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 is10units from the origin. Determine the coordinates of P

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- Jan 9th 2008, 05:15 PMyvonnesolving system by substitution Help
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 - Jan 9th 2008, 05:32 PMgalactus
#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. - Jan 9th 2008, 05:54 PMTwistedOne151Problems 1 and 2
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. - Jan 9th 2008, 06:03 PMyvonne
thank you