Question

PLEASE HELP ME I WILL MARK BRAINLYEST

Answer 1

                               Question 9:

Given the equations

$5x+2y=6;$

$4x-8y=0$

solving the system of the equations

$\begin{bmatrix}5x+2y=6\\ 4x-8y=0\end{bmatrix}$

$\mathrm{Multiply\:}5x+2y=6\mathrm{\:by\:}4\:\mathrm{:}\:\quad \:20x+8y=24$

$\mathrm{Multiply\:}4x-8y=0\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:20x-40y=0$

$\begin{bmatrix}20x+8y=24\\ 20x-40y=0\end{bmatrix}$

$20x-40y=0$

$-$

$\underline{20x+8y=24}$

$-48y=-24$

$\begin{bmatrix}20x+8y=24\\ -48y=-24\end{bmatrix}$

$-48y=-24$

$\frac{-48y}{-48}=\frac{-24}{-48}$

$y=\frac{1}{2}$

$y = 0.5$ ( 0.5 is the result of rounding 0.5 to the nearest 0.01)

$\mathrm{For\:}20x+8y=24\mathrm{\:plug\:in\:}y=\frac{1}{2}$

$20x+8\cdot \frac{1}{2}=24$

$20x+4=24$

$\frac{20x}{20}=\frac{20}{20}$

$x=1.00$ ( 1.00 is the result of rounding 1 to the nearest 0.01 )

So, the solution is:

(x, y) → (1.00, 0.5)

                            Question 10)

Similarly the question 10 can be solved.

Given the equations

$5x+2y=9;\:4x-3y=44$

solving

$\begin{bmatrix}5x+2y=9\\ 4x-3y=44\end{bmatrix}$

$\mathrm{Multiply\:}5x+2y=9\mathrm{\:by\:}4\:\mathrm{:}\:\quad \:20x+8y=36$

$\mathrm{Multiply\:}4x-3y=44\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:20x-15y=220$

$\begin{bmatrix}20x+8y=36\\ 20x-15y=220\end{bmatrix}$

$20x-15y=220$

$-$

$\underline{20x+8y=36}$

$-23y=184$

$\begin{bmatrix}20x+8y=36\\ -23y=184\end{bmatrix}$

$-23y=184$

$y=-8.00$   ( -8.00 Rounded to the nearest 0.01 or  the Hundredths Place )

$\mathrm{For\:}20x+8y=36\mathrm{\:plug\:in\:}y=-8$

$20x+8\left(-8\right)=36$

$20x=100$

$x=5.00$  ( 5.00 Rounded to the nearest 0.01 or  the Hundredths Place )

So, the solution is:

(x, y) → (5.00, -8.00)