The equation is 138=300-3x
now solve by first subtracting 300 from each side: -162=-3x
divide each side by -3: x=54
The wise man is 54 years old
In order for the expression to have infinitely many solutions, we need to have a situation like this:
![Ax+B = Cx+D,\qquad A=C,\quad B=D](https://tex.z-dn.net/?f=%20Ax%2BB%20%3D%20Cx%2BD%2C%5Cqquad%20A%3DC%2C%5Cquad%20B%3DD%20)
In other words, we need the left and right hand sides to be the same, so that the equality is actually an identity, and no matter which x-value you choose, the equality will hold.
In your case, if you mimic the template above, you have
![A=-90,\ B = Q,\ C = P,\ D = 45](https://tex.z-dn.net/?f=%20A%3D-90%2C%5C%20B%20%3D%20Q%2C%5C%20C%20%3D%20P%2C%5C%20D%20%3D%2045%20)
And since we want
, we want
![P=-90,\quad Q=45](https://tex.z-dn.net/?f=%20P%3D-90%2C%5Cquad%20Q%3D45%20)
Answer:
x = (ab - ac )/c
Step-by-step explanation:
x/a = b/c - 1 Multiply through by ac:
cx = ab - ac Divide both sides by c:
x = (ab - ac )/c.
First question:
The cube root of 626 is a number than, when cubed, returns 626. So, if it is between two integers n and n+1, it means that
. This is the case for 8 and 9, since you have
.
Second question:
The square root of 52 follows the same principles of the first question, but you have to square the numbers instead of cubing them. So, this time the number lies between 7 and 8, since
.
Third question:
To solve this kind of exercises, factor the number under the root, and use the root property
. You have
![\sqrt{44} = \sqrt{4\cdot 11} = \sqrt{4}\sqrt{11} = 2\sqrt{11}](https://tex.z-dn.net/?f=%20%5Csqrt%7B44%7D%20%3D%20%5Csqrt%7B4%5Ccdot%2011%7D%20%3D%20%5Csqrt%7B4%7D%5Csqrt%7B11%7D%20%3D%202%5Csqrt%7B11%7D%20)
Fourth question:
This exercise is similar to the third one, but again you change the order of the root. So, you have
![\sqrt[3]{108} = \sqrt[3]{2^23^3} = 3\sqrt[3]{4}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B108%7D%20%3D%20%5Csqrt%5B3%5D%7B2%5E23%5E3%7D%20%3D%203%5Csqrt%5B3%5D%7B4%7D%20)