriminant is equal to zero. Find the discriminant \( \Delta \) as a function of \( b \) \[ \Delta = ( 4 b )^2 - 4 (5)(b^2 - 4)\] Solve. \[ ( 4 b )^2 - 4 (5)(b^2 - 4) = 0 \] \[ -4b^2+80 = 0 \] Solve for \( b \) to obtain two solutions \[ b = \pm 2 \sqrt{5} \]

Problem 20

What is the remainder of the division \[ \dfrac{x^{100} - x^{99} - x + 1}{x^2 - 3x + 2} \] .

Solution:
Let \[ P(x) = x^{100} - x^{99} - x + 1 \quad \text{and} \quad D(x) = x^2 - 3x + 2 \] . The division of the two polynomials may be written as: \[ P(x) = D(x) Q(x) + r(x) \] where \( Q(x) \) is the quotient and \( r(x) \) is the remainder, which will have a degree equal to one or lower because the divisor \( D(x) \) has degree 2. Thus, we express the remainder as \( r(x) = ax + b \).

We now need to find the values of \( a \) and \( b \) that define the remainder.

Note that \( D(x) \) may be factored as follows: \[ D(x) = x^2 - 3x + 2 = (x - 1)(x - 2) \] Hence, we have: \[ P(x) = (x - 1)(x - 2) Q(x) + ax + b \] Using the zeros of \( D(x) \), we write: \[ P(1) = (1 - 1)(1 - 2) Q(1) + a(1) + b \] This simplifies to: \[ a + b = P(1) \] Similarly, we write: \[ P(2) = (2 - 1)(2 - 2) Q(2) + a(2) + b \] This simplifies to: \[ 2a + b = P(2) \] Now we need to evaluate \( P(1) \) and \( P(2) \). \[ P(1) = 1^{100} - 1^{99} - 1 + 1 = 0 \] Next, we rewrite \( P(x) \) as: \[ P(x) = x^{99}(x - 1) - x + 1 \] Hence, we calculate \( P(2) \): \[ P(2) = 2^{99}(2 - 1) - 2 + 1 = 2^{99} - 1 \] We now have a system of equations to solve for \( a \) and \( b \): \[ a + b = 0 \] \[ 2a + b = 2^{99} - 1 \] Solving the system, we get: \[ a = 2^{99} - 1 \quad \text{and} \quad b = 1 - 2^{99} \] Thus, the remainder is: \[ r(x) = (2^{99} - 1)x + 1 - 2^{99} \]

Problem 21

Evaluate the number represented by the infinite series \[ \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \]

Solution:
Let \( y = \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \).

Square both sides to obtain: \[ y^2 = \dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} \] We can write: \[ y^2 = \dfrac{1}{3} + y \] Solve the above quadratic equation to obtain: \[ y = \dfrac{3 + \sqrt{21}}{6} \quad \text{or} \quad y = \dfrac{3 - \sqrt{21}}{6} \] Since \( y \) is positive, the solution is: \[ \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \sqrt{\dfrac{1}{3} + \cdots}}} = y = \dfrac{3 + \sqrt{21}}{6} \]

Problem 22

Show that the 3 by 3 system of equations given below has no solutions. \[ 2x + y - 3z = 5 \] \[ -5x + 3y + 2z = 7 \] \[ 3x - 4y + z = 8 \]

Solution:
The coefficient matrix \(A\) is: \[ A = \begin{bmatrix} 2 & 1 & -3 \\ -5 & 3 & 2 \\ 3 & -4 & 1 \end{bmatrix} \] The constants matrix \(b\) is: \[ b = \begin{bmatrix} 5 \\ 7 \\ 8 \end{bmatrix} \]

Cramer's Rule gives the solution for each variable in terms of the determinants of matrices derived from the coefficient matrix \(A\) and modified by replacing columns with the constants from the right-hand side.

To find \(x\), replace the first column of \(A\) with the constants vector \(b\): \[ A_x = \begin{bmatrix} 5 & 1 & -3 \\ 7 & 3 & 2 \\ 8 & -4 & 1 \end{bmatrix} \] Then, the solution for \(x\) is: \[ x = \dfrac{\text{det}(A_x)}{\text{det}(A)} \] To find \(y\), replace the second column of \(A\) with the constants vector \(b\): \[ A_y = \begin{bmatrix} 2 & 5 & -3 \\ -5 & 7 & 2 \\ 3 & 8 & 1 \end{bmatrix} \] Then, the solution for \(y\) is: \[ y = \dfrac{\text{det}(A_y)}{\text{det}(A)} \] To find \(z\), replace the third column of \(A\) with the constants vector \(b\): \[ A_z = \begin{bmatrix} 2 & 1 & 5 \\ -5 & 3 & 7 \\ 3 & -4 & 8 \end{bmatrix} \] Then, the solution for \(z\) is: \[ z = \dfrac{\text{det}(A_z)}{\text{det}(A)} \]

Now, we need to compute the determinants of the matrices \(A\), \(A_x\), \(A_y\), and \(A_z\). \[ \text{det}(A) = \begin{vmatrix} 2 & 1 & -3 \\ -5 & 3 & 2 \\ 3 & -4 & 1 \end{vmatrix} \] Using cofactor expansion along the first row: \[ \text{det}(A) = 2 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} + (-3) \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} \] Calculating the 2x2 determinants: \[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 3(1) - 2(-4) = 3 + 8 = 11 \] \[ \begin{vmatrix} -5 & 2 \\ 3 & 1 \end{vmatrix} = (-5)(1) - (2)(3) = -5 - 6 = -11 \] \[ \begin{vmatrix} -5 & 3 \\ 3 & -4 \end{vmatrix} = (-5)(-4) - (3)(3) = 20 - 9 = 11 \] Substitute these values into the determinant formula: \[ \text{det}(A) = 2(11) - 1(-11) + (-3)(11) = 22 + 11 - 33 = 0 \]

So, \(\text{det}(A) = 0\).

Determinant of \(A_x\): \[ \text{det}(A_x) = \begin{vmatrix} 5 & 1 & -3 \\ 7 & 3 & 2 \\ 8 & -4 & 1 \end{vmatrix} \] We can calculate this using cofactor expansion along the first row: \[ \text{det}(A_x) = 5 \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} - 1 \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} + (-3) \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} \] Calculate the 2x2 determinants: \[ \begin{vmatrix} 3 & 2 \\ -4 & 1 \end{vmatrix} = 11, \quad \begin{vmatrix} 7 & 2 \\ 8 & 1 \end{vmatrix} = -9, \quad \begin{vmatrix} 7 & 3 \\ 8 & -4 \end{vmatrix} = -52 \] Substituting these values into the formula: \[ \text{det}(A_x) = 5(11) - 1(-9) + (-3)(-52) = 55 + 9 + 156 = 220 \] No need to continue: If one of the determnants of \(A_x\), \(A_y\) or \(A_z\) is equal to zero, then the system of equations has no solution.