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ISEE Upper Level Math : Equations

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

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All ISEE Upper Level Math Resources

6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Algebraic Concepts

Solve for x:

$\dpi{100} 3x^{2}+4=31$

Possible Answers:

$\dpi{100} \pm 3$

$\dpi{100} \pm 1$

$\dpi{100} \pm 9$

$\dpi{100} 9$

Correct answer:

$\dpi{100} \pm 3$

Explanation:

First, subtract 4 from both sides:

$\dpi{100} 3x^{2}+4-4=31-4$

$\dpi{100} 3x^{2}=27$

Next, divide both sides by 3:

$\dpi{100} \frac{3x^{2}}{3}=\frac{27}{3}$

$\dpi{100} x^{2}=9$

Now take the square root of both sides:

$\dpi{100} \sqrt{x^{2}}=\sqrt{9}$

$\dpi{100} x=\pm 3$

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Example Question #2 : Algebraic Concepts

Solve for $\dpi{100} x$ :

$\dpi{100} \frac{x}{3}+22 = 45$

Possible Answers:

$\dpi{100} 3$

$\dpi{100} 69$

$\dpi{100} 66$

$\dpi{100} 135$

Correct answer:

$\dpi{100} 69$

Explanation:

$\dpi{100} \frac{x}{3}+22 = 45$

$\dpi{100} \frac{x}{3}+22-22 = 45-22$

$\dpi{100} \frac{x}{3} = 23$

$\dpi{100} (3)\cdot \frac{x}{3} = 23\cdot (3)$

$\dpi{100} x=69$

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

Solve for :

| 3x + 9| = 24

Possible Answers:

$x = -5\textrm{ or }x = 5$

$x = -11 \textrm{ or }x = 5$

$x = -5 \textrm{ or }x = 11$

$x = 5 \textrm{ or }x = 11$

$x = -11 \textrm{ or }x = 11$

Correct answer:

$x = -11 \textrm{ or }x = 5$

Explanation:

Rewrite this as a compound statement and solve each separately:

| 3x + 9| = 24

$3x + 9 = - 24 \textrm{ or }3x + 9 = 24$

3x + 9 = - 24

3x + 9 -9= - 24-9

3x = - 33

$3x\div 3 = - 33 \div 3$

x = -11

3x + 9 = 24

3x + 9 -9= 24-9

3x = 15

$3x\div 3 = 15 \div 3$

x = 5

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Example Question #4 : Algebraic Concepts

Solve for :

$\left (x -4 \right )\left( x +3 \right )= \left (x-2 \right) \left ( x-1\right )$

Possible Answers:

$x = 1\textrm{ or }x = 4$

$x = 3\textrm{ or }x = 4$

$x = 1\textrm{ or }x = 7$

x = 7

x = 12

Correct answer:

x = 7

Explanation:

FOIL each of the two expressions, then solve:

$\left (x -4 \right )\left( x +3 \right )= \left (x-2 \right) \left ( x-1\right )$

$x ^{2} + 3 \cdot x -4\cdot x -4 \cdot 3 = x^{2} -1 \cdot x - 2 \cdot x + 2\cdot1$

$x ^{2} + 3 x -4 x -12 = x^{2} -1 x - 2x + 2$

$x ^{2} - x -12 = x^{2} - 3x + 2$

$x ^{2} - x -12 - x^{2} = x^{2} - 3x + 2 - x^{2}$

- x -12= - 3x + 2

Solve the resulting linear equation:

- x -12 +3x + 12 = - 3x + 2+3x+12

2x =14

$2x\div 2 =14 \div 2$

x = 7

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Example Question #5 : Equations

Solve for :

$x^{2} -10x = 24$

Possible Answers:

$x=-12 \textrm{ or }x=2$

$x=-2 \textrm{ or }x=12$

$x=-6\textrm{ or }x=-4$

$x=-4\textrm{ or }x=6$

$x=-6\textrm{ or }x=4$

Correct answer:

$x=-2 \textrm{ or }x=12$

Explanation:

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

$x^{2} -10x = 24$

$x^{2} -10x -24= 24-24$

$x^{2} -10x -24= 0$

Now factor the quadratic expression $x^{2} -10x -24$ into two binomial factors (x + ?)(x + ?) , replacing the question marks with two integers whose product is and whose sum is . These numbers are -12,2 , so:

(x -12)(x + 2) = 0

$x-12 = 0 \Rightarrow x = 12$

$x+2 = 0 \Rightarrow x = -2$

The solution set is $\left \{ -2,12\right \}$ .

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Example Question #5 : Algebraic Concepts

Solve for :

$\left ( x-3\right )(2x-1) = 18$

Possible Answers:

$x= - 5\textrm{ or } x= 1 \frac{1}{2}$

$x= -3 \textrm{ or } x= 5$

$x= -5 \textrm{ or } x= 3$

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

$x = 9\frac{1}{2} \textrm{ or }x=19$

Correct answer:

$x= - 1 \frac{1}{2} \textrm{ or } x= 5$

Explanation:

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then collecting all of the terms on the left side:

$\left ( x-3\right )(2x-1) = 18$

$x\cdot 2x -x\cdot 1 -3 \cdot 2x +3\cdot 1= 18$

$2x^{2} -x -6x +3= 18$

$2x^{2} -7x +3= 18$

$2x^{2} -7x +3-18= 18 -18$

$2x^{2} -7x -15= 0$

Use the method to factor the quadratic expression $2x^{2} -7x -15$ ; we are looking to split the linear term by finding two integers whose sum is and whose product is 2 (-15) = -30 . These integers are -10,3 , so:

$(2x^{2} -10x)+(3x -15)= 0$

2x (x -5)+3(x -5)= 0

$\left (2x +3 \right) \left ( x-5 \right )= 0$

Set each expression equal to 0 and solve:

2x +3= 0

2x = -3

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

x-5= 0

x=5

The solution set is $\left \{ - 1 \frac{1}{2}, 5\right \}$ .

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Example Question #7 : Equations

Solve for :

$x^{2} + 12 = 8x$

Give all solutions.

Possible Answers:

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

$x = 6 \textrm{ or }x = 8$

$x = -6 \textrm{ or }x =- 2$

$x = 2 \textrm{ or }x = 6$

$x = -6 \textrm{ or }x =6$

Correct answer:

$x = 2 \textrm{ or }x = 6$

Explanation:

Rewrite this quadratic equation in standard form:

$x^{2} + 12 = 8x$

$x^{2} -8x + 12 = 8x-8x$

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

Factor the expression on the left. We want two integers whose sum is and whose product is . These numbers are -6,-2 , so the equation becomes

(x-6)(x-2)=0 .

Set each factor equal to 0 separately, then solve:

$x - 6 = 0 \Rightarrow x = 6$

$x - 2 = 0 \Rightarrow x = 2$

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Example Question #6 : Algebraic Concepts

Solve for :

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

Possible Answers:

x =4

The equation has no solution.

The equation has the set of all real numbers as its solution set.

x = 2

x = 0

Correct answer:

x = 0

Explanation:

Simplify both sides, then solve:

4 (x + 6) - 3 (x + 8) = 2 (x + 4) + 2 (x - 4)

$4 \cdot x + 4 \cdot 6 - 3 \cdot x - 3 \cdot 8 = 2 \cdot x + 2 \cdot 4 +2 \cdot x - 2 \cdot 4$

4x + 24 - 3 x - 24 = 2x +8 +2 x -8

4x - 3 x+ 24 - 24 = 2x +2 x +8-8

$\left ( 4 - 3 \right ) x= \left ( 2 +2 \right ) x$

x= 4 x

x - 4x = 4 x - 4x

$\left (1 - 4 \right ) x = 0$

-3 x = 0

x = 0

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Example Question #2 : How To Find The Solution To An Equation

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

Possible Answers:

The equation has the set of all real numbers as its solution set.

x = -2

x = 6

The equation has no solution.

x = 0

Correct answer:

The equation has the set of all real numbers as its solution set.

Explanation:

Simplify both sides, then solve:

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

$3 \cdot x + 3 \cdot 2 + 4x = 5x + 2 \cdot x + 2 \cdot 3$

3x + 6 + 4x = 5x + 2x + 6

3x + 4x + 6 = 5x + 2x + 6

$\left ( 3 + 4 \right ) x + 6 = \left ( 5 + 2 \right ) x + 6$

7x + 6 =7 x + 6

This is an identically true statement, so the original equation has the set of all real numbers as its solution set.

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Example Question #8 : Algebraic Concepts

Solve for :

W = 5xp + 3q

Possible Answers:

$p = \frac{W-3q}{5x}$

$p = \frac{W- 5x }{3q}$

$p = \frac{W}{5x}-3q$

$p = W-\frac{3q}{5x}$

$p = \frac{W}{3q }-5x$

Correct answer:

$p = \frac{W-3q}{5x}$

Explanation:

5xp + 3q = W

5xp + 3q -3q = W-3q

5xp = W-3q

$5xp \div 5x =\left ( W-3q \right ) \div 5x$

$p =\frac{ W-3q }{5x}$

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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 #1 : Algebraic Concepts

Example Question #2 : Algebraic Concepts

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

Example Question #4 : Algebraic Concepts

Example Question #5 : Equations

Example Question #5 : Algebraic Concepts

Example Question #7 : Equations

Example Question #6 : Algebraic Concepts

Example Question #2 : How To Find The Solution To An Equation

Example Question #8 : Algebraic Concepts

All ISEE Upper Level Math Resources

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