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PSAT Math : Equations / Inequalities

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Example Questions

Example Question #2 : How To Find Out When An Equation Has No Solution

Consider the equation

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

Which of the following is true?

Possible Answers:

The equation has exactly one solution, which is positive.

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

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

The equation has no solution.

The equation has exactly one solution, which is negative.

Correct answer:

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

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-4)(2x+1)= 0

x-4 = 0

x= 4

2x+1 = 0

2x = -1

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

Substitution confirms that these are the solutions.

There are two solutions of unlike sign.

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Example Question #2 : Linear / Rational / Variable Equations

Which of the following equations has no solution?

Possible Answers:

6- |x + 2 | = 8

6- |x + 6 | = 8

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

6- |x + 8 | = 8

6- |x | = 8

Correct answer:

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

Explanation:

The problem is basically asking for what value of the equation

6- |x + A | = 8

has no solution.

We can simplify as folllows:

6- |x + A | - 6 = 8 - 6

- |x + A | = 2

|x + A | = - 2

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.

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Example Question #3 : Linear / Rational / Variable Equations

Consider the equation

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

Which of the following is true?

Possible Answers:

The equation has no real solutions.

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

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

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

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

Correct answer:

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

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)(y-2) = 0

$y+11 \Rightarrow y = -11$

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

There are two solutions of unlike sign.

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Example Question #1 : How To Find Out When An Equation Has No Solution

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

Possible Answers:

4x + 3(x-10) = 2x + 5 (x - 8)

4x + 3(x-10) = 2x + 5 (x - 4)

4x + 3(x-10) = 2x + 5 (x - 12)

4x + 3(x-10) = 2x + 5 (x - 10)

4x + 3(x-10) = 2x + 5 (x - 6)

Correct answer:

4x + 3(x-10) = 2x + 5 (x - 6)

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

4x + 3(x-10) = 2x + 5 (x - A)

have a solution set other than the empty set?

We can simplify as follows:

4x + 3x-30= 2x + 5 x - 5A

7x-30=7x - 5A

-30 = -5A

If -5A and are not equivalent expressions, the solution set is the empty set. If -5A and are equivalent expressions, the solution set is the set of all real numbers; this happens if and only if:

-5A = -30

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

A = 6

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

4x + 3(x-10) = 2x + 5 (x - 6)

which is the correct choice.

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Example Question #3 : Linear / Rational / Variable Equations

Which of the following equations has no real solutions?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We can examine each individually.

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

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

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

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

x-7 = 9,261

x = 9,268

This equation has a solution.

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

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

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

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

x-7 =- 9,261

x =- 9,254

This equation has a solution.

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

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

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

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

x-7 = 194,481

x = 194,488

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

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Example Question #2 : How To Find Out When An Equation Has No Solution

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

Possible Answers:

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

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

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

No solutions

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

Correct answer:

No solutions

Explanation:

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

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Example Question #1 : How To Find The Solution To A Rational Equation With Lcd

Rational_5

Possible Answers:

–2

–1

Correct answer:

Explanation:

Rational_2

Rational_3

Rational_4

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Example Question #81 : Equations / Inequalities

Linesmb1

Possible Answers:

–bm/(m²+ 1)

–b/(m + 1)

b/(m²+ 1)

bm/(m²+ 1)

–b/(m²– 1)

Correct answer:

b/(m²+ 1)

Explanation:

Linesmb5

Linesmb4

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Example Question #2 : How To Find The Solution To A Rational Equation With Lcd

In the equation below, , , and are non-zero numbers. What is the value of in terms of and ?

$\frac{1}{m^3}-\frac{1}{k^2}=\frac{1}{p}$

Possible Answers:

$m=p^{\frac{1}{2}}-k^{\frac{1}{3}}$

$m=(\frac{p+k^2}{pk^2})^{\frac{1}{3}}$

$m=\frac{p^2k^3}{p+k^2}$

$m=(\frac{pk^2}{p+k^2})^{\frac{1}{3}}$

$m=(\frac{p+k^2}{2})^3$

Correct answer:

$m=(\frac{pk^2}{p+k^2})^{\frac{1}{3}}$

Explanation:

Pkm_7-21-13

Pkm2_7-21-13

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Example Question #1 : How To Find The Solution To An Equation

If 6x = 42 and xk = 2, what is the value of k?

Possible Answers:

1/7

2/7

1/6

Correct answer:

2/7

Explanation:

Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.

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PSAT Math : Equations / Inequalities

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #2 : How To Find Out When An Equation Has No Solution

Example Question #2 : Linear / Rational / Variable Equations

Example Question #3 : Linear / Rational / Variable Equations

Example Question #1 : How To Find Out When An Equation Has No Solution

Example Question #3 : Linear / Rational / Variable Equations

Example Question #2 : How To Find Out When An Equation Has No Solution

Example Question #1 : How To Find The Solution To A Rational Equation With Lcd

Example Question #81 : Equations / Inequalities

Example Question #2 : How To Find The Solution To A Rational Equation With Lcd

Example Question #1 : How To Find The Solution To An Equation

All PSAT Math Resources

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