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

Nosol1

There is no solution

–3

–1/2

Explanation

Nosol2

Consider the equation

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

Which of the following is true?

The equation has exactly two real solutions, which are of unlike sign.

The equation has no real solutions.

The equation has exactly two real solutions, which are of like sign.

The equation has exacty one real solution, which is positive.

The equation has exacty one real solution, which is negative.

Explanation

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

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

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

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

$y^{2}+9y+18 = 40$

$y^{2}+9y+18- 40 = 40 - 40$

$y^{2}+9y-22 = 0$

$y+11 \Rightarrow y = -11$

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

There are two solutions of unlike sign.

Consider the equation

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

Which of the following is true?

The equation has exactly two solutions, which are of unlike sign.

The equation has exactly two solutions, which are of like sign.

The equation has exactly one solution, which is positive.

The equation has exactly one solution, which is negative.

The equation has no solution.

Explanation

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

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

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

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

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

$4x^{2}+3x -22 = x^{2}+7x -18 + x^{2}+3x$

$4x^{2}+3x -22 = 2x^{2}+10x -18$

$4x^{2}+3x -22 - 2x^{2}-10x +18 = 2x^{2}+10x -18 - 2x^{2}-10x +18$

$2x^{2}-7x -4 = 0$

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

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?

Explanation

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 real solutions?

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

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

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

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

Each of the equations given in the other choices has at least one real solution.

Explanation

We can examine each individually.

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

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

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

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

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.

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

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

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

$21^{4} =\left ( \sqrt[4]{x - 7 } \right )^{4}$

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

Which of the following equations has no solution?

Each of the equations in the other responses has no solution.

Explanation

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.

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

No solutions

$x<\frac{4}{3}$

$x>\frac{4}{3}$

$x<\frac{3}{4}$

$x>\frac{3}{4}$

Explanation

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

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

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

no possible solution

Explanation

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.

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