1. $\displaystyle Y=\sqrtX$
$\displaystyle Y=x-6$
2. $\displaystyle Y=|X|$
$\displaystyle Y=X+2$
3. Point $\displaystyle P$ is on the line $\displaystyle 3X+Y=26$ and it is 10 units from the origin. Determine the coordinates of P
1. $\displaystyle Y=\sqrtX$
$\displaystyle Y=x-6$
2. $\displaystyle Y=|X|$
$\displaystyle Y=X+2$
3. Point $\displaystyle P$ is on the line $\displaystyle 3X+Y=26$ and it is 10 units from the origin. Determine the coordinates of P
#3. Using the distance formula: $\displaystyle \sqrt{(x-0)^{2}+(y-0)^{2}}=\sqrt{x^{2}+y^{2}}=10$
But the equation of said line is $\displaystyle y=-3x+26$
So, we get:
$\displaystyle \sqrt{x^{2}+(-3x+26)^{2}}=10$
Solve for x, then sub that into the line equation to find y.
On problem 1, you have $\displaystyle y=\sqrt{x}$ and $\displaystyle y=x-6$, correct? Solving the first equation for x, we have $\displaystyle x=y^2$, and substituting this for x into the other equation gives $\displaystyle y=y^2-6$, thus $\displaystyle y^2-y-6=0$, which is an easy to solve quadratic in y.
For problem 2, we have an absolute value, so we consider two cases, that where the expression inside the absolute value is non-negative, and that where it is negative. First, subsituting |x| for y in the second equation to produce a single equation for x, we get $\displaystyle |x|=x+2$. Now we consider our two cases:
i). $\displaystyle x\ge0$:
We have $\displaystyle |x|=x$, and so our equation becomes $\displaystyle x=x+2$, which has no solution.
ii). x<0:
We have $\displaystyle |x|=-x$, and so our equation becomes $\displaystyle -x=x+2$, which is simple algebra to solve.
--Kevin C.