+2446 I'm going to assume that this is what the equation is supposed to be. I have two possible interpretations. I'll solve both
I'll solve for x using the first interpretation:
| \(\frac{x^2}{x}+1=x+\frac{2}{3}\) | I'll utilize an exponent rule to simplify \(\frac{x^2}{x}\) that says that \(\frac{a^b}{a^c}=a^{b-c}\) |
| \(\frac{x^2}{x}=x^{2-1}=x^1=x\) | Reinsert this back into the original equation. |
| \(x+1=x+\frac{2}{3}\) | Subtract x on both sides. |
| \(1=\frac{2}{3}\) | \(1\neq\frac{2}{3}\), so there is no value for x that satisfies this equation; in other words, there is no solution. |
I'll solve for the second interpretation:
| \(\frac{x^2}{x+1}=\frac{x+2}{3}\) | In order to solve this equation, we must cross multiply. Remember that if \(\text{If}\hspace{1mm}\frac{a}{b}=\frac{c}{d}\hspace{1mm}\text{,then}\hspace{1mm}ad=bc\) |
| \(3*x^2=(x+2)(x+1)\) | Let's simplify the left hand first, which is the easiest. |
| \(3x^2=(x+2)(x+1)\) | Multiply the two binomials together by distributing the x and the 2 to the x+1-term. |
| \((x+2)(x+1)=x(x+1)+2(x+1)=x^2+x+2x+2=x^2+3x+2\) | Reinsert the simplified version back into the equation. |
| \(3x^2=x^2+3x+2\) | Subtract x^2 from both sides of the equation. |
| \(2x^2=3x+2\) | Bring all the terms to the left side of the equation |
| \(2x^2-3x-2=0\) | Now, use any method to solve for x (factoring, completing the square, quadratic equation, etc.) This happens to be factorable, so I will use that method. |
To help visualize what I am doing, I'll attempt to draw an "x."
\ /
\ -4 /
\ /
\ /
\ /
A \/ B
/\
/ \
/ \
/ \
/ -3 \
/ \
The top number is the number of the product of the values of a and c in a quadratic and the bottom is simply b, the coefficient of the linear term of the equation. Our job is to find 2 numbers for A and B wherein they multiply to get -4 and their sum is -3. Let's create a table:
| Factors of -4 | Sum of Factors |
|---|---|
| \(4*-1\) | \(4-1=3\) |
| \(-1*4\) | \(-1+4=3\) |
| \(-4*1\) | \(-4+1=-3\) |
| \(1*-4\) | \(1-4=-3\) |
| \(2*-2\) | \(2-2=0\) |
| \(-2*2\) | \(-2+2=0\) |
Out of all these combinations for factors of 4, which one gives the sum of -3? -4 and 1 or 1 and -4. These are flips of each other. You can use either combination, though. I'll use -4 and 1.
\(2x^2-4x+x-2=0\)
What this method does is break up that b-term into 2 parts. Why is this useful? You'll see!
| \(2x^2-4x+x-2=0\) | Solve by grouping. I'll put parentheses aroung the groups I am solving. | ||||||||
| \((2x^2-4x)+(x-2)=0\) | Factor out the GCF of the first group, which is 2x. The other term, x+2, has a GCF of 1, so it cannot be simplified. | ||||||||
| \(2x^2-4x=2x(x-2)\) | Reinsert this back into the equation. | ||||||||
| \(2x(x-2)+(x-2)=0\) | Now, we will utilize the principle that \(ac+bc=c(a+b)\) | ||||||||
| \(2x(x-2)+(x-2)=2x(x-2)+1(x-2)=(2x+1)(x-2)\) | Ok, reinsert this back into the equation. | ||||||||
| \((2x+1)(x-2)=0\) | Set both factors equal to 0. | ||||||||
| Solve for both of the values for x. | ||||||||
Now, let's look at that original equation again:
\(\frac{x^2}{x+1}=\frac{x+2}{3}\)
The solutions work out in the original equation, so your answers are:
\(x=-\frac{1}{2}\hspace{1mm}\text{or}\hspace{1mm}x=2\)
.+2446 Let me try to do this. If I am not mistaken, this is the original equation:
\(2^{3x}*\frac{8^{x+3}}{16^{x-1}}=2^{\frac{x-1}{2}}\)
| \(2^{3x}*\frac{8^{x+3}}{16^{x-1}}=2^{\frac{x-1}{2}}\) | Our first step, I think, is to get rid of the fraction. I'm going to use the rule that \(\frac{1}{a^b}=a^{-b}\). |
| \(\frac{1}{16^{x-1}}=16^{-(x-1)}=16^{-x+1}\) | Doing this puts allows me to take this out of the fraction. Therefore, I am going to rewrite the current equation |
| \(2^{3x}*8^{x+3}*16^{-x+1}=2^{\frac{x-1}{2}}\) | Now, convert \(8^{x+3}\)into a form where it will be in base 2. Luckily for us, all these numbers can be in that form. |
| \(8^{x+3}=(2^3)^{x+3}\) | Okay, let's insert that back into the equation. |
| \(2^{3x}*(2^3)^{x+3}*16^{-x+1}=2^{\frac{x-1}{2}}\) | Now, convert \(16^{-x+1}\) into base 2, as well. |
| \(16^{-x+1}=(2^4)^{-x+1}\) | Insert that into the original equation again, too. |
| \(2^{3x}*(2^3)^{x+3}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\) | Now, I'll apply a rule on the term \((2^3)^{x+3}\) that says that \((a^b)^c=a^{b*c}\). Let's use it. |
| \((2^3)^{x+3}=2^{3(x+3)}\) | Insert it into the original equation. |
| \(2^{3x}*2^{3(x+3)}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\) | Now, I'll use another power rule that says that \(a^b*a^c=a^{b+c}\). I'll utilize this for \(2^{3x}*2^{3(x+3)}\) |
| \(2^{3x}*2^{3(x+3)}=2^{3x+3(x+3)}\) | Reinsert this back into the equation. |
| \(2^{3x+3(x+3)}*(2^4)^{-x+1}=2^{\frac{x-1}{2}}\) | Okay, now the only term left is \((2^4)^{-x+1}\). Just like before, we'll use an exponent rule that says that \((a^b)^c=a^{b*c}\). |
| \((2^4)^{-x+1}=2^{4(-x+1)}\) | Insert this back into the equation again. |
| \(2^{3x+3(x+3)}*2^{4(-x+1)}=2^{\frac{x-1}{2}}\) | Yet again, we'll utilize the same rule as before that says that \(a^b*a^c=a^{b+c}\). |
| \(2^{3x+3(x+3)}*2^{4(-x+1)}=2^{3x+3(x+3)+4(-x+1)}\) | Reinsert this into the equation again. |
| \(2^{3x+3(x+3)+4(-x+1)}=2^{\frac{x-1}{2}}\) | Now, we'll use another rule that says that\(a^{f(x)}=a^{g(x)},\text{then}\hspace{1mm}f(x)=g(x)\). This will reduce the equation to simply two-sided equation without exponents. |
| \(3x+3(x+3)+4(-x+1)=\frac{x-1}{2}\) | To clean this up, let's use the distribute property. |
| \(3x+3x+9-4x+4=\frac{x-1}{2}\) | Combine like terms on the left hand side of the equation. |
| \(2x+13=\frac{x-1}{2}\) | Multiply both sides by 2 to get rid of the pesky fraction. |
| \(4x+26=x-1\) | Subtract x on both sides. |
| \(3x+26=-1\) | Subtract 26 on both sides |
| \(3x=-27\) | Divide by 3 on both sides to finally isolate x. |
| \(x=-9\) | |
+2446 The equation for finding the area of a rectangle is as follows:
Let l = length of the rectangle
Let w = the width of the rectangle
\(A_{\text{rect.}}=lw\)
If you never know how to start a problem, utilizing a formula is probably the way to go. Plug in the known values into the equation and solve for the missing one:
| \(117=l*9\) | Solve for the width by dividing by 9 on both sides of the equation |
| \(13m=l\) | Of course, don't forget your units of measure! |
+2446 There's an excellent formula that I concocted 5 days ago to always give the equation of a parabola when only the directrix and focus is given. Here is the equation:
Let a = x-coordinate of focus
Let b = y-coordinate of focus
Let k= equation of line of the directrix
\(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\)
All you have to do is plug into this formula, simplify, and you're done. Let's try this together:
a=-2
b=4
k=6
| \(y=\frac{(x-a)^2}{2(b-k)}+\frac{b+k}{2}\) | Plug in the appropriate values that are given by the focus and directrix. |
| \(y=\frac{(x-(-2))^2}{2(4-6)}+\frac{4+6}{2}\) | Let's clean this up a bit, shall we? |
| \(y=\frac{(x+2)^2}{-4}+5\) | Technicaly, you could stop here and call it a day, but I am going to attempt to make it look even cleaner! I'll expand the \((x+2)^2\) |
| \(y=-\frac{x^2+4x+4}{4}+5\) | After expanding it, I'll change 5 into an improper fraction that I can add to the current fraction. |
| \(y=-\frac{x^2+4x+4}{4}+\frac{20}{4}\) | Since the fractions have common denominators. Add the fractions together, but you have to be very attentive to how you do this. Notice how there is a negative. I'm going to get rid of this because you can't combine a negative fraction with a positive one; it just doesn't work. |
| \(y=\frac{-(x^2+4x+4)}{4}+\frac{20}{4}\) | I'm distributing that negative because you cannot combine a negative and positive fraction. |
| \(y=\frac{-x^2-4x-4}{4}+\frac{20}{4}\) | Now, you can add the fractions together. Now you add the fractions together normally. |
| \(y=\frac{-x^2-4x+16}{4}\) | Break off the two last terms from the current fraction. You'll see why. |
| \(y=\frac{-x^2}{4}+\frac{-4x+16}{4}\) | The rightmost fraction can be simplified because a factor of 4 goes into both terms. Wow! |
| \(y=\frac{-x^2}{4}-x+4\) | |
| \(y=-\frac{1}{4}x^2-x+4\) | This is your final equation. |
No, I did not just pull that formula out of the air! I derived it myself. If you are wondering why this formula will work 100% of the time, I explained it, in detail, here where someone asked a similar question to yours.
