How to find out when an equation has no solution - PSAT Math

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Find the solution to the following equation if x = 3:

y = (4x2 - 2)/(9 - x2)

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Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

I. x = 0

II. x = –1

III. x = 1

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$\small h\left ( x\right )=\frac{28}{x+4}$$

$\small \textup{For which of the following values of},x,\textup{is the above function undefined?}$

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A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

$\small x+4=0$

$\small x=-4$

Consider the equation

$\frac{x+9}{x}$+\frac{x+3}{x-2}$ = $\frac{4x+11}{x}$

Which of the following is true?

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Multiply the equation on both sides by LCM $\frac{x(x-2)}{1}$ :

$\frac{x+9}{x}$+\frac{x+3}{x-2}$ = $\frac{4x+11}{x}$

$\frac{x(x-2)}{1}$ \cdot $\frac{x+9}{x}$+\frac{x(x-2)}{1}$ \cdot$\frac{x+3}{x-2}$ = $\frac{x(x-2)}{1}$ \cdot$\frac{4x+11}{x}$

$(x-2) \left ( x+9 \right ) + x( x+3 )= (x-2) (4x+11)$

$\left ( $x^{2}$+7x -18\right ) + ( $x^{2}$+3x )= $4x^{2}$+3x -22$

or

$x = -$\frac{1}{2}$$

Substitution confirms that these are the solutions.

There are two solutions of unlike sign.

Consider the equation

$\frac{4}{y+3}$ = $\frac{y+6}{10}$

Which of the following is true?

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Multiply both sides by LCD $\frac{10(y+3)}{1}$ :

$\frac{4}{y+3}$ = $\frac{y+6}{10}$

$\frac{4}{y+3}$\cdot $\frac{10(y+3)}{1}$ = $\frac{y+6}{10}$ \cdot $\frac{10(y+3)}{1}$

$4 \cdot 10 = (y+6) (y+3)$

$y+11 \Rightarrow y = -11$

or

$y-2 = 0 \Rightarrow y = 2$

There are two solutions of unlike sign.

All of the following equations have no solution except for which one?

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Since all of the equations have the same symbols save for one number, the problem is essentially as follows:

For what value of does the equation

have a solution set other than the empty set?

We can simplify as follows:

If and are not equivalent expressions, the solution set is the empty set. If and are equivalent expressions, the solution set is the set of all real numbers; this happens if and only if:

$-5A \div (-5)= -30 \div (-5)$

Therefore, the only equation among the given choices whose solution set is not the empty set is the equation

which is the correct choice.

Which of the following equations has no solution?

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The problem is basically asking for what value of the equation

has no solution.

We can simplify as folllows:

Since the absolute value of a number must be nonnegative, regardless of the value of , this equation can never have a solution. Therefore, the correct response is that none of the given equations has a solution.

Which of the following equations has no real solutions?

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We can examine each individually.

$14 - \sqrt[3]{x - 7 } = -7$

$14 - \sqrt[3]{x - 7 } + 7 + \sqrt[3]{x - 7 } = -7 + 7 + \sqrt[3]{x - 7 }$

$21 = \sqrt[3]{x - 7 }$

This equation has a solution.

$-14 - \sqrt[3]{x - 7 } = 7$

$-14 - \sqrt[3]{x - 7 } - 7 + \sqrt[3]{x - 7 } = 7 - 7 + \sqrt[3]{x - 7 }$

$-21 = \sqrt[3]{x - 7 }$

$\left (-21 \right $)^{3}$ =\left ( \sqrt[3]{x - 7 } \right $)^{3}$$

This equation has a solution.

$14 - \sqrt[4]{x - 7 } = -7$

$14 - \sqrt[4]{x - 7 } + 7 + \sqrt[4]{x - 7 } = -7 + 7 + \sqrt[4]{x - 7 }$

$21 = \sqrt[4]{x - 7 }$

This equation has a solution.

$-14 - \sqrt[4]{x - 7 } = 7$

$-14 - \sqrt[4]{x - 7 } - 7 + \sqrt[4]{x - 7 } = 7 - 7 + \sqrt[4]{x - 7 }$

$-21 = \sqrt[4]{x - 7 }$

This equation has no solution, since a fourth root of a number must be nonnegative.

The correct choice is $-14 - \sqrt[4]{x - 7 } = 7$ .

Solve $\left | 3-4x\right |<0$ .

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By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.

Consider the equation

$\frac{4}{y+3}$ = $\frac{y+6}{10}$

Which of the following is true?

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Multiply both sides by LCD $\frac{10(y+3)}{1}$ :

$\frac{4}{y+3}$ = $\frac{y+6}{10}$

$\frac{4}{y+3}$\cdot $\frac{10(y+3)}{1}$ = $\frac{y+6}{10}$ \cdot $\frac{10(y+3)}{1}$

$4 \cdot 10 = (y+6) (y+3)$

$y+11 \Rightarrow y = -11$

or

$y-2 = 0 \Rightarrow y = 2$

There are two solutions of unlike sign.

Find the solution to the following equation if x = 3:

y = (4x2 - 2)/(9 - x2)

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Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

I. x = 0

II. x = –1

III. x = 1

Tap to see back →

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$\small h\left ( x\right )=\frac{28}{x+4}$$

$\small \textup{For which of the following values of},x,\textup{is the above function undefined?}$

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A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

$\small x+4=0$

$\small x=-4$

Consider the equation

$\frac{x+9}{x}$+\frac{x+3}{x-2}$ = $\frac{4x+11}{x}$

Which of the following is true?

Tap to see back →

Multiply the equation on both sides by LCM $\frac{x(x-2)}{1}$ :

$\frac{x+9}{x}$+\frac{x+3}{x-2}$ = $\frac{4x+11}{x}$

$\frac{x(x-2)}{1}$ \cdot $\frac{x+9}{x}$+\frac{x(x-2)}{1}$ \cdot$\frac{x+3}{x-2}$ = $\frac{x(x-2)}{1}$ \cdot$\frac{4x+11}{x}$

$(x-2) \left ( x+9 \right ) + x( x+3 )= (x-2) (4x+11)$

$\left ( $x^{2}$+7x -18\right ) + ( $x^{2}$+3x )= $4x^{2}$+3x -22$

or

$x = -$\frac{1}{2}$$

Substitution confirms that these are the solutions.

There are two solutions of unlike sign.

All of the following equations have no solution except for which one?

Tap to see back →

Since all of the equations have the same symbols save for one number, the problem is essentially as follows:

For what value of does the equation

have a solution set other than the empty set?

We can simplify as follows:

If and are not equivalent expressions, the solution set is the empty set. If and are equivalent expressions, the solution set is the set of all real numbers; this happens if and only if:

$-5A \div (-5)= -30 \div (-5)$

Therefore, the only equation among the given choices whose solution set is not the empty set is the equation

which is the correct choice.

Which of the following equations has no solution?

Tap to see back →

The problem is basically asking for what value of the equation

has no solution.

We can simplify as folllows:

Since the absolute value of a number must be nonnegative, regardless of the value of , this equation can never have a solution. Therefore, the correct response is that none of the given equations has a solution.

Which of the following equations has no real solutions?

Tap to see back →

We can examine each individually.

$14 - \sqrt[3]{x - 7 } = -7$

$14 - \sqrt[3]{x - 7 } + 7 + \sqrt[3]{x - 7 } = -7 + 7 + \sqrt[3]{x - 7 }$

$21 = \sqrt[3]{x - 7 }$

This equation has a solution.

$-14 - \sqrt[3]{x - 7 } = 7$

$-14 - \sqrt[3]{x - 7 } - 7 + \sqrt[3]{x - 7 } = 7 - 7 + \sqrt[3]{x - 7 }$

$-21 = \sqrt[3]{x - 7 }$

$\left (-21 \right $)^{3}$ =\left ( \sqrt[3]{x - 7 } \right $)^{3}$$

This equation has a solution.

$14 - \sqrt[4]{x - 7 } = -7$

$14 - \sqrt[4]{x - 7 } + 7 + \sqrt[4]{x - 7 } = -7 + 7 + \sqrt[4]{x - 7 }$

$21 = \sqrt[4]{x - 7 }$

This equation has a solution.

$-14 - \sqrt[4]{x - 7 } = 7$

$-14 - \sqrt[4]{x - 7 } - 7 + \sqrt[4]{x - 7 } = 7 - 7 + \sqrt[4]{x - 7 }$

$-21 = \sqrt[4]{x - 7 }$

This equation has no solution, since a fourth root of a number must be nonnegative.

The correct choice is $-14 - \sqrt[4]{x - 7 } = 7$ .

Solve $\left | 3-4x\right |<0$ .

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By definition, the absolute value of an expression can never be less than 0. Therefore, there are no solutions to the above expression.