The idea is to "isolate" the variable, in this case "x", on one side. In this case, you would start by multiplying both sides of the equation by "x".
5. If 56 is x, then 336-x-x-x-x-x=0. This can be shortened to 336-5x=0, and by adding 5x to each side we get 336=5x. If we substitute the number back in, we get 336=5x52.
Sorry but i dont know te answer to this :( x/a= y/b therefore x = ay/b
b divided by 2
b/5 = 31 Therefore, b = 31 x 5 b = 155
B/5 = 6 means that 6 x 5 = B 6 x 5 = 30 Therefore, B = 30
5. If 56 is x, then 336-x-x-x-x-x=0. This can be shortened to 336-5x=0, and by adding 5x to each side we get 336=5x. If we substitute the number back in, we get 336=5x52.
Sorry but i dont know te answer to this :( x/a= y/b therefore x = ay/b
b divided by 2
y = mx + b
56 x 6 = 336
b/5 = 31 Therefore, b = 31 x 5 b = 155
336 , INV, x^3 = 6.952 in3
B/5 = 6 means that 6 x 5 = B 6 x 5 = 30 Therefore, B = 30
28 x 12 = 336
4 divided by x equals 10 = 0.4
|x-3|/2=10 |x-3|=20 |a|=b means a=b or a=-b Then : x-3=20 or x-3=-20 So: x=23 or x=-17
180176 first multiply, then subtract...