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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #51 : Solving Equations

Solve for :

$2 = \frac{ 2x - 4 }{ 3 }$

Possible Answers:

$\frac{ 1}{2}$

Correct answer:

Explanation:

In order to solve the equation, we have to isolate the variable. We do this by performing the same operation to either side of the equation.

$2 = \frac{ 2x - 4 }{ 3 }$

First multiply both sides by 3.

6 = 2x - 4 .

Next, add 4 to both sides.

10 = 2x

Finally, divide both sides by 2.

5 = x

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Example Question #2382 : Algebra Ii

Solve for :

$\frac{ 20 }{ x - 7 } = 4$

Possible Answers:

$1.\overline{81}$

6.75

220

Correct answer:

Explanation:

Before we start with inverse opperations, we need x to be out of the denominator:

$\frac{ 20 }{x - 7 } = 4$

First multiply both sides by x- 7.

20 = 4(x-7)

Next, divide both sides by 4.

5 = x - 7

Finally, add 7 to both sides.

12 =x

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Example Question #2383 : Algebra Ii

Solve the following equation for :

x^2-5x+6=0

Possible Answers:

x=2

x=-2 and x=-3

No real solutions

x=2 and x=3

Correct answer:

x=2 and x=3

Explanation:

Solve the following equation for x:

x^2-5x+6=0

Begin by factoring the given equation.

We want something of the following form:

$(x\pm a)(x\pm b)=0$

Where a and b multiply to get 6 and add to get .

A good place to start would be listing the multiples of 6:1,2,3,6

2 and 3 seem like a good place to start. Since we need negative 5 and positive 6, keeping 2 and 3 negative seem like a good bet.

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

Because this equation is set equal to zero, we can use the zero product property rule to set each half of our factored equation equal to zero and then solve.

x-2=0

x=2

x-3=0

x=3

So we have 2 answers: x=2 and x=3

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Example Question #2384 : Algebra Ii

Solve this equation: $4x+.25=8x-\frac{31}{4}$

Possible Answers:

x=.75

x=.25

x=1

x=2

x=4

Correct answer:

x=2

Explanation:

$4x+.25=8x-\frac{31}{4}$

Combine terms by subtracting:

$.25=4x-\frac{31}{4}$

Convert non terms to decimals or to fractions:

$\frac{1}{4}=4x-\frac{31}{4}$

We choose to convert to a fraction because it is easier.

Combine fractions by adding:

$\frac{1}{4}+\frac{31}{4}=4x$

$\frac{32}{4}=4x$

Simplify and solve for :

8=4x

x=2

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Example Question #2385 : Algebra Ii

Solve: $e^{x+2} = e^{-x}$

Possible Answers:

ln(-1)

2ln(2)

$\frac{1}{e}$

Correct answer:

Explanation:

Since the bases are alike in this problem, taking the log of both sides will eliminate the base terms.

Take the natural log of both sides.

$ln(e^{x+2}) = ln(e^{-x})$

The equation becomes:

x+2=-x

Solve for .

2x+2=0

2x=-2

x=-1

The answer is:

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Example Question #2386 : Algebra Ii

Solve the equation for

5(-2x-4)-3(4x+5)=97

Possible Answers:

132

Correct answer:

Explanation:

5(-2x-4)-3(4x+5)=97

-10x-20-12x-15=97

-22x-35=97

-22x=132

Divide both sides by

x = -6

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Example Question #191 : Equations

Solve the system of equations.

x-3y=-22

4x+2y=-4

Possible Answers:

(-4,6)

(-4,-6)

(4,-6)

(-4,0)

Correct answer:

(-4,6)

Explanation:

Use elimination, multiply the top equation by -4 so that you can eliminate the X's.

x-3y=-22

4x+2y=-4

__________________

-4x+12y=88

4x+2y=-4

Combine these two equations, and then you have;

14y=84

y=6

Plug in y into one of the original equations and solve for x.

x-3(6)=-22

x-18=-22

x=-4

Your solution is (-4,6) .

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Example Question #542 : Basic Single Variable Algebra

Find a solution to the following equation:

45-13x=84

Possible Answers:

x=-3

x=-6

x=3

x=13

Correct answer:

x=-3

Explanation:

Find a solution to the following equation:

45-13x=84

We can solve for x using basic algebra.

Begin by subtracting 45:

-13x=84-45

-13x=39

Now, divide by -13 to find the final answer:

$x=39\div-13=-3$

Making our answer:

x=-3

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Example Question #62 : Solving Equations

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

Possible Answers:

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

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

The answer is not present

x=2

x=-2

Correct answer:

x=-2

Explanation:

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

Isolate the term with x:

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

Simplify:

$\frac{2}{3}x=-\frac{4}{3}$

Isolate x entirely:

$\frac{3}{2}*\frac{2}{3}x=-\frac{4}{3}*\frac{3}{2}$

$\mathbf{x=-2}$

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Example Question #2389 : Algebra Ii

$x=\sqrt{8+2x}$

Possible Answers:

Cannot be solved

$x=2\hspace{1mm}and\hspace{1mm}x=-4$

$x=2\hspace{1mm}and\hspace{1mm}x=4$

$x=-2\hspace{1mm}and\hspace{1mm}x=4$

$x=-2\hspace{1mm}and\hspace{1mm}x=-4$

Correct answer:

$x=-2\hspace{1mm}and\hspace{1mm}x=4$

Explanation:

$x=\sqrt{8+2x}$

Square both sides to remove the radical:

x^2=8+2x

Move all terms to the left side and set equal to 0:

x^2-2x-8=0

Factor the quadratic and set the factors equal to 0:

(x+2)(x-4)=0

Solve the factors:

$(x+2)=0\rightarrow \mathbf{x=-2}$

$(x-4)=0\rightarrow \mathbf{x=4}$

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Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #51 : Solving Equations

Example Question #2382 : Algebra Ii

Example Question #2383 : Algebra Ii

Example Question #2384 : Algebra Ii

Example Question #2385 : Algebra Ii

Example Question #2386 : Algebra Ii

Example Question #191 : Equations

Example Question #542 : Basic Single Variable Algebra

Example Question #62 : Solving Equations

Example Question #2389 : Algebra Ii

All Algebra II Resources

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