# Two surds problems - one involving simultaneous equations..

• Aug 7th 2008, 08:36 AM
struck
Two surds problems - one involving simultaneous equations..
1. Two sides of a triangle are 5sqrt(3) and 10sqrt(2) respectively, while the hypotenuse is unknown. Here I tried finding the hypotenuse using pythagoras theorem.

x ^ 2 = (5sqrt(3))^2 + (10sqrt(2)) ^ 2

The answer on should be x = 5sqrt(11) but I get 5sqrt(3) + 10 (Headbang)

2. Two simultaneous equations are as follows:

Code:

7x - (3sqrt(5))y = 9sqrt(5)          (1) (2sqrt(5)x + y = 34                    (2)
I tried rearranging (1) to y = (9sqrt(5) - 7x) / 3sqrt(5) but I cannot reach an answer at all :( ..

P.S. Is there any tutorial on how to generate real maths symbols on this forum?
• Aug 7th 2008, 08:58 AM
Soroban
Hello, struck!

Quote:

1. Two sides of a right triangle are $\displaystyle 5\sqrt{3}$ and $\displaystyle 10\sqrt{2}$
Find the hypotenuse.

$\displaystyle x^2 \:= \:(5\sqrt{3})^2 + (10\sqrt{2})^2$

$\displaystyle x^2 \;=\;5^2(\sqrt{3})^2 + 10^2(\sqrt{2})^2$

$\displaystyle x^2 \;=\;25(3) + 100(2) \;=\;75 + 200 \;=\;275$

$\displaystyle x \;=\;\sqrt{275} \;=\;\sqrt{25\!\cdot\!11} \;=\;\boxed{5\sqrt{11}}$

Quote:

$\displaystyle 2)\;\;\begin{array}{cccc}7x - (3\sqrt{5})y &= &9\sqrt{5} & {\color{blue}[1]} \\ (2\sqrt{5})x + y &=& 34 & {\color{blue}[2]} \end{array}$

$\displaystyle \begin{array}{cccc}\text{Multiply {\color{blue}[2]} by }3\sqrt{5}\!: & 30x + 3\sqrt{5}\,y & = & 102\sqrt{5} \\ \text{Add {\color{blue}[1]}:} & 7x - 3\sqrt{5}\,y &=& 9\sqrt{5} \end{array}$

And we have: .$\displaystyle 37x \:=\:111\sqrt{5} \quad\Rightarrow\quad \boxed{x \:=\:3\sqrt{5}}$

Substitute into [2]: .$\displaystyle (2\sqrt{5})(3\sqrt{5}) + y \:=\:34 \quad\Rightarrow\quad 30 + y \:=\:34 \quad\Rightarrow\quad \boxed{y \:=\:4}$

• Aug 7th 2008, 11:38 PM