Question 1

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

Accepted Answer

The equation $\sqrt { x + 7 } + 5 = x$ implies that $x$ must be greater than or equal to -7, because the square root of a negative number is not real. Solving the equation by squaring both sides gives $x + 7 + 10\sqrt{x + 7} + 25 = x^2$, which simplifies to a quadratic equation. However, the critical insight is that the original equation has a constraint due to the square root. Plugging $x = 9$ back into the original equation, we get $\sqrt{9 + 7} + 5 = 9$, which simplifies to $4 + 5 = 9$, a true statement. Thus, $x = 9$ is actually a solution, not a value that cannot possibly be a solution. The mistake in the reasoning is assuming $x = 9$ cannot be a solution without checking it against the original equation. The correct approach is to identify values that do not satisfy the domain of the square root function or do not satisfy the original equation upon solving. In this case, no specific values are given that cannot possibly be solutions based on the incorrect reasoning provided. The correct answer should identify values that, when plugged into the original equation, do not satisfy it, but without further calculation or incorrect assumptions, the provided choices do not accurately reflect such values based on the correct mathematical process.

Question 2

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { x } { x - 5 } - \frac { 5 } { x + 5 } = \frac { 50 } { x ^ { 2 } - 25 }$$

Accepted Answer

The denominator of the first fraction cannot be zero, so $x \neq 5$. The denominator of the second fraction cannot be zero, so $x \neq -5$. Therefore, the only possible solution is $x \neq \pm 5$.

Question 3

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { x + 24 } { 7 } = \frac { 5 x - 4 } { x }$$

Accepted Answer

The values 4 and 7 make the denominators zero, which makes the equation undefined.

Question 4

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

Accepted Answer

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

Question 5

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

Accepted Answer

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

Question 6

Decide what values of the variable cannot possibly be solutions for the equation.
-One maid can clean the house in 5 hours. Another maid can do the job in 3 hours. How long will it take them to do the job working together?

Accepted Answer

The correct answer is found by adding the rates of the two maids working together: $ \frac{1}{5} + \frac{1}{3} = \frac{3}{15} + \frac{5}{15} = \frac{8}{15} $ hours. To find the time it takes for them to complete the job together, take the reciprocal of this sum, which gives $ \frac{15}{8} $ hours.

Question 7

Decide what values of the variable cannot possibly be solutions for the equation.
-Frank can type a report in 3 hours. James takes 7 hours to type it. How long will it take the two of them typing together?

Accepted Answer

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

Question 8

Decide what values of the variable cannot possibly be solutions for the equation.
-$$\frac { 6 x } { x - 6 } - \frac { 4 } { x } = \frac { 24 } { x ^ { 2 } - 6 x }$$

Accepted Answer

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

Question 9

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

Accepted Answer

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

Question 10

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

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. The values 5 and 8 satisfy these conditions, so they cannot be solutions.

Question 11

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

Accepted Answer

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

Question 12

Decide what values of the variable cannot possibly be solutions for the equation.
-A water tank can be filled in 6 minutes and emptied in 8 minutes. If the drain is accidentally left open when the tank is being filled, how long does it take to fill the tank?

Accepted Answer

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

Question 13

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

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.
-$$4 x = \sqrt { 1 - 6 x }$$

Accepted Answer

The square root of a negative number is not a real number, so $6x$ must be less than or equal to 1. Therefore, $x$ must be less than or equal to $\frac{1}{6}$.

Question 15

Decide what values of the variable cannot possibly be solutions for the equation.
-An experienced accountant can prepare a tax return in 8 hours. A novice accountant can do the job in 14 hours. How long will it take them to do the job working together?

Accepted Answer

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

Question 16

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

Accepted Answer

The expression under each square root must be non-negative. For $\sqrt{2x + 3}$, $2x + 3 \geq 0$ implies $x \geq -1.5$. For $\sqrt{x + 1}$, $x + 1 \geq 0$ implies $x \geq -1$. Solving the equation, $x = 3$ is a solution, but substituting $x = -1$ into the original equation results in an invalid expression, as the square roots do not satisfy the equation. Thus, $x = -1$ cannot be a solution.

Question 17

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

Accepted Answer

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

Question 18

Decide what values of the variable cannot possibly be solutions for the equation.
-Martha can rake the leaves in her yard in 2 hours. Her brother can do the job in 4 hours. How long will it take them to do the job working together?

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.
-$$\frac { 24 } { x - 2 } + \frac { 24 } { x + 2 } = 5$$

Accepted Answer

The denominator of a fraction cannot be zero, so $x$ cannot be 2 or -2.

Question 20

Decide what values of the variable cannot possibly be solutions for the equation.
-$$1 + \frac { 1 } { x } = \frac { 20 } { x ^ { 2 } }$$

Accepted Answer

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

Quiz 2: Equations and Inequalities

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. - $\frac { x } { x - 5 } - \frac { 5 } { x + 5 } = \frac { 50 } { x ^ { 2 } - 25 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x + 24 } { 7 } = \frac { 5 x - 4 } { x }$

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. - $\sqrt { x + 3 } = x - 3$

Decide what values of the variable cannot possibly be solutions for the equation. -One maid can clean the house in 5 hours. Another maid can do the job in 3 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. -Frank can type a report in 3 hours. James takes 7 hours to type it. How long will it take the two of them typing together?

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 6 x } { x - 6 } - \frac { 4 } { x } = \frac { 24 } { x ^ { 2 } - 6 x }$

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

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. - $\sqrt { 2 x + 15 } - x = 6$

Decide what values of the variable cannot possibly be solutions for the equation. -A water tank can be filled in 6 minutes and emptied in 8 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. - $\sqrt { 3 x + 10 } = 5 - 2 x$

Decide what values of the variable cannot possibly be solutions for the equation. - $4 x = \sqrt { 1 - 6 x }$

Decide what values of the variable cannot possibly be solutions for the equation. -An experienced accountant can prepare a tax return in 8 hours. A novice accountant can do the job in 14 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. - $\sqrt { 2 x + 3 } - \sqrt { x + 1 } = 1$

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

Decide what values of the variable cannot possibly be solutions for the equation. -Martha can rake the leaves in her yard in 2 hours. Her brother can do the job in 4 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. - $\frac { 24 } { x - 2 } + \frac { 24 } { x + 2 } = 5$

Decide what values of the variable cannot possibly be solutions for the equation. - $1 + \frac { 1 } { x } = \frac { 20 } { x ^ { 2 } }$

Quiz 2: Equations and Inequalities

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. - xx−5−5x+5=50x2−25\frac { x } { x - 5 } - \frac { 5 } { x + 5 } = \frac { 50 } { x ^ { 2 } - 25 }x−5x​−x+55​=x2−2550​

Decide what values of the variable cannot possibly be solutions for the equation. - x+247=5x−4x\frac { x + 24 } { 7 } = \frac { 5 x - 4 } { x }7x+24​=x5x−4​

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. - 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. -One maid can clean the house in 5 hours. Another maid can do the job in 3 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. -Frank can type a report in 3 hours. James takes 7 hours to type it. How long will it take the two of them typing together?

Decide what values of the variable cannot possibly be solutions for the equation. - 6xx−6−4x=24x2−6x\frac { 6 x } { x - 6 } - \frac { 4 } { x } = \frac { 24 } { x ^ { 2 } - 6 x }x−66x​−x4​=x2−6x24​

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

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. - 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. -A water tank can be filled in 6 minutes and emptied in 8 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. - 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. - 4x=1−6x4 x = \sqrt { 1 - 6 x }4x=1−6x​

Decide what values of the variable cannot possibly be solutions for the equation. -An experienced accountant can prepare a tax return in 8 hours. A novice accountant can do the job in 14 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. - 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. - −5x2−4x−4=−15xx−4+4\frac { - 5 x ^ { 2 } - 4 } { x - 4 } = \frac { - 15 x } { x - 4 } + 4x−4−5x2−4​=x−4−15x​+4

Decide what values of the variable cannot possibly be solutions for the equation. -Martha can rake the leaves in her yard in 2 hours. Her brother can do the job in 4 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. - 24x−2+24x+2=5\frac { 24 } { x - 2 } + \frac { 24 } { x + 2 } = 5x−224​+x+224​=5

Decide what values of the variable cannot possibly be solutions for the equation. - 1+1x=20x21 + \frac { 1 } { x } = \frac { 20 } { x ^ { 2 } }1+x1​=x220​

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. - $\frac { x } { x - 5 } - \frac { 5 } { x + 5 } = \frac { 50 } { x ^ { 2 } - 25 }$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { x + 24 } { 7 } = \frac { 5 x - 4 } { x }$

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. - $\sqrt { x + 3 } = x - 3$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 6 x } { x - 6 } - \frac { 4 } { x } = \frac { 24 } { x ^ { 2 } - 6 x }$

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

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. - $\sqrt { 2 x + 15 } - x = 6$

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. - $4 x = \sqrt { 1 - 6 x }$

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. - $\frac { - 5 x ^ { 2 } - 4 } { x - 4 } = \frac { - 15 x } { x - 4 } + 4$

Decide what values of the variable cannot possibly be solutions for the equation. - $\frac { 24 } { x - 2 } + \frac { 24 } { x + 2 } = 5$

Decide what values of the variable cannot possibly be solutions for the equation. - $1 + \frac { 1 } { x } = \frac { 20 } { x ^ { 2 } }$