Algebra II : Equations

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #381 : Basic Single Variable Algebra

Set up the equation:  Two times a number is equal to one less than the number.

Possible Answers:

\displaystyle 2x-1=1-x

\displaystyle 2(x-1)=x-1

\displaystyle 2x=1-x

\displaystyle 2x-1=x-1

\displaystyle 2x=x-1

Correct answer:

\displaystyle 2x=x-1

Explanation:

Start by breaking up the details.  Let \displaystyle x be the unknown number.

Two times a number:  \displaystyle 2x

Is equal to:  \displaystyle =

One less than the number:  \displaystyle x-1

Combine all the terms.

The answer is:  \displaystyle 2x=x-1

Example Question #41 : Equations

Set up the following equation:  Four times the quantity of two less than a number is eight times the number.

Possible Answers:

\displaystyle 4(x-2)=8(x-2)

\displaystyle 4(x-2)=8x-2

\displaystyle 4(x-2)=8x

\displaystyle 4(2-x)=8x

\displaystyle 4x-2 = 8x

Correct answer:

\displaystyle 4(x-2)=8x

Explanation:

To solve this question, break up the wording by parts.

Quantity of two less than a number:  \displaystyle (x-2)

Four times the quantity of two less than a number:  \displaystyle 4(x-2)

Is:  \displaystyle =

Eight times the number:  \displaystyle 8x

Combine the terms.

The answer is:  \displaystyle 4(x-2)=8x

Example Question #41 : Setting Up Equations

A photo company charges a \displaystyle \$25 set-up fee to print photos and then \displaystyle \$0.75 per photo printed. Translate this into an expression.

Possible Answers:

\displaystyle 25+.75

\displaystyle .75+25p

\displaystyle 25.75p

\displaystyle .75p+25

\displaystyle 1.25p

Correct answer:

\displaystyle .75p+25

Explanation:

To translate this into an expression, it's helpful to identify what the constant is in the problem. In other words, what is not changing no matter how many photos you print? In this case, it's the set-up fee \displaystyle (25). Then, we identify where our variable will be coming into play. We don't know how many photos will be printed, so let's call the number of photos \displaystyle p. We know that it costs \displaystyle .75 per photo printed so that becomes \displaystyle .75p. Add those together to get your final expression: \displaystyle .75p+25.

Example Question #41 : Setting Up Equations

Set up an equation in terms of a single variable that can be used to solve the problem below.

The area of a rectangular pool is \displaystyle 24 ft^{2}. The length of the pool is two feet less than twice the width. Find the dimensions of the pool.

Possible Answers:

\displaystyle 2W^{2}-2W=24

None of the above

\displaystyle 2W^{2}+2W=24

\displaystyle 2W-2W^{2}=24

\displaystyle 2W-2=24

Correct answer:

\displaystyle 2W^{2}-2W=24

Explanation:

The area of a rectangular is length times width. So,

If \displaystyle A=LW and \displaystyle A=24. Then,

\displaystyle LW=24

The length of the pool is two feet less than twice the width, which means:

\displaystyle L=2W-2

Substitute \displaystyle L in \displaystyle LW=24 with \displaystyle 2W-2:

\displaystyle (2W-2)W=24

so 

\displaystyle 2W^{2}-2W=24

Example Question #42 : Setting Up Equations

Write the following equation:

Shirts are 15 dollars and jeans are 30 dollars.

The total amount of money you made from selling shirts and jeans is at least 200 dollars.

Possible Answers:

\displaystyle \small 15s+30j\geq 200

\displaystyle \small 15s+30j=200

\displaystyle \small 15s+30j\leq 200

\displaystyle \small 15s+30j< 200

\displaystyle \small 15s+30j> 200

Correct answer:

\displaystyle \small 15s+30j\geq 200

Explanation:

You know that shirts sell for 15 dollars and jeans sell for 30 dollars.

The tricky part is knowing that you sold at least 200 dollars worth of merchandise.

This means that you could have sold more than 200 dollars, but no less than that.

This means that your answer is

\displaystyle \small 15s+30j\geq 200

Example Question #41 : Equations

Corey buys potatoes for $2.25 per pound and quinoa for $5.75 per pound. If Corey buys p pounds of potatoes and q pounds of quinoa for $37.25, which of the equations below represents his transaction?

Possible Answers:

\displaystyle 37.25+2.25p=5.75q

\displaystyle 8.00(p+q)=37.25

\displaystyle 5.75q-2.25p=37.25

\displaystyle 2.25p+5.75q=37.25

\displaystyle 5.75p+2.25q=37.25

Correct answer:

\displaystyle 2.25p+5.75q=37.25

Explanation:

The cost for p pounds of potatoes is 2.25p and the cost for q pounds of quinoa is 5.75q. The total amount Corey pays is the sum of the cost for potatoes and the cost for quinoa.

2.25p + 5.75q = 37.25

Example Question #43 : Setting Up Equations

Write in slope-intercept form

\displaystyle y+2 = \frac{1}{4}( x-4)

Possible Answers:

\displaystyle y = \frac{x}{3}-4

\displaystyle y+2=\frac{1}{4}x-2

\displaystyle y= \frac{4}{x}-2

\displaystyle y= \frac{-x}{4}+3

\displaystyle y = \frac{x}{4}-3

Correct answer:

\displaystyle y = \frac{x}{4}-3

Explanation:

Recall the form for slope-intercept as follows.

\displaystyle y=mx+b

where \displaystyle m represents the slope and \displaystyle b is the y-intercept.

Given the equation in this question, first distribute the one fourth to both terms inside the parentheses.

\displaystyle y+2 = \frac{1}{4}(x-4) = \frac{x}{4}-1    

Next, subtract two from both sides to isolate y.

\displaystyle y+2=\frac{x}{4}-1

     \displaystyle {\color{Red} -2}            \displaystyle {\color{Red} -2}

This results in the slope-intercept form of the equation.

\displaystyle y=\frac{x}{4}-3

       

Example Question #42 : Setting Up Equations

Beetle A runs in a straight line for 30cm before bumping into Beetle B, who then runs for another 90cm at a rate 3 times faster than Beetle A.

What is the rate of Beetle B?

Possible Answers:

\displaystyle 1.5\frac{cm}{sec}

\displaystyle 3\frac{cm}{sec}

Unable to determine from the information provided

\displaystyle 4.5\frac{cm}{sec}

\displaystyle 2\frac{cm}{sec}

Correct answer:

\displaystyle 1.5\frac{cm}{sec}

Explanation:

Knowing:

  • \displaystyle distance=rate*time
  • \displaystyle x=rate_{BeetleA}
  • \displaystyle time_A+time_B=time_{total}

Beetle A: 

\displaystyle time_A=\frac{distance}{rate}=\frac{30cm}{x}

Beetle B: 

\displaystyle time_B=\frac{distance}{rate}=\frac{90cm}{3x}

Combined:

\displaystyle time_A+time_B=time_{total}

\displaystyle \frac{30cm}{x}+\frac{90cm}{3x}=40sec

Multiply each term by LCD (3x) to get:

\displaystyle 90cm+90cm=120sec*x

\displaystyle \frac{180cm}{120}=x

\displaystyle x=\frac{3cm}{2sec}=1.5\frac{cm}{sec}

 

 

Example Question #42 : Setting Up Equations

Set up the equation:  The square root of six less than twice a number is equal to nine.

Possible Answers:

\displaystyle \sqrt{2x}-6=9

\displaystyle \sqrt{2x-6}=9

\displaystyle 2\sqrt{x}-6=9

\displaystyle 2\sqrt{x-6}=9

\displaystyle \sqrt{2(x-6)}=9

Correct answer:

\displaystyle \sqrt{2x-6}=9

Explanation:

Split up the sentence into parts.

Twice a number:  \displaystyle 2x

Six less than twice a number:  \displaystyle 2x-6

The square root of six less than twice a number:  \displaystyle \sqrt{2x-6}

Is equal to nine:  \displaystyle \sqrt{2x-6}=9

The answer is:  \displaystyle \sqrt{2x-6}=9

Example Question #171 : Algebraic Functions

Set up the equation:  The sum of two times a number and forty is equal to sixteen.

Possible Answers:

\displaystyle 2x+20=16

\displaystyle (x+40)^2=16

\displaystyle 2x+40=16

\displaystyle 2(x+40)=16

\displaystyle x^2+40=16

Correct answer:

\displaystyle 2x+40=16

Explanation:

Break up the problem into parts.

Two times a number:  \displaystyle 2x

 The sum of two times a number and forty:  \displaystyle 2x+40

Is equal to sixteen:  \displaystyle 16

Combine the terms to form the equation.

The answer is:  \displaystyle 2x+40=16

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