Question 1

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$$
A) $\left\{ \frac { 2 } { 3 } , - 6 \right\}$

B) $\left\{ - \frac { 2 } { 3 } , 6 \right\}$

C) $\left\{ - \frac { 3 } { 2 } , - 6 \right\}$

D) $\{ 2,6 \}$

Accepted Answer

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Question 2

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { x + 35 } { 10 } = \frac { 5 x - 5 } { x }$$
A) $\left\{ \frac { 50 } { 13 } \right\}$

B) $\{ - 5 , - 10 \}$

C) $\{ 5,10 \}$

D) $\left\{ \frac { 95 } { 49 } \right\}$

Accepted Answer

The value $ x = 0 $ makes the denominator zero in the original equation, but it is not listed in the options. The values $ x = 5 $ and $ x = 10 $ make the denominator zero in the equation, making them invalid solutions.

Question 3

Decide what values of the variable cannot possibly be solutions for the equation.
-$\sqrt { 2 x + 3 } - \sqrt { x + 1 } = 1$
A) $\{ 3 \}$
B) $\{ - 3 , - 1 \}$
C) $\varnothing$
D) $\{ 3 , - 1 \}$

Accepted Answer

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Question 4

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { 2 x + 5 } - \sqrt { x - 2 } = 3$$
A) $\{ - 2 \}$
B) $\{ 2,38 \}$
C) $\{ 2 \}$
D) $\{ 3,8 \}$

Accepted Answer

The expression under the square root must be non-negative. For $\sqrt{x - 2}$, $x$ must be at least 2. Thus, $x = 2$ is a boundary value and cannot be a solution because it makes the equation undefined. Additionally, substituting $x = 38$ into the equation does not satisfy it.

Question 5

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { 3 x + 10 } = 5 - 2 x$$
A) $\left\{ \frac { 3 } { 4 } \right\}$

B) $\left\{ \frac { 5 } { 4 } , 9 \right\}$

C) $\{ 5 \}$

D) $\left\{ \frac { 3 } { 4 } , 5 \right\}$

Accepted Answer

Squaring both sides of the equation $\sqrt{3x + 10} = 5 - 2x$ gives $3x + 10 = (5 - 2x)^2$. Solving this equation may lead to extraneous solutions due to the squaring process. The original equation implies $3x + 10 \geq 0$, which restricts $x$ to values that make the expression under the square root non-negative. However, the choice of $x = \frac{3}{4}$ leads to a contradiction when substituted back into the original equation, making it an extraneous solution. The other choices do not directly result from analyzing the given equation for impossible values without further solving and checking for extraneous solutions.

Question 6

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 7 x } { x - 7 } - \frac { 4 } { x } = \frac { 28 } { x ^ { 2 } - 7 x }$$
A) $\left\{ \frac { 4 } { 7 } \right\}$

B) $\left\{ \frac { 7 } { 4 } \right\}$

C) $\left\{ 0 , \frac { 4 } { 7 } \right\}$

D) $\left\{ \frac { 4 } { 7 } , - \frac { 4 } { 7 } \right\}$

Accepted Answer

The denominator of the first fraction cannot be 0, so $x \neq 7$. The denominator of the second fraction cannot be 0, so $x \neq 0$.

Question 7

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { x + 3 } = x - 3$$
A) $\{ 1,13 \}$
B) $\{ 1,6 \}$
C) $\{ 6 \}$
D) $\{ 6,13 \}$

Accepted Answer

The square root of a number is defined for non-negative inputs, so $x + 3 \geq 0$, which means $x \geq -3$. Squaring both sides of the equation gives $x + 3 = (x - 3)^2$, which simplifies to $x^2 - 7x + 6 = 0$. Factoring gives $(x - 1)(x - 6) = 0$, so $x = 1$ or $x = 6$. However, substituting $x = 6$ back into the original equation gives a false statement ($\sqrt{9} = 3$ is true, but substituting $x = 1$ gives $\sqrt{4} = -2$, which is false because the square root of 4 is 2, not -2). Therefore, $x = 1$ cannot be a solution, leaving $x = 6$ as the only value that cannot be a solution.

Question 8

Decide what values of the variable cannot possibly be solutions for the equation.
-A water tank can be filled in 9 minutes and emptied in 10 minutes. If the drain is accidentally left open when the tank is being filled, how long does it take to fill the tank? 
A) $\frac { 1 } { 19 } \mathrm {~min}$

B) $\frac { 1 } { 90 } \mathrm {~min}$

C) $\frac { 90 } { 19 } \mathrm {~min}$

D) $90 \mathrm {~min}$

Accepted Answer

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Question 9

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { x + 7 } + 5 = x$$
A) $\{ 2 \}$
B) $\{ 9,18 \}$
C) $\{ 9 \}$
D) $\{ 2,9 \}$

Accepted Answer

The equation $\sqrt { x + 7 } + 5 = x$ implies that $x$ must be greater than or equal to -7 for the square root to be defined. Solving the equation by squaring both sides gives $x^2 - x - 2 = 0$, which simplifies to $(x-2)(x+1) = 0$, yielding solutions $x = 2$ and $x = -1$. However, substituting $x = 2$ back into the original equation gives a contradiction ($\sqrt{9} + 5 \neq 2$), so $x = 2$ cannot be a solution. The correct answer is the set that does not include 2 but includes 9, as 9 is a value that, when substituted back into the equation, does not satisfy the original equation, making $\{9\}$ the set of values that cannot possibly be solutions.

Question 10

Decide what values of the variable cannot possibly be solutions for the equation.
-$$x = \sqrt { 2 x + 24 }$$
A) $\{ 6 \}$
B) $\{ 6 , - 4 \}$
C) $\varnothing$
D) $\{ 2 \}$

Accepted Answer

The equation $x = \sqrt{2x + 24}$ implies $x$ must be non-negative because it is equal to a square root. Solving $x = \sqrt{2x + 24}$ leads to $x^2 = 2x + 24$, which simplifies to $x^2 - 2x - 24 = 0$. Factoring gives $(x - 6)(x + 4) = 0$, so $x = 6$ or $x = -4$. However, substituting $x = -4$ into the original equation results in a negative value under the square root, which is not possible. Therefore, $x = -4$ is not a valid solution, leaving $x = 6$ as the only possible solution.

Question 11

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 6 } { x + 5 } = 1 - \frac { 1 } { x - 5 }$$
A) $\varnothing$
B) $\{ 0,7 \}$
C) $\{ 0 , - 5 \}$
D) $\{ 1,5 \}$

Accepted Answer

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Question 12

Decide what values of the variable cannot possibly be solutions for the equation.
-Martha can rake the leaves in her yard in 6 hours. Her brother can do the job in 8 hours. How long will it take them to do the job working together? 
A) $\frac { 24 } { 7 } \mathrm { hr }$

B) $\frac { 48 } { 2 } \mathrm { hr }$

C) $\frac { 1 } { 14 } \mathrm { hr }$

D) $\frac { 1 } { 48 } \mathrm { hr }$

Accepted Answer

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Question 13

Decide what values of the variable cannot possibly be solutions for the equation.
-One maid can clean the house in 8 hours. Another maid can do the job in 6 hours. How long will it take them to do the job working together? 
A) $\frac { 24 } { 7 } \mathrm { hr }$

B) $\frac { 48 } { 2 } \mathrm { hr }$

C) $\frac { 1 } { 14 } \mathrm { hr }$

D) $\frac { 1 } { 48 } \mathrm { hr }$

Accepted Answer

The answer of Decide what values of the variable cannot...

Question 14

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { - 5 x ^ { 2 } - 6 } { x - 4 } = \frac { - 15 x } { x - 4 } + 6$$
A) $\left\{ - \frac { 6 } { 5 } , 3 \right\}$

B) $\left\{ \frac { 6 } { 5 } , - 3 \right\}$

C) $\{ 3 \}$

D) $\left\{ \frac { - 5 \pm \sqrt { 105 } } { 3 } \right\}$

Accepted Answer

The denominator of the left-hand side cannot be zero, so $x \neq 4$.

Question 15

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { 3 x + 1 } = 3 + \sqrt { x - 4 }$$
A) $\{ 5,8 \}$
B) $\varnothing$
C) $\{ - 5 , - 8 \}$
D) $\{ - 1 \}$

Accepted Answer

The expression under the square root must be non-negative. For $\sqrt{3x + 1}$, $3x + 1 \geq 0$ implies $x \geq -\frac{1}{3}$. For $\sqrt{x - 4}$, $x - 4 \geq 0$ implies $x \geq 4$. Therefore, $x$ must be at least 4. Testing $x = 5$ and $x = 8$ in the equation shows they do not satisfy it, so they cannot be solutions.

Question 16

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { x } { x - 8 } - \frac { 8 } { x + 8 } = \frac { 128 } { x ^ { 2 } - 64 }$$
A) $\varnothing$

B) $\left\{ \frac { 1 } { 8 } \right\}$

C) $\{ \pm 8 \}$

D) $\{ \pm 8 i \}$

Accepted Answer

The denominator of the first fraction cannot be zero, so $x \ne 8$. The denominator of the second fraction cannot be zero, so $x \ne -8$.

Question 17

Decide what values of the variable cannot possibly be solutions for the equation.
-$$16 x = \sqrt { 1 - 24 x }$$
A) $\left\{ - \frac { 1 } { 8 } \right\}$

B) $\left\{ \frac { 1 } { 16 } \right\}$

C) $\left\{ \frac { 1 } { 32 } \right\}$

D) $\left\{ \frac { 1 } { 8 } \right\}$

Accepted Answer

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Question 18

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\sqrt { 2 x + 15 } - x = 6$$
A) $\{ - 7 \}$
B) $\{ - 3 \}$
C) $\{ - 7 , - 3 \}$
D) $\varnothing$

Accepted Answer

The answer of Decide what values of the variable cannot...

Question 19

Decide what values of the variable cannot possibly be solutions for the equation.
-Frank can type a report in 7 hours. James takes 2 hours to type it. How long will it take the two of 312) them typing together? 
A) $\frac { 14 } { 5 } \mathrm { hr }$

B) $\frac { 1 } { 9 } \mathrm { hr }$

C) $\frac { 1 } { 14 } \mathrm { hr }$

D) $\frac { 14 } { 9 } \mathrm { hr }$

Accepted Answer

The answer of Decide what values of the variable cannot...

Question 20

Decide what values of the variable cannot possibly be solutions for the equation.
-An experienced accountant can prepare a tax return in 13 hours. A novice accountant can do the job in 19 hours. How long will it take them to do the job working together? 
A) $\frac { 247 } { 32 } \mathrm { hr }$

B) $\frac { 1 } { 247 } \mathrm { hr }$

C) $\frac { 1 } { 32 } \mathrm { hr }$

D) $\frac { 247 } { 6 } \mathrm { hr }$

Accepted Answer

The experienced accountant can do $\frac { 1 } { 13 } $ of the job in 1 hour, and the novice accountant can do $\frac { 1 } { 19 } $ of the job in 1 hour. Working together, they can do $\frac { 1 } { 13 } + \frac { 1 } { 19 } = \frac { 32 } { 247 } $ of the job in 1 hour. So, it will take them $\frac { 247 } { 32 } $ hours to do the job.

Quiz 2: Equations and Inequalities

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x + 35 } { 10 } = \frac { 5 x - 5 } { x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 3 } - \sqrt { x + 1 } = 1$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 5 } - \sqrt { x - 2 } = 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 3 x + 10 } = 5 - 2 x$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 7 x } { x - 7 } - \frac { 4 } { x } = \frac { 28 } { x ^ { 2 } - 7 x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { x + 3 } = x - 3$

Decide what values of the variable cannot possibly be solutions for the equation. -A water tank can be filled in 9 minutes and emptied in 10 minutes. If the drain is accidentally left open when the tank is being filled, how long does it take to fill the tank?

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { x + 7 } + 5 = x$

Decide what values of the variable cannot possibly be solutions for the equation. - $x = \sqrt { 2 x + 24 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 6 } { x + 5 } = 1 - \frac { 1 } { x - 5 }$

Decide what values of the variable cannot possibly be solutions for the equation. -Martha can rake the leaves in her yard in 6 hours. Her brother can do the job in 8 hours. How long will it take them to do the job working together?

Decide what values of the variable cannot possibly be solutions for the equation. -One maid can clean the house in 8 hours. Another maid can do the job in 6 hours. How long will it take them to do the job working together?

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { - 5 x ^ { 2 } - 6 } { x - 4 } = \frac { - 15 x } { x - 4 } + 6$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 3 x + 1 } = 3 + \sqrt { x - 4 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x } { x - 8 } - \frac { 8 } { x + 8 } = \frac { 128 } { x ^ { 2 } - 64 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $16 x = \sqrt { 1 - 24 x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 15 } - x = 6$

Decide what values of the variable cannot possibly be solutions for the equation. -Frank can type a report in 7 hours. James takes 2 hours to type it. How long will it take the two of 312) them typing together?

Decide what values of the variable cannot possibly be solutions for the equation. -An experienced accountant can prepare a tax return in 13 hours. A novice accountant can do the job in 19 hours. How long will it take them to do the job working together?

Quiz 2: Equations and Inequalities

Decide what values of the variable cannot possibly be solutions for the equation. - 8x−2+8x+2=3\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3x−28​+x+28​=3

Decide what values of the variable cannot possibly be solutions for the equation. - x+3510=5x−5x\frac { x + 35 } { 10 } = \frac { 5 x - 5 } { x }10x+35​=x5x−5​

Decide what values of the variable cannot possibly be solutions for the equation. - 2x+3−x+1=1\sqrt { 2 x + 3 } - \sqrt { x + 1 } = 12x+3​−x+1​=1

Decide what values of the variable cannot possibly be solutions for the equation. - 2x+5−x−2=3\sqrt { 2 x + 5 } - \sqrt { x - 2 } = 32x+5​−x−2​=3

Decide what values of the variable cannot possibly be solutions for the equation. - 3x+10=5−2x\sqrt { 3 x + 10 } = 5 - 2 x3x+10​=5−2x

Decide what values of the variable cannot possibly be solutions for the equation. - 7xx−7−4x=28x2−7x\frac { 7 x } { x - 7 } - \frac { 4 } { x } = \frac { 28 } { x ^ { 2 } - 7 x }x−77x​−x4​=x2−7x28​

Decide what values of the variable cannot possibly be solutions for the equation. - x+3=x−3\sqrt { x + 3 } = x - 3x+3​=x−3

Decide what values of the variable cannot possibly be solutions for the equation. -A water tank can be filled in 9 minutes and emptied in 10 minutes. If the drain is accidentally left open when the tank is being filled, how long does it take to fill the tank?

Decide what values of the variable cannot possibly be solutions for the equation. - x+7+5=x\sqrt { x + 7 } + 5 = xx+7​+5=x

Decide what values of the variable cannot possibly be solutions for the equation. - x=2x+24x = \sqrt { 2 x + 24 }x=2x+24​

Decide what values of the variable cannot possibly be solutions for the equation. - 6x+5=1−1x−5\frac { 6 } { x + 5 } = 1 - \frac { 1 } { x - 5 }x+56​=1−x−51​

Decide what values of the variable cannot possibly be solutions for the equation. -Martha can rake the leaves in her yard in 6 hours. Her brother can do the job in 8 hours. How long will it take them to do the job working together?

Decide what values of the variable cannot possibly be solutions for the equation. -One maid can clean the house in 8 hours. Another maid can do the job in 6 hours. How long will it take them to do the job working together?

Decide what values of the variable cannot possibly be solutions for the equation. - −5x2−6x−4=−15xx−4+6\frac { - 5 x ^ { 2 } - 6 } { x - 4 } = \frac { - 15 x } { x - 4 } + 6x−4−5x2−6​=x−4−15x​+6

Decide what values of the variable cannot possibly be solutions for the equation. - 3x+1=3+x−4\sqrt { 3 x + 1 } = 3 + \sqrt { x - 4 }3x+1​=3+x−4​

Decide what values of the variable cannot possibly be solutions for the equation. - xx−8−8x+8=128x2−64\frac { x } { x - 8 } - \frac { 8 } { x + 8 } = \frac { 128 } { x ^ { 2 } - 64 }x−8x​−x+88​=x2−64128​

Decide what values of the variable cannot possibly be solutions for the equation. - 16x=1−24x16 x = \sqrt { 1 - 24 x }16x=1−24x​

Decide what values of the variable cannot possibly be solutions for the equation. - 2x+15−x=6\sqrt { 2 x + 15 } - x = 62x+15​−x=6

Decide what values of the variable cannot possibly be solutions for the equation. -Frank can type a report in 7 hours. James takes 2 hours to type it. How long will it take the two of 312) them typing together?

Decide what values of the variable cannot possibly be solutions for the equation. -An experienced accountant can prepare a tax return in 13 hours. A novice accountant can do the job in 19 hours. How long will it take them to do the job working together?

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 8 } { x - 2 } + \frac { 8 } { x + 2 } = 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x + 35 } { 10 } = \frac { 5 x - 5 } { x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 3 } - \sqrt { x + 1 } = 1$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 5 } - \sqrt { x - 2 } = 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 3 x + 10 } = 5 - 2 x$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 7 x } { x - 7 } - \frac { 4 } { x } = \frac { 28 } { x ^ { 2 } - 7 x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { x + 3 } = x - 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { x + 7 } + 5 = x$

Decide what values of the variable cannot possibly be solutions for the equation. - $x = \sqrt { 2 x + 24 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 6 } { x + 5 } = 1 - \frac { 1 } { x - 5 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { - 5 x ^ { 2 } - 6 } { x - 4 } = \frac { - 15 x } { x - 4 } + 6$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 3 x + 1 } = 3 + \sqrt { x - 4 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x } { x - 8 } - \frac { 8 } { x + 8 } = \frac { 128 } { x ^ { 2 } - 64 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $16 x = \sqrt { 1 - 24 x }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\sqrt { 2 x + 15 } - x = 6$