# Algebra question

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• Jun 14th 2007, 08:36 AM
jono
Algebra question
Hey, would appreciate any help on this, haven't had any hope with it at all

we are given the equations
x^2 = (y^2) +4
(x+0.95)^2 = ((y+1.05) ^2) +4
2.1y = 1.9x - 0.2

The question is to form a quadratic for x and then solve it and then find out terms for y.

I expanded out the second equation to y^2 - x^2 + 2.1y - 1.9x = -4.2
h/e everytime i try and replace a y value i end up with -4.2 = -4.2 or something similar.
It was based on 2 right angle triangles, first with sides 2, x, y
second with sides 2 , x + 0.95, y + 1.05

Thanks for any help in advance
• Jun 14th 2007, 11:39 AM
CaptainBlack
Quote:

Originally Posted by jono
Hey, would appreciate any help on this, haven't had any hope with it at all

we are given the equations
x^2 = (y^2) +4
(x+0.95)^2 = ((y+1.05) ^2) +4
2.1y = 1.9x - 0.2

The question is to form a quadratic for x and then solve it and then find out terms for y.

I expanded out the second equation to y^2 - x^2 + 2.1y - 1.9x = -4.2
h/e everytime i try and replace a y value i end up with -4.2 = -4.2 or something similar.
It was based on 2 right angle triangles, first with sides 2, x, y
second with sides 2 , x + 0.95, y + 1.05

Thanks for any help in advance

Substitute the third into the first to get:

\$\displaystyle
20 x^2 + 19 x - 442=0
\$

(Note in general three equations in two unknowns has no solution, though that does not
mean that these equations have no solution)

RonL
• Jun 14th 2007, 02:08 PM
jono
Thanks for the help :o

hope im not being too much bother but could anyone explain how to substitute the third into the first? im not entirely sure how :o
• Jun 14th 2007, 02:16 PM
tukeywilliams
\$\displaystyle 2.1y = 1.9x-0.2 \$

So

\$\displaystyle y \approx 0.90x - 0.095 \$

and \$\displaystyle x^{2} = y^{2} + 4 = (0.90x - 0.095)^{2} + 4 \$ \$\displaystyle \approx 19.0000x^{2}+17.10000x -400.902500 \$(slightly different from CaptainBlack depending on how you round).

From the following: \$\displaystyle (0.90x - 0.095)^{2} + 4 \$ we multiply by 100 to get whole numbers (as CaptainBlack did)