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ISEE Upper Level Math : How to find the solution to an equation

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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 #51 : Algebraic Concepts

Solve the equation for :

x^2-14x+49=0

Possible Answers:

$x=7 \ or\ x=5$

x=7

x=6

$x=7 \ or\ x=8$

x=5

Correct answer:

x=7

Explanation:

If a, b, c are real numbers with $a\neq 0$ and if ax^2+bx+c=0 , then we have

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

The expression b^2-4ac is called the discriminant of the quadratic equation, and we say D=b^2-4ac . We can write

$x=\frac{-b\pm \sqrt{D}}{2a}$ .

In this problem we have $a=1,\ b=-14,\ and\ c=49.$

$D=b^2-4ac=(14)^2-4\times 1\times (49)=196-196=0$

$x=\frac{-b}{2a}=\frac{-(-14)}{2\times 1}=7$

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Example Question #52 : How To Find The Solution To An Equation

What is the value of in terms of and ?

$\frac{1}{3}x=3y-4z$

Possible Answers:

$y=\frac{2x+12z}{9}$

$y=\frac{x+12z}{12}$

$y=\frac{x+12z}{9}$

$y=\frac{x+12z}{6}$

$y=\frac{x+8z}{9}\textup{}$

Correct answer:

$y=\frac{x+12z}{9}$

Explanation:

$\frac{1}{3}x=3y-4z$

First multiply both sides by :

x=3(3y-4z)=9y-12z

Add 12z to both sides:

x+12z=9y

Now divide both sides by :

$y=\frac{x+12z}{9}$

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Example Question #52 : Algebraic Concepts

If x=2 is a possible answer for the following equation, give

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

Possible Answers:

a=4 or a=2

a=0 or a=4

a=0 or a=1

a=2 or a=-2

a=0 or a=2

Correct answer:

a=0 or a=4

Explanation:

Rewrite the absolute value equation as two separate equations, one positive and the other negative, then solve each equation separately:

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

-(x-a)=2 or $x-a=-2\Rightarrow 2-a=-2$

$\Rightarrow -a+2-2=-2-2$

$\Rightarrow -a=-4\Rightarrow a=4$

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Example Question #52 : Algebraic Concepts

Solve for :

$\left | 7t+1 \right|=15$

Possible Answers:

t=2 or t=-16

t=-2 or $t=-\frac{16}{7}$

t=2 or $t=-\frac{16}{7}$

t=2 or t=-7

t=2 or t=-2

Correct answer:

t=2 or $t=-\frac{16}{7}$

Explanation:

We should rewrite the equation as a compound statement and solve each part:

$\left | 7t+1 \right|=15\Rightarrow 7t+1=\pm 15$

7t+1=15 or 7t+1=-15

$7t+1=15\Rightarrow 7t+(1-1)=(15-1)\Rightarrow 7t=14\Rightarrow t=\frac{14}{7}=2$

$7t+1=-15\Rightarrow 7t+(1-1)=(-15-1)\Rightarrow 7t=-16\Rightarrow t=-\frac{16}{7}$

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Example Question #52 : Algebraic Concepts

Solve the equation for :

$\left | t \right |+2t=15$

Possible Answers:

t=-5 or t=15

t=5 or t=-15

t=-5 or t=-15

t=10 or t=15

t=5 or t=15

Correct answer:

t=5 or t=15

Explanation:

We need to rewrite the absolute value equation as two separate equations, then solve each equation separately:

$t+2t=15\Rightarrow 3t=15\Rightarrow t=\frac{15}{3}=5$

$-t+2t=15\Rightarrow t=15$

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Example Question #51 : Equations

Solve the equation for :

$\left |3t \right |-1=t+3$

Possible Answers:

t=2 or t=-1

t=1 or t=-1

t=-2 or t=1

t=-2 or t=-1

t=2 or t=1

Correct answer:

t=2 or t=-1

Explanation:

We should rewrite the absolute value equation as two separate equations, one positive and the other negative, then solve each equation separately:

$3t-1=t+3\Rightarrow 3t-1+1=t+3+1\Rightarrow 3t=t+4$

$\Rightarrow 3t-t=4\Rightarrow 2t=4\Rightarrow t=2$

$-3t-1=t+3\Rightarrow -3t-t=3+1\Rightarrow -4t=4\Rightarrow t=-1$

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Example Question #53 : Algebraic Concepts

Solve the absolute value equation for :

$\left |2t+\frac{3}{4} \right |=\frac{1}{}2$

Possible Answers:

$t=-\frac{1}{8}$ or $t=-\frac{5}{8}$

$t=\frac{1}{8}$ or $t=-\frac{5}{8}$

$t=\frac{1}{4}$ or $t=-\frac{5}{4}$

$t=\frac{1}{8}$ or $t=\frac{5}{8}$

$t=-\frac{1}{8}$ or $t=\frac{5}{8}$

Correct answer:

$t=-\frac{1}{8}$ or $t=-\frac{5}{8}$

Explanation:

We need to rewrite the absolute value equation as two separate equations, one positive and the other negative, then solve each equation separately:

$2t+\frac{3}{4}=\frac{1}{2}\Rightarrow 2t+\frac{3}{4}-\frac{3}{4}=\frac{1}{2}-\frac{3}{4}\Rightarrow 2t=\frac{1\times 2-3}{4}\Rightarrow 2t=-\frac{1}{4}$

$\Rightarrow t=-\frac{1}{8}$

$2t+\frac{3}{4}=-\frac{1}{2}\Rightarrow 2t+\frac{3}{4}-\frac{3}{4}=-\frac{1}{2}-\frac{3}{4}\Rightarrow 2t=\frac{-1\times 2-3}{4}\Rightarrow 2t=-\frac{5}{4}$

$\Rightarrow t=-\frac{5}{8}$

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Example Question #54 : How To Find The Solution To An Equation

Find :

$4\left | 2x-1 |-2=10$

Possible Answers:

x=2 or x=-1

x=-2 or x=-1

x=2 or x=1

x=-3 or x=-1

x=-2 or x=1

Correct answer:

x=2 or x=-1

Explanation:

First we simplify the equation:

$4\left | 2x-1 \right |-2=10\Rightarrow 4\left | 2x-1 \right |-2+2=10+2$

$\Rightarrow 4 \left |2x-1 \right |=12\Rightarrow \left |2x-1 \right |=\frac{12}{4}$

$\left |2x-1 \right |=3$

Now we need to rewrite the absolute value equation as two separate equations, one positive and the other negative, then solve each equation separately:

$2x-1=3\Rightarrow 2x-1+1=3+1\Rightarrow 2x=4\Rightarrow x=2$

$2x-1=-3\Rightarrow 2x-1+1=-3+1\Rightarrow 2x=-2\Rightarrow x=-1$

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Example Question #51 : How To Find The Solution To An Equation

Solve for :

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

Possible Answers:

x=2

x=-4

x=0 or x=-4

No solution

x=0

Correct answer:

No solution

Explanation:

An absolute value equation has no solution if the absolute value expression equals a negative number, because an absolute value can never be negative.

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Example Question #56 : How To Find The Solution To An Equation

Find in terms of :

$\left |3x+7 \right |=3t^2$

Possible Answers:

x=t^2+5 or $x=-t^2+\frac{1}{3}$

x=t^2-5 or $x=-t^2+\frac{1}{3}$

x=t^2-5 or $x=t^2+\frac{1}{3}$

x=t^2-5 or $x=-t^2-\frac{1}{3}$

x=t^2-5 or $x=t^2-\frac{1}{3}$

Correct answer:

x=t^2-5 or $x=-t^2+\frac{1}{3}$

Explanation:

First we simplify the equation:

Now we need to rewrite the absolute value equation as two separate equations, one positive and the other negative, then solve each equation separately:

$3x+7=3t^2-8\Rightarrow 3x+7-7=3t^2-8-7\Rightarrow 3x=3t^2-15\Rightarrow x=\frac{3t^2-15}{3}$

x=t^2-5

$3x+7=-3t^2+8\Rightarrow 3x+7-7=-3t^2+8-7\Rightarrow 3x=-3t^2+1$

$\Rightarrow x=\frac{-3t^2+1}{3}\Rightarrow x=-t^2+\frac{1}{3}$

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ISEE Upper Level Math : How to find the solution to an equation

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #51 : Algebraic Concepts

Example Question #52 : How To Find The Solution To An Equation

Example Question #52 : Algebraic Concepts

Example Question #52 : Algebraic Concepts

Example Question #52 : Algebraic Concepts

Example Question #51 : Equations

Example Question #53 : Algebraic Concepts

Example Question #54 : How To Find The Solution To An Equation

Example Question #51 : How To Find The Solution To An Equation

Example Question #56 : How To Find The Solution To An Equation

All ISEE Upper Level Math Resources

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