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ISEE Upper Level Math : Equations

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Example Questions

Example Question #151 : How To Find The Solution To An Equation

Solve for in this equation:

$\frac{5-2}{\sqrt{x}}=\frac{3}{4}$

Possible Answers:

Correct answer:

Explanation:

In order to solve for $\frac{5-2}{\sqrt{x}}=\frac{3}{4}$ , cross multiplication must be used. Applying this, the equestion to be solved will be:

$4(5-2)=3\sqrt{x}$

$4\cdot3=3\sqrt{x}$

Next, each side is divided by 3.

$4={\sqrt{x}}$

Next, each side is squared.

x=16

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Example Question #152 : How To Find The Solution To An Equation

Solve for :

$x=\frac{125}{5x}$

Possible Answers:

$\pm 20$

$\pm 5$

Correct answer:

$\pm 5$

Explanation:

$x=\frac{125}{5x}$

The first step to solve for x is to multiply each side by 5x. This results in:

$5x^{2}=125$

Divide each side by 5.

$x^{2}=25$

Next, take the square root of each side. This results in:

$x=\pm 5$

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Example Question #151 : How To Find The Solution To An Equation

What is the solution to the equation 2(x-5)=3x+12 ?

Possible Answers:

x=-5

x=-1

x=22

x=11

x=-22

Correct answer:

x=-22

Explanation:

To start, use the distributive property on the left side: 2(x-5)=2x-10 , so that you then have 2x-10=3x+12 . Then, combine like terms to get 1x=-22 so that your final answer is x=-22.

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Example Question #151 : How To Find The Solution To An Equation

Which of the following equations has as its solution set $\left \{ - 16, 20 \right \}$ ?

Possible Answers:

$\left | x - 3 \right | = 17$

$\left | x - 2 \right | = 18$

$\left | x - 4 \right | = 16$

$\left | x - 5 \right | = 15$

Correct answer:

$\left | x - 2 \right | = 18$

Explanation:

The absolute value of a nonnegative number is the number itself; the absolute value of a negative number is its positive opposite.

By substitution, 20 can be seen to be a solution of each of the equations in the four choices.

$\left | x - 2 \right | = 18$

$\left | 20 - 2 \right | = 18$

$\left |18 \right | = 18$ - true.

20 can be confirmed as a solution to the other three equations similarly. Therefore, the question is essentially to choose the equation with as a solution. Substituting for in each equation:

$\left | x - 2 \right | = 18$

$\left | -16 - 2 \right | = 18$

|-18| = 18 - true. This is the correct choice.

As for the other three:

$\left | -16 - 3 \right | = 17$

|-19| = 17

19 = 17 - false.

The other two equations can be similarly proved to not have as a solution.

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Example Question #855 : Isee Upper Level (Grades 9 12) Mathematics Achievement

$\frac{1}{x} = 12$

What is one tenth of ?

Possible Answers:

$1 \frac{1}{5}$

$\frac{5}{6}$

$\frac{1}{120}$

120

Correct answer:

$\frac{1}{120}$

Explanation:

$\frac{1}{x} = 12$ ; taking the reciprocal of both sides,

$x = \frac{1}{\left (\frac{1}{x} \right )} = \frac{1}{12}$

One tenth of is

$\frac{1}{10} \cdot x = \frac{1}{10} \cdot \frac{1}{12} = \frac{1}{120}$

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Example Question #151 : How To Find The Solution To An Equation

$\frac{1}{x+2} = 12$

What is one sixth of ?

Possible Answers:

$12\frac{1}{2}$

$\frac{25}{72}$

$- \frac{23}{72}$

$- 11\frac{1}{2}$

Correct answer:

$- \frac{23}{72}$

Explanation:

Take the reciprocal of both sides of the equation, then solve:

$\frac{1}{x+2} = 12$

$x+2 = \frac{1}{12}$

$x+2 - 2 = \frac{1}{12} - 2$

$x = \frac{1}{12} - \frac{24}{12} = \frac{1-24}{12} = -\frac{23}{12}$

One sixth of this is

$\frac{1}{6} x = \frac{1}{6}\left ( -\frac{23}{12} \right ) = -\frac{23}{6 \cdot 12} = - \frac{23}{72}$

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Example Question #151 : Equations

Give the solution set of the equation

$\left | 2x- 7\right | + 49 = 20$

Possible Answers:

$\left \{ 11, 18\right \}$

The equation has no solution

$\left \{ -18, -11 \right \}$

The set of all real numbers

Correct answer:

The equation has no solution

Explanation:

$\left | 2x- 7\right | + 49 = 20$

$\left | 2x- 7\right | + 49 - 49 = 20 - 49$

$\left | 2x- 7\right | = -29$

Since it is impossible for the absolute value of a number to be negative, the equation has no solution.

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Example Question #158 : How To Find The Solution To An Equation

How many real solutions does the equation

$8x^{3} + 12x^{2} = 18 x + 27$

have?

Possible Answers:

Three

None

One

Two

Correct answer:

Two

Explanation:

$8x^{3} + 12x^{2} = 18 x + 27$

$8x^{3} + 12x^{2} - 18 x - 27 = 18 x + 27 - 18 x - 27$

$8x^{3} - 18 x + 12x^{2}- 27 = 0$

$(8x^{3} - 18 x )+ (12x^{2}- 27) = 0$

$2x (4x^{2} - 9 )+ 3 (4x^{2}- 9) = 0$

$(2x + 3 )(4x^{2} - 9 ) = 0$

$(2x + 3 ) [ (2x)^{2} - 3^{2} ]= 0$

(2x + 3 ) (2x + 3 ) (2x - 3 ) = 0

One of two things must happen - either

2x+3 = 0

2x-3 = 0

This gives this equation two real solutions - the solutions to these two equations.

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Example Question #159 : How To Find The Solution To An Equation

Give the solution set of the equation

' $12x^{2} + 26x = 56$

Possible Answers:

$\left \{ -3\frac{1}{2}, 1 \frac{1}{3 } \right \}$

$\left \{ -2, 2 \frac{1}{3 } \right \}$

$\left \{ -1 \frac{1}{3 }, 3\frac{1}{2} \right \}$

$\left \{ -2 \frac{1}{3 } , 2 \right \}$

Correct answer:

$\left \{ -3\frac{1}{2}, 1 \frac{1}{3 } \right \}$

Explanation:

$12x^{2} + 26x = 56$

$12x^{2} + 26x - 56 = 56 - 56$

$12x^{2} + 26x - 56 = 0$

$2 (6x^{2} + 13x - 28 ) = 0$

The quadratic trinomial can be factored using the method by looking for two integers whose sum is 13 and whose product is $6 \times (-28 ) = -168$ . Throught trial and error, we see that these integers are and 21, so we continue:

$2 (6x^{2} -8x+21 x - 28 ) = 0$

$2\left [ (6x^{2} -8x)+( 21 x - 28 ) \right ]= 0$

$2\left [2x (3x -4)+7( 3x - 4) \right ]= 0$

2(2x+7)(3x-4) = 0

One of the factors must be equal to 0, so either:

2x+7 = 0

2x = -7

$x = -\frac{7}{2} = -3\frac{1}{2}$

3x-4 = 0

3x = 4

$x = \frac{4}{3} = 1 \frac{1}{3 }$

The correct choice is $\left \{ -3\frac{1}{2}, 1 \frac{1}{3 } \right \}$ .

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Example Question #861 : Isee Upper Level (Grades 9 12) Mathematics Achievement

List all real solutions of the equation

$x^{3}+4x^{2}= 16x+ 64$

Possible Answers:

No real solutions

$\left \{ -4\right \}$

$\left \{ -4, 4\right \}$

$\left \{ 4\right \}$

Correct answer:

$\left \{ -4, 4\right \}$

Explanation:

$x^{3}+4x^{2}= 16x+ 64$

$x^{3}+4x^{2} - 16x - 64 = 16x+ 64 - 16x - 64$

$x^{3}+4x^{2} - 16x - 64 = 0$

$(x^{3}+4x^{2} )- (16x + 64 )= 0$

$x^{2} (x + 4 )- 16 (x + 4 )= 0$

$(x^{2} - 16) (x + 4 )= 0$

$(x^{2} - 4^{2}) (x + 4 )= 0$

(x - 4 )(x + 4 ) (x + 4 )= 0

$(x - 4 )(x + 4 ) ^{2}= 0$

By the Zero Product Principle:

x- 4 = 0 , in which case x = 4 ,

x+ 4 = 0 , in which case x = -4 .

The correct choice is $\left \{ -4, 4\right \}$ .

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Equations

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #151 : How To Find The Solution To An Equation

Example Question #152 : How To Find The Solution To An Equation

Example Question #151 : How To Find The Solution To An Equation

Example Question #151 : How To Find The Solution To An Equation

Example Question #855 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #151 : How To Find The Solution To An Equation

Example Question #151 : Equations

Example Question #158 : How To Find The Solution To An Equation

Example Question #159 : How To Find The Solution To An Equation

Example Question #861 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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