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ISEE Upper Level Math : Equations

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Example Questions

Example Question #31 : How To Find The Solution To An Equation

Solve the system of equations:

3x+y=10

x+3y=-2

Possible Answers:

x=4;y=2

x=-4;y=2

x=4;y=-2

x=-4;y=-2

Correct answer:

x=4;y=-2

Explanation:

Solve for in the first equation:

y=10-3x

Substitute this expression into the first equation:

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

30-8x=-2

-8x=-32

x=4

Substitute this value into the first equation:

3(4)+y=10

12+y=10

y=-2

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Example Question #32 : How To Find The Solution To An Equation

Solve for :

$x -6 \sqrt {x}+9 = 0$

Possible Answers:

x = -9

The equation has no solution.

$x = -9\textrm{ or }x = 9$

$x =0\textrm{ or }x = 9$

x = 9

Correct answer:

x = 9

Explanation:

Substitute $t = \sqrt {x}$ and, subsequently, $t^{2} = x$ :

$x -6 \sqrt {x}+9 = 0$

$t^{2}-6t+9 = 0$

Factor as (t + ?) (t + ?) , replacing the two question marks with integers whose product is and whose sum is . These integers are -3,-3 :

(t-3)(t-3)= 0

Since this is a perfect square, we can rewrite this simply as:

t - 3 = 0

t = 3

Replace $\sqrt{x}$ for :

$\sqrt{x} = 3$

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

x = 9 , which is the only solution.

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Example Question #33 : How To Find The Solution To An Equation

Give the -intercept of the line of the equation 8x + 9y = 720 .

Possible Answers:

(0, 9)

(0, 90)

(0, 8)

(0, 80)

(0, 72)

Correct answer:

(0, 80)

Explanation:

Substitute x = 0 :

8x + 9y = 720

$8 \cdot 0 + 9y = 720$

0 + 9y = 720

9y = 720

$9y\div 9 = 720 \div 9$

y = 80

The -intercept is (0,80) .

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Example Question #34 : How To Find The Solution To An Equation

Which of the following is the solution set of this equation?

$\left | 2t - \frac{1}{4} \right | = \frac{5}{8}$

Possible Answers:

$t = -1.75\textrm{ or }t = 1.75$

$t = 0.1875 \textrm{ or }t = 0.4375$

$t = -0.75\textrm{ or }t = 1.75$

$t = -0.1875 \textrm{ or }t = 0.4375$

$t = -0.4375\textrm{ or }t = 0.4375$

Correct answer:

$t = -0.1875 \textrm{ or }t = 0.4375$

Explanation:

Rewrite the fractions in decimal form:

$\left | 2t - \frac{1}{4} \right | = \frac{5}{8}$

$\left | 2t - 0.25 \right | = 0.625$

Now rewrite this as a compound statement and solve each part:

$2t - 0.25=- 0.625 \textrm{ or }2t - 0.25= 0.625$

2t - 0.25=- 0.625

2t - 0.25+0.25=- 0.625+0.25

2t = -0.375

$2t \div 2 = -0.375\div 2$

t = -0.1875

2t - 0.25= 0.625

2t - 0.25+ 0.25= 0.625+ 0.25

2t = 0.875

$2t \div 2 = 0.875 \div 2$

t = 0.4375

The solution set is $\left \{ -0.1875, 0.4375 \right \}$ .

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Example Question #35 : How To Find The Solution To An Equation

Solve for :

$x + 8\sqrt{x} + 15 = 0$

Possible Answers:

$x = -25\textrm{ or }x = 25$

$x = -25\textrm{ or }x = -9$

x = -25

x = -9

The equation has no solution

Correct answer:

The equation has no solution

Explanation:

This can be quickly answered by making an observation.

$x + 8\sqrt{x} + 15 = 0$

$x + 8\sqrt{x} + 15 - 15 = 0- 15$

$x + 8\sqrt{x} = - 15$

Since appears as a radicand, it must hold that $x \geq 0$ . This means that $x + 8\sqrt{x}$ must be a nonnegative number, and that, subsequently, $x + 8\sqrt{x} = - 15$ has no solution.

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Example Question #32 : Equations

Solve for , giving all real solutions:

$x^{4} - 6x^{2} -16 = 0$

Possible Answers:

The equation has no real solution.

$x = -\sqrt{8} \textrm{ or }x = -\sqrt{2} \textrm{ or }x = \sqrt{2}\textrm{ or }x = \sqrt{8}$

$x = \sqrt{2} \textrm{ or }x = \sqrt{8}$

$x = -\sqrt{2} \textrm{ or }x = \sqrt{2}$

$x = -\sqrt{8} \textrm{ or }x = \sqrt{8}$

Correct answer:

$x = -\sqrt{8} \textrm{ or }x = \sqrt{8}$

Explanation:

Substitute $t = x^{2}$ and, subsequently, $t ^{2} =\left ( x^{2} \right )^{2} = x ^{4}$ .

$x^{4} - 6x^{2} -16 = 0$

$t^{2} - 6t-16 = 0$

Factor as (t + ?) (t + ?) , replacing the two question marks with integers whose product is and whose sum is . These integers are -8,2 , so

(t -8) (t +2) = 0 .

Break this up into two equations, replacing $x^{2}$ for in each:

t -8 = 0

t = 8

$x ^{2}= 8$

$x = \pm \sqrt{8}$

t+2= 0

t = -2

$x ^{2}= -2$ ,

which has no real solution.

The solution set is $x = \pm \sqrt{8}$ .

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Example Question #36 : How To Find The Solution To An Equation

Give the -intercept of the line of the equation: $\frac{}{}$ $\frac{2}{5}x + \frac{4}{7}y = 8$

Possible Answers:

$\left (3 \frac{1}{5} ,0\right )$

(20,0)

$\left (4 \frac{4}{7} ,0\right )$

(14,0)

The line has no -intercept.

Correct answer:

(20,0)

Explanation:

Substitute y = 0 :

$\frac{2}{5}x + \frac{4}{7}y = 8$

$\frac{2}{5}x + \frac{4}{7} \cdot 0 = 8$

$\frac{2}{5}x + 0 = 8$

$\frac{2}{5}x = 8$

$\frac{2}{5}x \div\frac{2}{5} = 8\div\frac{2}{5}$

$x = \frac{8}{1}\cdot\frac{5}{2} = \frac{4}{1}\cdot\frac{5}{1} = 20$

The -intercept is (20,0) .

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Example Question #37 : How To Find The Solution To An Equation

Solve for :

$\left | 5 - \frac{2}{3} t \right | = 15$

Possible Answers:

$t = -6 \frac{2}{3} \textrm{ or }t = 13 \frac{1}{3}$

$t = -15 \textrm{ or }t = 30$

$t = -30 \textrm{ or }t = 30$

$t = -15 \textrm{ or }t = 15$

$t = -6 \frac{2}{3} \textrm{ or }t = 6 \frac{2}{3}$

Correct answer:

$t = -15 \textrm{ or }t = 30$

Explanation:

$\left | 5 - \frac{2}{3} t \right | = 15$ can be rewritten as a compound equation:

$5-\frac{2}{3} t = -15 \textrm{ or } 5-\frac{2}{3} t = 15$

Solve them separately:

$5 - \frac{2}{3} t = 15$

$5 - \frac{2}{3} t -5 = 15 -5$

$- \frac{2}{3} t = 10$

$- \frac{2}{3} t \cdot\left ( - \frac{3} {2} \right )= 10 \cdot\left ( - \frac{3} {2} \right )$

$t = - \frac{30} {2} = -15$

$5-\frac{2}{3} t = -15$

$5 - \frac{2}{3} t -5 = -15 -5$

$- \frac{2}{3} t = -20$

$- \frac{2}{3} t\cdot\left ( - \frac{3} {2} \right ) = -20\cdot\left ( - \frac{3} {2} \right )$

$t =\frac{60} {2} = 30$

The solution set is $\left \{ -15, 30\right \}$ .

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Example Question #38 : How To Find The Solution To An Equation

Give the -intercept of the line of the equation 8x + 9y = 720 .

Possible Answers:

(8, 0)

(72, 0)

(90, 0)

(80,0)

(9, 0)

Correct answer:

(90, 0)

Explanation:

Substitute y = 0 :

8x + 9y = 720

$8x + 9 \cdot 0 = 720$

8x +0 = 720

8x = 720

$8x \div 8= 720 \div 8$

x = 90

The -intercept is (90, 0) .

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Example Question #741 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Solve for :

$x - 4 \sqrt {x} - 12 = 0$

Possible Answers:

x = -4

$x = -4\textrm{ or }x = 4$

x = 36

$x = -4\textrm{ or }x = 36$

$x = 4\textrm{ or }x = 36$

Correct answer:

x = 36

Explanation:

Substitute $t = \sqrt {x}$ and, subsequently, $t^{2} = x$ :

$x - 4 \sqrt {x} - 12 = 0$

$t^{2} - 4 t - 12 = 0$

Factor as (t + ?) (t + ?) , replacing the two question marks with integers whose product is and whose sum is . These integers are -6, 2 :

(t - 6) (t + 2) = 0

Break this up into two equations, replacing $\sqrt {x}$ for in each:

t - 6 = 0

t = 6

$\sqrt {x} = 6$

$(\sqrt {x})^{2} = 6^{2}$

x = 36

t + 2 = 0

t = -2

$\sqrt {x} = -2$

This has no solution, since $\sqrt {x}$ must be nonnegative.

x = 36 is the only solution.

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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 #31 : How To Find The Solution To An Equation

Example Question #32 : How To Find The Solution To An Equation

Example Question #33 : How To Find The Solution To An Equation

Example Question #34 : How To Find The Solution To An Equation

Example Question #35 : How To Find The Solution To An Equation

Example Question #32 : Equations

Example Question #36 : How To Find The Solution To An Equation

Example Question #37 : How To Find The Solution To An Equation

Example Question #38 : How To Find The Solution To An Equation

Example Question #741 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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