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SAT II Math I : Mathematical Relationships

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Example Questions

Example Question #41 : Other Mathematical Relationships

$3 \sqrt{a} + 4 \sqrt{b} = 9$

$2 \sqrt{a} +5 \sqrt{b} = -1$

Evaluate .

Possible Answers:

$\textup{The system has no solution}$

b = 9

b= 49

b = -9

b = -49

Correct answer:

$\textup{The system has no solution}$

Explanation:

The system has no solution. The clue can be seen in the second equation

$2 \sqrt{a} +5 \sqrt{b} = -1$ .

The square root of a number (assuming it is real) must be a nonnegative number. Consequently, by the closure of the nonnegative numbers under multiplication, it holds that $2 \sqrt{a}$ and $5 \sqrt{b}$ are nonnegative as well. Also, by the closure of the nonnegative numbers under multiplication, $2 \sqrt{a} +5 \sqrt{b}$ must also be a positive quantity. Therefore,

$2 \sqrt{a} +5 \sqrt{b} = -1$

cannot have a solution; consequently, neither can the system.

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Example Question #45 : Other Mathematical Relationships

$\frac{4 }{a-5} + \frac{3 }{b-5} = 27$

$\frac{5 }{a-5} + \frac{2 }{b-5} = 25$

Evaluate .

Possible Answers:

$b= \frac{1}{3}$

$b= \frac{1}{5}$

$b= 5 \frac{1}{5}$

$b= 5 \frac{1}{3}$

No solution

Correct answer:

$b= 5 \frac{1}{5}$

Explanation:

Rewrite the two equations by setting $x = \frac{1}{a-5 }$ and $y = \frac{1}{b-5}$ and substituting:

$\frac{4 }{a-5} + \frac{3 }{b-5} = 27$

$4 \cdot \frac{1 }{a-5} + 3 \cdot \frac{1 }{b-5} = 27$

4x+3y = 27

$\frac{5 }{a-5} + \frac{2 }{b-5} = 25$

$5 \cdot \frac{1 }{a-5} + 2 \cdot \frac{1 }{b-5} = 27$

5x+2y = 25

In terms of and , this is a two-by-two linear system:

4x+3y = 27

5x+2y = 25

Rewrite this as an augmented matrix as follows:

$\begin{bmatrix} 4& 3\\ 5& 2 \end{matrix} \begin{vmatrix} 27\\ 25 \end{bmatrix}$

Perform the following row operations to make the left two columns the identity matrix $\begin{bmatrix}1 & 0\\ 0 &1 \end{bmatrix}$ :

$R1 \leftrightarrow R2$

$\begin{bmatrix} 5& 2 \\ 4& 3 \end{matrix} \begin{vmatrix} 25 \\ 27 \end{bmatrix}$

$R1-R2 \rightarrow R1$

$\begin{bmatrix} 5 -4& 2 -3 \\ 4& 3 \end{matrix} \begin{vmatrix} 25 -27 \\ 27 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 4& 3 \end{matrix} \begin{vmatrix} -2 \\ 27 \end{bmatrix}$

$R2 - 4R1 \rightarrow R2$

$\begin{bmatrix} 1& -1 \\ 4 - 4(1)& 3- 4(-1) \end{matrix} \begin{vmatrix} -2 \\ 27-4(-2) \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 0& 7 \end{matrix} \begin{vmatrix} -2 \\ 35 \end{bmatrix}$

$R2 \div 7 \rightarrow R2$

$\begin{bmatrix} 1& -1 \\ 0 \div 7 & 7 \div 7 \end{matrix} \begin{vmatrix} -2 \\ 35 \div 7 \end{bmatrix}$

$\begin{bmatrix} 1& -1 \\ 0 & 1 \end{matrix} \begin{vmatrix} -2 \\ 5 \end{bmatrix}$

$R1 +R2 \rightarrow R1$

$\begin{bmatrix} 1+ 0& -1+1 \\ 0 & 1 \end{matrix} \begin{vmatrix} -2+5 \\ 5 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{matrix} \begin{vmatrix} 3 \\ 5 \end{bmatrix}$

x = 3 and y = 5 .

Substituting $\frac{1}{b-5}$ back for :

$\frac{1}{b-5} = 5$

Taking the reciprocal of both sides:

$b-5 = \frac{1}{5}$

$b-5 + 5 = \frac{1}{5} + 5$

$b= 5 \frac{1}{5}$ .

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Example Question #46 : Other Mathematical Relationships

Evaluate :

$3 \sqrt [3] {a} + 4 \sqrt [3]{b} = 9$

$2 \sqrt [3]{a} +5 \sqrt [3]{b} = -1$

Possible Answers:

a = - 343

a = 27

No solution

a = 343

a = -27

Correct answer:

a = 343

Explanation:

Rewrite the two equations by setting $x = \sqrt [3] {a}$ and $y = \sqrt [3] {b}$ and substituting:

$3 \sqrt [3] {a} + 4 \sqrt [3]{b} = 9$

3 x+ 4 y = 9

$2 \sqrt [3]{a} +5 \sqrt [3]{b} = -1$

2 x +5 y= -1

In terms of and , this is the system of linear equations:

3 x + 4 y = 9

2x+ 5y = -1

Rewrite this as an augmented matrix as follows:

$\begin{bmatrix} 3 & 4 \\ 2 & 5 \end{matrix} \begin{vmatrix} 9 \\ -1 \end{bmatrix}$

Perform the following row operations to make the left two columns the identity matrix $\begin{bmatrix}1 & 0\\ 0 &1 \end{bmatrix}$ :

$R1 -R2 \rightarrow R1$

$\begin{bmatrix} 3-2 & 4-5 \\ 2 & 5 \end{matrix} \begin{vmatrix} 9-(-1) \\ -1 \end{bmatrix}$

$\begin{bmatrix} 1 & -1 \\ 2 & 5 \end{matrix} \begin{vmatrix} 10 \\ -1 \end{bmatrix}$

$R2 - 2R1 \rightarrow R2$

$\begin{bmatrix} 1 & -1 \\ 2 - 2(1) & 5-2(-1) \end{matrix} \begin{vmatrix} 10 \\ -1 -2(10)\end{bmatrix}$

$\begin{bmatrix} 1 & -1 \\ 0 & 7 \end{matrix} \begin{vmatrix} 10 \\ -21 \end{bmatrix}$

$R2 \div 7 \rightarrow R2$

$\begin{bmatrix} 1 & -1 \\ 0 \div 7 & 7\div 7 \end{matrix} \begin{vmatrix} 10 \\ -21 \div 7 \end{bmatrix}$

$\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{matrix} \begin{vmatrix} 10 \\ -3 \end{bmatrix}$

$R1 +R2 \rightarrow R1$

$\begin{bmatrix} 1 +0 & -1 +1\\ 0 & 1 \end{matrix} \begin{vmatrix} 10+ (-3) \\ -3 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{matrix} \begin{vmatrix} 7 \\ -3 \end{bmatrix}$

x= 7 and y = -3 .

Setting $x= \sqrt [3] {a}$ and solving for by cubing both sides:

$\sqrt [3] {a} = 7$

$(\sqrt [3] {a}) ^{3} =7^{3}$

a = 343

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Example Question #42 : Other Mathematical Relationships

Which of the following is a cube root of sixty-four?

Possible Answers:

$2 + 2 \sqrt{5}$

$2 + 2 i \sqrt{5}$

$2+ 2 \sqrt{3}$

$2+ 2 i \sqrt{3}$

None of these

Correct answer:

None of these

Explanation:

Let be a cube root of 64. The question is to find a solution of the equation

$x^{3} = 64$ .

One way to solve this is to subtract 64 from both sides:

$x^{3} - 64 = 64 - 64$

$x^{3} - 64 =0$

64 is a perfect cube, so, as the difference of cubes, the left expression can be factored:

$x^{3} - 4 ^{3} =0$

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

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

We can set both factors equal to zero and solve:

x - 4 = 0

x = 4

4 is a cube root of 64; however, this is not one of the choices.

Setting

$x^{2}+4x + 16 = 0$ ,

we can make use of the quadratic formula, setting a = 1, b= 4, c= 16 in the following:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

$x = \frac{-(4) \pm \sqrt{4^{2}-4(1)(16)}}{2(1)}$

$x = \frac{-4 \pm \sqrt{16-64}}{2}$

$x = \frac{-4 \pm \sqrt{-48 }}{2}$

$x = \frac{-4 \pm i \sqrt{48 }}{2}$

$x = \frac{-4 \pm i \sqrt{16} \cdot \sqrt{3}}{2}$

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

$x = -2 \pm 2 i \sqrt{3}$

$- 2 - 2 i \sqrt{3}$ and $-2 + 2 i \sqrt{3}$ are the other two cube roots of ; neither is a choice. Therefore, none of the four choices given are cube roots.

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SAT II Math I : Mathematical Relationships

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

All SAT II Math I Resources

Example Questions

Example Question #41 : Other Mathematical Relationships

Example Question #45 : Other Mathematical Relationships

Example Question #46 : Other Mathematical Relationships

Example Question #42 : Other Mathematical Relationships

All SAT II Math I Resources

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