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3 years ago

9

REALLY URGENT, PLEASE HELP

1 answer:

ASHA 777 [7]3 years ago

3 0

This can be solved by using simultaneous equations

If a cheap ball is X and an expensive one is y,

14x+6y=47

21x+10 y=72.5

Now all you have to do is solve for X and y.

Where X is the price of a cheap ball and y is the price of an expensive ball ⚽

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3 years ago

Which of the points is a solution to the following system of equations? -5x- 3/2y=15 3x+5/6 y=-44/3

tatiyna

<em>Note: Your question sounds a little unclear, but I am assuming that your system of equations is:</em>

$-5x-\:\frac{3}{2}y=15$

$3x+\frac{5}{6}y=-\frac{44}{3}$

It would anyways clear your concept because the procedure to find the solutions remains the same for any set of a system of equations.

Answer:

The solution of the system of equations be:

$x=-\frac{57}{2},\:y=85$

Step-by-step explanation:

Given the system of equations

$-5x-\:\frac{3}{2}y=15$

$3x+\frac{5}{6}y=-\frac{44}{3}$

solving the system of equations

$\begin{bmatrix}-5x-\frac{3}{2}y=15\\ 3x+\frac{5}{6}y=-\frac{44}{3}\end{bmatrix}$

$\mathrm{Multiply\:}-5x-\frac{3}{2}y=15\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:-15x-\frac{9}{2}y=45$

$\mathrm{Multiply\:}3x+\frac{5}{6}y=-\frac{44}{3}\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:15x+\frac{25}{6}y=-\frac{220}{3}$

so the system of equations becomes

$\begin{bmatrix}-15x-\frac{9}{2}y=45\\ 15x+\frac{25}{6}y=-\frac{220}{3}\end{bmatrix}$

adding the equations

$15x+\frac{25}{6}y=-\frac{220}{3}$

$+$

$\underline{-15x-\frac{9}{2}y=45}$

$-\frac{1}{3}y=-\frac{85}{3}$

so

$\begin{bmatrix}-15x-\frac{9}{2}y=45\\ -\frac{1}{3}y=-\frac{85}{3}\end{bmatrix}$

solving $-\frac{1}{3}y=-\frac{85}{3}$ for y

$-\frac{1}{3}y=-\frac{85}{3}$

Multiply both sides by -3

$\left(-\frac{1}{3}y\right)\left(-3\right)=\left(-\frac{85}{3}\right)\left(-3\right)$

$y=85$

$\mathrm{For\:}-15x-\frac{9}{2}y=45\mathrm{\:plug\:in\:}y=85$

$-15x-\frac{9}{2}\cdot \:85=45$

$\mathrm{Add\:}\frac{765}{2}\mathrm{\:to\:both\:sides}$

$-15x-\frac{765}{2}+\frac{765}{2}=45+\frac{765}{2}$

$-15x=\frac{855}{2}$

$\mathrm{Divide\:both\:sides\:by\:}-15$

$\frac{-15x}{-15}=\frac{\frac{855}{2}}{-15}$

$x=-\frac{57}{2}$

Therefore, the solution of the system of equations be:

$x=-\frac{57}{2},\:y=85$

7 0

3 years ago