# [SOLVED] Need help with a few questions

• Mar 16th 2010, 07:28 AM
NewtoMath
[SOLVED] Need help with a few questions
2. For the simultaneous equation ax+by = a^2 + 2ab - b^2

and bx+ay = a^2 + b^2

I know there should be some kinda shortcut but I can't figure it out

3. ABCD is a parallelogram with coordinates A(2,2), B(1,5,4), C(6,6)

Find the coordinates of point of intersection of the diagonals?
• Mar 16th 2010, 08:31 AM
Soroban
Hello, NewtoMath!

Quote:

Solve by elimination: .$\displaystyle \begin{array}{cccc} (a+b)x+cy &=& bc & [1]\\ (b+c)y +ax &=& \text{-}ab & [2]\end{array}$
Where did you mess up? . . . You didn't use Elimination.

$\displaystyle \begin{array}{cccc} \text{Multiply [1] by }a & a(a+b)x + acy &=& abc \\ \text{Multiply [2] by }\text{-}c & \text{-}c(b+c)x - acy &=& abc \end{array}$

Add: .$\displaystyle \bigg[a(a+b)-c(b+c)\bigg]x \;=\;2abc$

In the brackets, we have:
. . $\displaystyle a^2+ab - bc - c^2 \;=\;a^2-c^2 + ab-bc \;=\;(a-c)(a+c) + b(a-c) \;=\;(a-c)(a+b+c)$

So we have: .$\displaystyle (a-c)(a+b+c)x \:=\:2abc \quad\Rightarrow\quad\boxed{ x \;=\;\frac{2abc}{(a-c)(a+b+c)}}$

Now substitute into [1] and solve for $\displaystyle y.$

Quote:

$\displaystyle (2)\;\;\begin{array}{cccc}ax+by &=& a^2 + 2ab - b^2 & [1] \\bx+ay &=& a^2 + b^2 & [2] \end{array}$

Note: If $\displaystyle a=b$, the system has infinite solutions.
. . . . . We will assume that $\displaystyle a \neq b.$

$\displaystyle \begin{array}{ccccc} \text{Multiply [1] by }a & a^2x + aby &=& a(a^2+2ab - b^2) \\ \text{Multiply [2] by }\text{-}b & \text{-}b^2x - aby &=& \text{-}b(a^2+b^2) \end{array}$

Add: .$\displaystyle (a^2-b^2)x \:=\:a^3 + 2a^2b - ab^2 - a^2b - b^3$

. . The right side is: .$\displaystyle a^3 + a^2b - ab^2 - b^3 \;=\;a^2(a+b)-b^2(a+b) \;=\;(a+b)(a^2-b^2)$

So we have: .$\displaystyle (a^2-b^2)x \;=\;(a+b)(a^2-b^2) \quad\Rightarrow\quad\boxed{ x \:=\:a+b}$

Substitute into [2]: .$\displaystyle b(a+b) + ay \:=\:a^2+b^2 \quad\Rightarrow\quad ay \:=\:a^2 + b^2 - b(a+b)$

. . . . . $\displaystyle ay \:=\:a^2-ab \:=\:a(a-b) \quad\Rightarrow\quad\boxed{ y \:=\:a-b}$

• Mar 16th 2010, 03:38 PM
NewtoMath
Hey thanks Soroban!

Quote:

Originally Posted by Soroban
Hello, NewtoMath!

Where did you mess up? . . . You didn't use Elimination.

So why was my method wrong?

(a+b)x+cy = bc AND (b+c)y +ax = -ab

After removing the brackets the equations became...

ax + cy + bx = bc

ax + cy + by = -ab

After subtracting should become bx-by = bc + ab no?
• Mar 16th 2010, 04:14 PM
bigwave
Quote:

2. For the simultaneous equation ax+by = a^2 + 2ab - b^2

and bx+ay = a^2 + b^2

I know there should be some kinda shortcut but I can't figure it out
set both eq = 0 then set them equal to each other

$\displaystyle a^2+2 a b-b^2-a x-a b = b x+a y-a^2-b^2$

collect and factor
$\displaystyle a (2 a+b-x) = a y+b x$

however the only answer I see is a=0 and b=0
• Mar 16th 2010, 11:28 PM
NewtoMath
Quote:

Originally Posted by bigwave
however the only answer I see is a=0 and b=0[/FONT]

Thanks bigwave. I'd really like to know what was wrong with my method tho so I don't repeat the mistake on an exam or something (Surprised)
• Mar 17th 2010, 01:47 AM
HallsofIvy
There was nothing "wrong" with your method except that there is no reason to do it! Whatever method you use to solve simultaneous equations, your objective is to eliminate one of the unknowns to get a single equation in a single unknown. Subtracting the two equations as you do does not eliminate anything.
• Mar 17th 2010, 05:20 AM
NewtoMath
Quote:

Originally Posted by HallsofIvy
There was nothing "wrong" with your method except that there is no reason to do it! Whatever method you use to solve simultaneous equations, your objective is to eliminate one of the unknowns to get a single equation in a single unknown. Subtracting the two equations as you do does not eliminate anything.

Ohh so I just turned the two equations into an equation with infinite solutions with x and y still being there.

Thanks, that's all I needed to know regarding that.

Would you know to then solve simultaneous equations with 3 or 4 variables?
• Mar 18th 2010, 04:57 AM
NewtoMath
Can someone help me solve this equation?

A parabola has x intercepts at -5 and -1 and passes through the point, 3,6 find the equation?

If possible could I solve this with the y = a(x - h)2 + k formula?
• Mar 18th 2010, 05:06 AM
HallsofIvy
Quote:

Originally Posted by NewtoMath
Can someone help me solve this equation?

A parabola has x intercepts at -5 and -1 and passes through the point, 3,6 find the equation?

If possible could I solve this with the y = a(x - h)2 + k formula?

You could- though it is not the simplest way: saying that it has x-intercepts at -5 and -1 mean that the graph passes through (-5, 0) and (-1, 0). That, together with the fact that it passes through (3, 6) tells you that
a) when x= -5, y= 0 so $\displaystyle 0= a(-5- h)^2+ k$
b) when x= -1, y= 0 so $\displaystyle 0= a(-1-h)^2+ k$
c) when x= 3, y= 6 so $\displaystyle 6= a(3- h)^2+ k$

But, since none of the given points is the vertex, it would probably be simpler to use the general form $\displaystyle y= ax^2+ bx+ c$ so that your equations are
a)$\displaystyle a(-5)^2+ b(-5)+ c= 0$
b) $\displaystyle a(-1)^2+ b(-1)+ c= 0$
c) $\displaystyle a(3)^2+ b(3)+ c= 0$

Simplest of all is to use the fact that since x= -5 and x= -1 make the function equal to 0, x-(-5)= x+ 5 and x-(-1)= x+ 1 must be factors:
y= a(x+1)(x+5). Now set x= 3, y= 6 to get 6= a(3+1)(3+5) and solve for a.
• Mar 20th 2010, 07:21 AM
NewtoMath
Thanks again Ivy, I think I'll be able to handle most parabola questions in my class with your methods (Whew)

This one is probably hard to explain w/o pen paper demonstration, but P & Q are points of intersection at y/2 + x/3 = 1 which becomes (0, 2), (3, 0). Gradient of QR is 1/2, where R is point x-coordinate 2a, a>0

The question is find y-coord of R in terms of a but more importantly how do I draw this up to work from?
• Mar 20th 2010, 04:26 PM
NewtoMath
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