Unit 4: Cross Multiplication

This is an example of a more challenging problem, following the cross multiplication method.

Step 1: 5x ÷ x^2-25+10 ÷ x+5=20x ÷ x^2+8x+15

My first step was to create my original equation and write it down.

Step 2: x^2-25

My next step is to start to find the LCM of the denominators by using the difference of 2 squares method to factor the equation above. In order to use the difference of two squares method I had to make a^2=x^2 and b^2=25.After connecting the proper terms I had to then break them down by finding the square root of each. When I solved for the square root of each term I got a=x and b= ±5. I then was able to form the equation (x+5)(x-5).

Step 3: x^2+8x+15

Next, I had to factor out the denominator of the last fraction by using the (mn)(m+n) method. First, I had to find what two numbers could multiply together to get 15 and add together to get 8. I thought about the factors of both numbers and determined that 5 and 3 were the numbers that could amount to both the 15 and 8. Once I determined that, I wrote them out into the equation: (x+5)(x+3).

Step 4: (x+5)(x-5)(x+3)

I then had to follow the proper steps to form my LCM. I took all three pairs of denominators: (x+5)(x-5), x+5, (x+5)(x+3) and I started to look for the terms that they have in common. The only one that each one of them share with the other is x+5. Knowing that I was able to add one x+5 to the LCM. I only added one because one of the rules of the LCM method states that if the same term is in all three fractions once then you can have only one x+5 represent all of them. I then realized that they did not share anything else so I carried the other two terms down. Once I carried everything down I was left with the set of terms above, which represent the LCM of this problem.

Step 5: 5x ÷ x^2-25*(x+5)(x-5)(x+3)+10 ÷ x+5*(x+5)(x-5)(x+3)=20x ÷ x^2+8x+15*(x+5)(x-5)(x+3)

The next step was for me to multiply each individual fraction by the entire LCM. I first multiplied the 5x ÷ x^2-25 by the LCM and I got 5x(x+3). Next, I multiplied the 10 ÷ x+5 by LCM and got 10(x-5)(x+3). Last, I multiplied the 20x ÷ x^2+8x+15 by the LCM and got 20x(x-5). Once I did each of those things I ended up with the equation: 5x(x+3) + 10(x-5)(x+3) = 20x(x-5).

Step 6: 5x*x + 5x*3 + 10(x*x + 3*x - 5*x - 5*3) = 20x*x - 20 *-5

My next step was to Distribute/ FOIL/ and Combine like terms. I started by distributing the 5x into each term in the first equations and when I did, I ended up with 5x2+15.I then moved on to FOIL the (x-5)(x+3) and I got (x2+3x-5x-15)and I carried down the 10 that was connected to it by multiplication, leaving me with the equation:10(x2+3x-5x-15). Once that step was completed I moved on to distributing the 20x into the last equation. After multiplying it by the (x-5), I was left with the equation: 20x2-100x.At the end of it, I was left with the new equation: 5x2+ 15x+10(x2+3x-5x-15)=20x2-100x.

Step 7: 5x^2+ 15x+10(x^2-2x-15)=20x^2-100x

My next minor step was to identify the like terms in the middle equation and combine them. The terms that I spotted were 3x and -5x so I combined them by subtracting them. Once I subtracted them I got -2x. With this I rewrote my equation as the one above.

Step 8: 5x^2+ 15x+10x^2-20x-150=20x^2-100x

Next, I distributed the 10 into the middle equation in order to connect all of the terms by addition or subtraction with no multiplication. I started by multiplying 10 by the x^2 in order to get 10x^2. After that I multiplied the 10 by the -2x in order to get -20x. Last, I multiplied the 10 by the -15 and got -150. At the end, I was left with the equation above.

Step 9: 15x^2-5x-150=20x^2-100x

The next step was combining like terms within the equation and I did so by combining the 5x^2 and the 10x^2 by addition in order to get 15x^2. After combining those terms, I combined the 15x and -20x by subtraction in order to get -5x. Lastly, I carried everything else down in order to get the final equation as shown above.

Step 10: 15x^2-5x-150+20x^2-100x=0

I then had to bring all of the terms to one side in order to equal them to 0. In order to do that I had to find the highest number with the highest exponent which was the 20x^2 to determine which side to move over to the other. When I did that I started to move the terms over by subtracting the -15x^2 from the 20x^2, adding the 5x to the -100x, and adding the 150 to the end. Once I did that I got the equation above.

Step 11: 5(x^2-19x+30)

My next step is to take out the GCF between the terms in the equation. I start off by determining that they all share the common number of 5, so I divided each term by that number. I started by dividing the 5x^2 by the 5 which gave me x^2 and then I moved on to dividing the -95x by 5 which gave me -19x. Once I did that, I divided the 150 by the 5 which gave me 30. After completing those steps, I formed the equation shown above.

Step 12: mn/m+n

Next, I moved on to trying to factor out the 5(x^2-19x+30) by finding out what two numbers could multiply together to get 30 and add together to get -19. After awhile of trying to find two numbers that could result in those things, I found that it was not possible. This then led me to the answer that this equation cannot be factored.

Step 13: x=-b± √ b^2-4ac ÷2a

After determining that the equation could not be factored using the mn/m+n method, I used my knowledge of the Quadratic Formula in order to simply the equation even more. I started by identifying the a,b,c of the equation. I determined that a=1, b=-19, and c=30. I then moved to substituting each letter in the equation by its numerical value. After substitution I ended up with the equation:x=--19± √ (-19)^2-4(1)(30) ÷2(1). After simplifying further I ended up with the equation: x=19±√241÷2. I got this by changing the two negative signs in front of the 19 into a positive sign since two negatives in math always equal a positive. After that I determined what -1-19^2 was by multiplying 19 by itself and then I multiplied 4*30 in order to get 120. From there I subtracted the 120 from the 361 and that left me with the 241.

Step 14: x=19±√241÷2

My next step was to find out what the √241 was and I did so by using a calculator. When I did I got the answer of 15.5.

Step 15: x=19±15.5÷2

I then had to follow the proper steps of the by adding 19 and 15.5 ÷2 and subtracting 19 and 15.5 ÷2. When I added 19 and 15.5, I got 34.5 and when I subtracting 19 and 15.5 I got 3.5. I then had to divide both of those numbers by 2 in order to get the two x values of 17.25 and 1.75.