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Extraneous solutions of an equation are EXTRA solutions that work in subsequent equations that are obtained in trying to solve an equation, but that do NOT work in the original equation!! These extraneous solutions are caused by squaring both sides of an equation or by multiplying both sides of an equation by an expression containing a variable.

For example, if you took the simple equation x=3, and square both sides, you would get x^2=9, which has two solutions, x=3 and x=-3. The second solution x=-3 is an extraeous solution in that it does NOT solve the original equation.

I have a TON of extraneous solutions on my website, which you can find by clicking on my tutor name "rapaljer" anywhere in algebra.com. Then look for my MATH IN LIVING COLOR pages, and choose Intermediate Algebra. Look for Section 2.07 on Fractional Equations, the last two problems that are posted there.

Also look for Section 3.07 Radical Equations! This one is LOADED with extraneous solutions!! I hope this helps!!

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Let x = distance from the wall
x+3 = length of the ladder

By Theorem of Pythagoras:



Subtract x^2 from each side:
81=6x+9
72 = 6x
x= 12 feet = distance from wall
x+3 = 15 feet = length of the ladder

Check:



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Let x = side of the square.
= area of the square
= perimeter of the square




or

The only answer that is greater than 1 is x= 3.

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Subtract 2 from each side:
2e^5a= 6

Divide by 2:
e^5a = 3

Take the ln of each side:
ln(e^5a) = ln 3
5a = ln 3

Divide by 5:

a = 0.2197 approximately

If you need additional help with logarithms and exponential functions, please see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look down the homepage for a topic called "Logarithms." I have an entire chapter on Logarithms in a format that my students UNANIMOUSLY tell me it easier to understand than the traditional textbooks. In addition to this, please see my MATH IN LIVING COLOR PAGES (College Algebra, Chapter 4). Let me know if this helps you!!!

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Since both of these equations are in slope-intercept form, I recommend the slope-intercept method of graphing them. If you need additional explanations on this, I have a good Lesson PLan on this method here in algebra.com, or go to my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look on my homepage for "Basic Intermediate and College Algebra: One Step at a Time." Select "Basic Algebra" and look for Chapter 5 on Graphing.

Now to solve the prlblem. The first graph y=3x is actually y=3x+ 0, so it has a y-intercept of 0 and a slope of 3. Plot the point (0,0). From this point move UP 3 units and RIGHT 1 unit and plot another point. Connect the two points with the first line.

FOr the second graph y=-1x+4, the y intercept is 4, so go UP 4 units on the y axis and plot the first point. Then, with your pencil on this point, move DOWN 1 unit and RIGHT 1 unit, and plot the second point. Connect the dots, and you have the second line. It should look like this:



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Let x = Monday's food
x-4 = Tuesday's food
x+3 = Wednesday's food
_____ = Thursday's food


x + x-4 + x+3 + _______ = 87

You apparently forgot to tell us how much the otter ate on Thursday. We can't solve it unless we know that!!

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Divide both sides of the inequality by -9. Remember that very important rule, that when you divide or multiply both sides of an inequality by a NEGATIVE number, you must REVERSE the inequality sign!





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One solution means you have two lines that intersect in a single point.
No solution means that the two lines are parallel.
Infinitely many solutions means that the two lines are actually the same line.

Is this a take home test???

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She is NOT correct to divide her answer by 10.

When she multiplied both sides of the equation by the same number, this retains an EQUIVALENT equation. This is an equation that has the SAME solution as the original equation. So the solution that she gets for the new equation is actually the SAME solution as the solution for the original equation. She should NOT divide the answers by 10!!

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What problemm???

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43-(-12)= 43 + 12 = 55

-12+(-5)-3-(-8)
-12-5-3+8
-17-3+8
-20+8
-12





That's enough for one question!! Post the rest in another question.

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Do you mean the opposite of 5x^2-3x+7? By that you probably mean
-(5x^2 -3x + 7). It may help to think of this as
-1*(5x^2-3x+7), which would be according to the Distributive Property
-1*5x^2 + -1*(-3x) + -1*7
-5x^2 + 3x -7

If you ADD the original number plus its opposite, then you get zero. I think the problem was to find the opposite.

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2x+5y=7
3x-2y=1

To make the y coefficients subtract out, multiply both sides of the first equation times 2, and the second equation times 5:

2(2x+5y=7)
5(3x-2y=1)

4x+10y=14
15x-10y=5

Add these equations together:
19x=19
x=1

Substitute x=1 into the first equation:
2x+5y=7
2*1+5y=7
2+5y=7
5y=5
y=1

Substitute both values back into the second equation to check:
3x-2y=1
3*1-2*1=1
3-2=1 It checks!!

For more detailed explanations, examples, and exercises, including solutions in LIVING COLOR on my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look for the pages on my home page, either "Basic, Intermediate, and College Algebra" or "Math in Living Color". In either of these pages, select "Basic Algebra" and then "Chapter 4." Look for the appropriate sections, for an easier to understand explanation than traditional textbooks!!

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2x+5y=7
3x-2y=1

To make the y coefficients subtract out, multiply both sides of the first equation times 2, and the second equation times 5:

2(2x+5y=7)
5(3x-2y=1)

4x+10y=14
15x-10y=5

Add these equations together:
19x=19
x=1

Substitute x=1 into the first equation:
2x+5y=7
2*1+5y=7
2+5y=7
5y=5
y=1

Substitute both values back into the second equation to check:
3x-2y=1
3*1-2*1=1
3-2=1 It checks!!

For more detailed explanations, examples, and exercises, including solutions in LIVING COLOR on my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look for the pages on my home page, either "Basic, Intermediate, and College Algebra" or "Math in Living Color". In either of these pages, select "Basic Algebra" and then "Chapter 4." Look for the appropriate sections, for an easier to understand explanation than traditional textbooks!!

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2x+5y=7
3x-2y=1

To make the y coefficients subtract out, multiply both sides of the first equation times 2, and the second equation times 5:

2(2x+5y=7)
5(3x-2y=1)

4x+10y=14
15x-10y=5

Add these equations together:
19x=19
x=1

Substitute x=1 into the first equation:
2x+5y=7
2*1+5y=7
2+5y=7
5y=5
y=1

Substitute both values back into the second equation to check:
3x-2y=1
3*1-2*1=1
3-2=1 It checks!!

For more detailed explanations, examples, and exercises, including solutions in LIVING COLOR on my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look for the pages on my home page, either "Basic, Intermediate, and College Algebra" or "Math in Living Color". In either of these pages, select "Basic Algebra" and then "Chapter 4." Look for the appropriate sections, for an easier to understand explanation than traditional textbooks!!

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y=4x+2

There are two steps:
Step 1: Interchange the x and y variables!
x= 4y+2

Step 2: Solve for y!
x-2=4y

Divide both sides by 4:



or

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Just add these two equations together:
3x+y=1
2x-y=-6

5x=-5
x=-1

Substitute back into the first equation:
3x+y=1
3(-1)+Y=1
-3+y=1
y=4

Check in the second equation:
2x-y=-6
2(-1)-(4)=-6
-2-4=-6 It checks!!

(-1, 4)


For an easier explanation of this topic, please see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. On my homepage, look for "Basic, Intermediate, and College Algebra: One Step at a Time." Select "Basic Algebra" and look for the appropriate sections in Chapter 4 of my Basic Algebra book. It's all FREE, just like algebra.com!!

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(a^2b^2)^3

If a=1 and b=2,
(1^2*2^2)^3
(1*4)^3
4^3
64

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Let r(x)=2x and s(x)=x^3-3.
Find s(r(2)).

r(2) = 2*2=4

s(r(2))
=s(4)
=4^3-3
=64-3
=61

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You need to state the problem!! We don't have your book!!

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There may be several ways to simplify this, but the easiest way is to divide 3 into the 6 and 7 into 14:






If you need additional help with square roots, see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. On my homepage, look for the link "Basic, Intermediate, and College Algebra: One Step at a Time." Select "Basic Algebra" and look in Chapter 5 for topics on Square Roots. In addition to full explanations, examples, and exercises written so ordinary students can understand them, many of the exercises are solved in my MATH IN LIVING COLOR pages.

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The first step is to take out all the common factors. In this case, you have common factors of 2 and x.

2x(4x^2 - 3x -1)

Next, see if the trinomial factors (and it does!):
2x(4x +1)(x-1)

If you need additional explanation and help with factoring common factors OR factoring trinomials, check out my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Look for the link on my main page that says, "Basic, Intermediate, and College Algebra: One Step at a Time." Then select "Basic Algebra" and look in Chapter 2 for the factoring sections. There you will find detailed explanations, examples, and exercises written in a format that is easier to understand than the traditional textbooks.

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How about a fraction? Maybe a ratio of two integers? Maybe a rational number!!

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Do you mean ???

If so, then the trick is to multiply numerator and denominator times the same thing as the denominator but with the opposite sign in the middle. This is called the "conjugate" of the denominator. In other words multiply numerator and denominator times . It looks like this:


When you multiply out the denominator you get
.
.

If you need extra explanation on SQUARE ROOTS, go to my website by clicking on my tutor name "Rapaljer" anywhere in algebra.com. On my Homepage, look for about the third link that says "Basic, Intermediate, and College Algebra: One Step at a Time." Click on this, then click on "Basic Algebra." Go to "Chapter 5 (Square Roots)". The explanations, examples, and exercises are integrated and easier to understand than traditional textbooks. Answers to all problems are provided, and many of the exercises are solved in the "MATH IN LIVING COLOR" pages. Check it out! Like Algebra.com, it's all free!!

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We don't have your book, so you have to state the problem!!

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Please see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com. Take about the third link on my homepage which is the one that says "Basic, Intermediate, and College Algebra: One Step at a Time." Select "Basic Algebra", then go to "Chapter 2" and look for the sections dealing with Factoring. The first lesson MUST be Factoring the Common Factor. Then after you have done that, do the section called "Factoring Trinomials." I have tried to write very "user-friendly" explanations integrated with examples and exercises in a workbook format. Also look for my pages that are called "Math in Living Color," which are my "full color" solutions to the most difficult sections in the "One Step" explanations. If you like this, send me an Email.

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Rewrite this equation as

h =-16t^2 + 40t

It is well-known that the highest point on a parabolic graph y=ax^2 +bx +c occurs at the value of .

In this equation for h=-16t^2 + 40t, the maximum value of h occurs when



seconds

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Divide each of the 3 parts of this inequality by 4 in order to solve for x.


, which means all numbers on a numberline that are BETWEEN -9 and 3, but not including the endpoints.

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If secA =x, then to solve for A you must do the INVERSE operation to both sides in order to "undo" the sec. The inverse operation is called the "Arccos" or sometimes it is written "sec^-1". So the answer is

A = Arcsec x or A =sec^-1 x

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I'm sorry to be so late answering this, but I thought it was a nice question, and I didn't see it earlier. If you have other questions about square roots and their operations, check out my website by clicking on my tutor name "rapaljer" anywhere in algebra.com, on my Homepage find the link that says "Basic, Intermediate, and College Algebra: One Step at a Time." Then select "Basic Algebra" and go to Chapter 5. The whole explanation is there written so ordinary folks can understand it. At least that's what my students tell me!! Best of all, it's all FREE!

Now, here is your problem and solution:

sqrt75 - sqrt12 + sqrt3 = 5 sqrt3 - 2 sqrt3 + sqrt3 = 3 sqrt3

This is just like combining like terms, but I think you should have ended up with 4 sqrt3. I explain it like this. If it had been "cats" instead of square root of 3, you would have had
5 cats - 2 cats + cat = 3 cats + cat = 4 cats

If it had been x terms instead of square root of 3 or instead of "cats", it would have looked like this:
5 x - 2x + x = 3x + x = 4x


So you are correct in what you said about the invisible 1:
sqrt75 - sqrt12 + sqrt3
= 5 sqrt3 - 2 sqrt3 + sqrt3
= 3 sqrt3 + 1 sqrt3
= 4 sqrt3

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