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SAT II Math I : Single-Variable Algebra

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Example Questions

Example Question #101 : Single Variable Algebra

Give the set of all real solutions of the equation $8 x -2 x ^{\frac{1}{2}} - 3 = 0$ .

Possible Answers:

$\left \{ \frac{1}{4} , \frac{9}{16} \right \}$

$\left \{ - \frac{9}{16} , \frac{9}{16} \right \}$

$\left \{ - \frac{1}{4} , \frac{1}{4} \right \}$

$\left \{ - \frac{1}{4} , \frac{9}{16} \right \}$

None of these

Correct answer:

None of these

Explanation:

Set $y = x^{\frac{1}{2}}$ . Then $x = \left (x^{\frac{1}{2}} \right )^{2} = y ^{2}$ .

$8 x -2 x ^{\frac{1}{2}} - 3 = 0$ can be rewritten as

$8 \left (x^{\frac{1}{2}} \right )^{2} -2 x ^{\frac{1}{2}} - 3 = 0$

Substituting $y^{2}$ for and for $x^{\frac{1}{2}}$ , the equation becomes

$8y^{2} -2 y - 3 = 0$ ,

a quadratic equation in .

This can be solved using the method. We are looking for two integers whose sum is and whose product is 8 (-3) = -24 . Through some trial and error, the integers are found to be and 4, so the above equation can be rewritten, and solved using grouping, as

$8y^{2}-6y + 4 y - 3 = 0$

2y ( 4 y - 3)+1 ( 4 y - 3) = 0

(2y +1) ( 4 y - 3) = 0

By the Zero Product Principle, one of these factors is equal to zero:

Either:

2y + 1 = 0

2y + 1 - 1 = 0 - 1

2y = - 1

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

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

Substituting back:

$x^{\frac{1}{2}} = -\frac{1}{2}$ , or $\sqrt{x} = - \frac{1}{2}$

However, this does not hold for any real value of . No solution is yielded here.

Or:

4y - 3 = 0

4y - 3+ 3 = 0+ 3

4 y = 3

$\frac{4 y}{4} = \frac{3}{4}$

$y= \frac{3}{4}$

Substituting back:

$x^{\frac{1}{2}} = \frac{3}{4}$ , or $\sqrt{x} = \frac{3}{4}$

$(\sqrt{x} )^{2} = \left (\frac{3}{4} \right )^{2}$

$x = \frac{9}{16}$ ,

the only solution. This is not among the choices.

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Example Question #101 : Single Variable Algebra

Give the solution set of the inequality

$\frac{2x-5}{x-7} > 0$

Possible Answers:

$\left ( 2\frac{1}{2}, 7 \right )$

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 2\frac{1}{2}, 7 \right ) \cup \left ( 7, \infty\right )$

$\left (-\infty, 2\frac{1}{2} \right )$

$\left ( 2\frac{1}{2}, 7 \right )\cup \left ( 7, \infty\right )$

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

Correct answer:

$\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

Explanation:

Two numbers of like sign have a positive quotient.

Therefore, $\frac{2x-5}{x-7} > 0$ has as its solution set the set of points at which 2x-5 and x- 7 are both positive or both negative.

To find this set of points, we identify the zeroes of both expressions.

2x- 5 = 0

2x= 5

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

x-7 = 0

x = 7

Since $\frac{2x-5}{x-7}$ is nonzero we have to exclude $x = 2\frac{1}{2}$ ; x = 7 is excluded anyway since it would bring about a denominator of zero. We choose one test point on each of the three intervals $\left (-\infty, 2\frac{1}{2} \right ) , \left ( 2\frac{1}{2}, 7 \right ) , \left ( 7, \infty\right )$ and determine where the inequality is correct.

$\left (-\infty, 2\frac{1}{2} \right )$

Choose x = 0 :

$\frac{2 \cdot 0-5}{0-7} = \frac{-5}{-7} = \frac{5}{7} > 0$ - True.

$\left ( 2\frac{1}{2}, 7 \right )$

Choose x=3 :

$\frac{2 \cdot 3-5}{3-7} = \frac{6}{-4} =- \frac{3}{2} > 0$ - False.

$\left ( 7, \infty\right )$

Choose x = 8 :

$\frac{2 \cdot 8-5}{8-7} = \frac{11}{1} =11> 0$ - True.

The solution set is $\left (-\infty, 2\frac{1}{2} \right ) \cup \left ( 7, \infty\right )$

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Example Question #251 : Sat Subject Test In Math I

Solve for x.

$2x-7\geq 1$

Possible Answers:

$x\geq 5$

$x\geq 4$

$x\leq 4$

$x\geq -3$

x> 4

Correct answer:

$x\geq 4$

Explanation:

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, add 7 to each side.

$2x-7+7\geq 1+7$

$2x\geq 8$

Now, divide both sides by 2.

$x\geq 4$

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Example Question #102 : Single Variable Algebra

Solve for x.

$3x+2\geq -7$

Possible Answers:

$x\leq -3$

$x\geq 3$

$x\geq 1$

$x\geq -1$

$x\geq -3$

Correct answer:

$x\geq -3$

Explanation:

Solving inequalities is very similar to solving an equation. We must start by isolating x by moving the terms farthest from it to the other side of the inequality. In this case, subtract 2from each side.

$3x+2-2\geq -7-2$

$3x\geq -9$

Now, divide both sides by 2.

$x\geq -3$

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Example Question #261 : Sat Subject Test In Math I

Solve the following inequality:

3x+7<9

Possible Answers:

x<-4

x<3

x< 9

$x< \frac{2}{3}$

x<7

Correct answer:

$x< \frac{2}{3}$

Explanation:

To solve for an inequality, you solve like you would for a single variable expression and get by itself.

First, subtract from both sides to get,

3x<2 .

Then divide both sides by and your final answer will be,

$x< \frac{2}{3}$ .

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Example Question #101 : Single Variable Algebra

Solve the inequality: 3-(3-2x)<10

Possible Answers:

x<5

$x>-\frac{1}{5}$

$x<\frac{1}{5}$

x<-5

x>5

Correct answer:

x<5

Explanation:

Simplify the left side.

3-3+2x<10

The inequality becomes:

2x<10

Divide by two on both sides.

$\frac{2}{2}x<\frac{10}{2}$

The answer is: x<5

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Example Question #1 : Solving Inequalities

Solve the inequality: $2x+6\leq 9x-3$

Possible Answers:

$x\geq-\frac{3}{11}$

$x\leq-\frac{9}{7}$

$x\leq-\frac{3}{11}$

$x\geq-\frac{9}{7}$

$x\geq\frac{9}{7}$

Correct answer:

$x\geq\frac{9}{7}$

Explanation:

Subtract on both sides.

$2x+6-2x\leq 9x-3-2x$

$6\leq 7x-3$

Add 3 on both sides.

$6+3\leq 7x-3+3$

$9\leq 7x$

Divide by 7 on both sides.

$\frac{9}{7}\leq \frac{7x}{7}$

$\frac{9}{7}\leq x$

The answer is: $x\geq\frac{9}{7}$

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Example Question #5 : How To Divide Monomial Quotients

Simplify the expression.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

Possible Answers:

$\small \frac{z^3}{x^2y^5}$

$\small \frac{xy^3}{z^3}$

$\small \frac{z^3}{xy^3}$

$\small \frac{x^2y^5}{z^3}$

Correct answer:

$\small \frac{xy^3}{z^3}$

Explanation:

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}$

Because we are only multiplying terms in the numerator, we can disregard the parentheses.

$\small \frac{(x^2y^2)(x^3y^4z)}{x^4y^3z^4}=\frac{x^2y^2x^3y^4z}{x^4y^3z^4}$

To combine like terms in the numerator, we add their exponents.

$\small \frac{x^5y^6z}{x^4y^3z^4}$

To combine like terms between the numerator and denominator, subtract the denominator exponent from the numerator exponent.

$x^{5-4}y^{6-3}z^{1-4}=xy^3z^{-3}$

Remember that any negative exponents stay in the denominator.

$\frac{xy^3}{z^3}$

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Example Question #101 : Single Variable Algebra

Give the value of that makes the polynomial $9x^{2}+ Nx + 100$ the square of a linear binomial.

Possible Answers:

N = 81

N = 30

N = 90

None of the other responses gives a correct answer.

N = 60

Correct answer:

N = 60

Explanation:

A quadratic trinomial is a perfect square if and only if takes the form

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$9x^{2}+ Nx + 100 = (3x)^{2} + Nx + 10^{2}$ , so

A = 3x and B = 10 .

For $9x^{2}+ Nx + 100$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 3x \cdot 10 = 60x$ ,

so N = 60 . This is the correct choice.

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Example Question #3 : Simplifying Expressions

Factor:

$27Q^{3}- 270Q^{2}+900Q -1,000$

Possible Answers:

$(3Q-10)(9Q^{2}+ 30 Q +100)$

$(3Q-10)(3Q+10)^{2}$

The polynomial is prime.

$(3Q - 10)^{3}$

$(3Q-10)(9Q^{2}- 30 Q +100)$

Correct answer:

$(3Q - 10)^{3}$

Explanation:

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot 90Q^{2}+3\cdot 300Q -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot 9Q^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

This can be factored out as the cube of a difference, where A = 3Q, B = 10 :

$A^{3}- 3A^{2}B+3AB^{2} -B^{3} = (A - B)^{3}$

Therefore,

$27Q^{3}- 270Q^{2}+900Q -1,000$

$=\left (3Q \right )^{3}- 3 \cdot \left (3Q \right )^{2}\cdot 10+3\cdot 3Q \cdot 10^{2} -10^{3}$

$= (3Q - 10)^{3}$

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SAT II Math I : Single-Variable Algebra

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #101 : Single Variable Algebra

Example Question #101 : Single Variable Algebra

Example Question #251 : Sat Subject Test In Math I

Example Question #102 : Single Variable Algebra

Example Question #261 : Sat Subject Test In Math I

Example Question #101 : Single Variable Algebra

Example Question #1 : Solving Inequalities

Example Question #5 : How To Divide Monomial Quotients

Example Question #101 : Single Variable Algebra

Example Question #3 : Simplifying Expressions

All SAT II Math I Resources

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