(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous
First, let me do the Mathematical part of that, and then I shall explain the theory behind it.
Mathematical part:
We are going to multiply 513 with 46. So the two partial products that we are going to choose are 40 and 6.
Multiply 513 with 6 first.
513
x46
--------------------------
18 (as 6*3 = 18)
60 (as 6*10 = 60; In 513, the digit at tenths place is 1, so 1*10=10)
3000 (as 6*500 = 3000; In 513, 5 is at hundredth place, so 5*100=500)
120 (as 40*3 = 120; since 4 is at the tenth place, so 4*10=40)
400 (as 40*10 = 400)
20000 (as 40*500 = 20000)
--------------------------
23598 (Add all of them)
Theory:
As you can see above that we have chosen the two partial products individually which are 6 and 40. Since 4 in 46 is in tenth place, we have to consider it 40 (since 4*10 = 40). One by one, we first multiply 6 with 513. Then we move to the tenth place, and multiply 513 with 40. At the end, we have added all the results we found after multiplication.
Check: If we check the multiplication result by using the calculator, we would get the same result (23598).
Another Method (instant):
513 * (40+6) = (513*40) + (513*6) = 23598.
All we have to do is divide the distance of the old sidewalk by 3.14:
65 / 3.14 ≈ 20.7006
That means that, rounded to the nearest meter, the new sidewalk is:
21 meters long!
(-2,-5)
The first goal with equations like these is to get one of the variables to go away. You can do that by adding/subtracting one equation from the other. Then you solve for the one variable. Then you plug that answer into one of the original equations and solve.
4 + 3x = 12
3x = 12 - 4
3x = 8
x = 8/3
x = 2,6 grams
