document.write( "Question 1101744: Y=-2x+4
\n" ); document.write( "Y-3x=-11
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\n" ); document.write( "\n" ); document.write( "I also have a problem that I cant figure out but if you cant help thats fine.
\n" ); document.write( "\"The product of two numbers is 60. One number exceeds another by 7. Find the two numbers.\" \n" ); document.write( "

Algebra.Com's Answer #716401 by richwmiller(17219) $\"\"$ $\"About$

You can put this solution on YOUR website!
The product =60 not the sum
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\n" ); document.write( "x^2-7x-60=0
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression $\"x%5E2-7x-60\"$ , we can see that the first coefficient is $\"1\"$ , the second coefficient is $\"-7\"$ , and the last term is $\"-60\"$ .

Now multiply the first coefficient $\"1\"$ by the last term $\"-60\"$ to get $\"%281%29%28-60%29=-60\"$ .

Now the question is: what two whole numbers multiply to $\"-60\"$ (the previous product) and add to the second coefficient $\"-7\"$ ?

To find these two numbers, we need to list all of the factors of $\"-60\"$ (the previous product).

Factors of $\"-60\"$ :

1,2,3,4,5,6,10,12,15,20,30,60

-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to $\"-60\"$ .

1*(-60) = -60
2*(-30) = -60
3*(-20) = -60
4*(-15) = -60
5*(-12) = -60
6*(-10) = -60
(-1)*(60) = -60
(-2)*(30) = -60
(-3)*(20) = -60
(-4)*(15) = -60
(-5)*(12) = -60
(-6)*(10) = -60

Now let's add up each pair of factors to see if one pair adds to the middle coefficient $\"-7\"$ :

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First Number	Second Number	Sum
1	-60	1+(-60)=-59
2	-30	2+(-30)=-28
3	-20	3+(-20)=-17
4	-15	4+(-15)=-11
5	-12	5+(-12)=-7
6	-10	6+(-10)=-4
-1	60	-1+60=59
-2	30	-2+30=28
-3	20	-3+20=17
-4	15	-4+15=11
-5	12	-5+12=7
-6	10	-6+10=4

From the table, we can see that the two numbers $\"5\"$ and $\"-12\"$ add to $\"-7\"$ (the middle coefficient).

So the two numbers $\"5\"$ and $\"-12\"$ both multiply to $\"-60\"$ and add to $\"-7\"$

Now replace the middle term $\"-7x\"$ with $\"5x-12x\"$ . Remember, $\"5\"$ and $\"-12\"$ add to $\"-7\"$ . So this shows us that $\"5x-12x=-7x\"$ .

$\"x%5E2%2Bhighlight%285x-12x%29-60\"$ Replace the second term $\"-7x\"$ with $\"5x-12x\"$ .

$\"%28x%5E2%2B5x%29%2B%28-12x-60%29\"$ Group the terms into two pairs.

$\"x%28x%2B5%29%2B%28-12x-60%29\"$ Factor out the GCF $\"x\"$ from the first group.

$\"x%28x%2B5%29-12%28x%2B5%29\"$ Factor out $\"12\"$ from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.

$\"%28x-12%29%28x%2B5%29\"$ Combine like terms. Or factor out the common term $\"x%2B5\"$

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Answer:

So $\"x%5E2-7%2Ax-60\"$ factors to $\"%28x-12%29%28x%2B5%29\"$ .

In other words, $\"x%5E2-7%2Ax-60=%28x-12%29%28x%2B5%29\"$ .

Note: you can check the answer by expanding $\"%28x-12%29%28x%2B5%29\"$ to get $\"x%5E2-7%2Ax-60\"$ or by graphing the original expression and the answer (the two graphs should be identical).

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