# Thread: 9 ?'s that i really need help on!!!

1. ## 9 ?'s that i really need help on!!!

questions are on attachment!

2. #1:

Let x=amount invested at 10%.

Then 3500-x is the amount invested at 15%.

The difference between the two interest rates is \$85.

$.15(3500-x)-.10x=85$

3. Hello, Lane!

Here are a few more . . .

Solve: . $\begin{array}{cc}3x - 4y \,= \,7 \\ x + 2y \,=\,9\end{array}$

With no method specified, I'll use Elimination.

We have: . $\begin{array}{cc}3x - 4y \:= \:7 \\ x + 2y \:=\:9\end{array}\;\begin{array}{cc}(1)\\(2)\end{a rray}$

Multiply (2) by 2: . $2x + 4y \:=\:18$
. . . . . . Add (1): . $3x - 4y \:=\:7$

And we have: . $5x\,=\,25\quad\Rightarrow\quad \boxed{x = 5}$

Substitute into (2): . $5 + 2y\:=\:9\quad\Rightarrow\quad 2y \,=\,4\qiad\Rightarrow\quad \boxed{y = 2}$

Solve with Cramer's Rule: . $\begin{array}{cc}6x - 4y \:=\:6 \\ 3x \:= \:2y - 5\end{array}$

We have: . $\begin{array}{cc}6x-4y\:=\:7\\3x-2y\:=\:\text{-}5\end{array}$

$D\:=\:\begin{vmatrix}6 & \text{-}4 \\ 3 & \text{-}2\end{vmatrix} \;= \;(6)(\text{-}2) - (\text{-}4)(3) \;= \;-12 + 12 \;= \;0$ . . . No solution!

6. Solve: . $\begin{array}{cc}xy\:=\:10 \\ x^2 + 4y^2\:=\:36\end{array}\;\begin{array}{cc}(1)\\(2) \end{array}$

Another trick question!

From (1), we have: . $y = \frac{10}{x}$

Substitute into (2): . $x^2 + 4\left(\frac{10}{x}\right)^2\;=\;36 \quad\Rightarrow\quad x^2 + \frac{400}{x^2}\:=\:36$

Multiply by $x^2:\;\;x^4 + 400 \:=\:36x^2\quad\Rightarrow\quad x^4 - 36x^2 + 400 \:= \:0$

Quadratic Formula: . $x^2\:=\:\frac{-(\text{-}36) \pm \sqrt{(\text{-}36)^2 - 4(1)(400)}}{2(1)}$

. . . $x^2\;=\;\frac{36 \pm\sqrt{-304}}{2}$ . . . No real root

Therefore, the system has no real solution.