Note: Your question sounds a little unclear, but I am assuming that your system of equations is:
[tex]-5x-\:\frac{3}{2}y=15[/tex]
[tex]3x+\frac{5}{6}y=-\frac{44}{3}[/tex]
- 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:
[tex]x=-\frac{57}{2},\:y=85[/tex]
Step-by-step explanation:
Given the system of equations
[tex]-5x-\:\frac{3}{2}y=15[/tex]
[tex]3x+\frac{5}{6}y=-\frac{44}{3}[/tex]
solving the system of equations
[tex]\begin{bmatrix}-5x-\frac{3}{2}y=15\\ 3x+\frac{5}{6}y=-\frac{44}{3}\end{bmatrix}[/tex]
[tex]\mathrm{Multiply\:}-5x-\frac{3}{2}y=15\mathrm{\:by\:}3\:\mathrm{:}\:\quad \:-15x-\frac{9}{2}y=45[/tex]
[tex]\mathrm{Multiply\:}3x+\frac{5}{6}y=-\frac{44}{3}\mathrm{\:by\:}5\:\mathrm{:}\:\quad \:15x+\frac{25}{6}y=-\frac{220}{3}[/tex]
so the system of equations becomes
[tex]\begin{bmatrix}-15x-\frac{9}{2}y=45\\ 15x+\frac{25}{6}y=-\frac{220}{3}\end{bmatrix}[/tex]
adding the equations
[tex]15x+\frac{25}{6}y=-\frac{220}{3}[/tex]
[tex]+[/tex]
[tex]\underline{-15x-\frac{9}{2}y=45}[/tex]
[tex]-\frac{1}{3}y=-\frac{85}{3}[/tex]
so
[tex]\begin{bmatrix}-15x-\frac{9}{2}y=45\\ -\frac{1}{3}y=-\frac{85}{3}\end{bmatrix}[/tex]
solving [tex]-\frac{1}{3}y=-\frac{85}{3}[/tex] for y
[tex]-\frac{1}{3}y=-\frac{85}{3}[/tex]
Multiply both sides by -3
[tex]\left(-\frac{1}{3}y\right)\left(-3\right)=\left(-\frac{85}{3}\right)\left(-3\right)[/tex]
[tex]y=85[/tex]
[tex]\mathrm{For\:}-15x-\frac{9}{2}y=45\mathrm{\:plug\:in\:}y=85[/tex]
[tex]-15x-\frac{9}{2}\cdot \:85=45[/tex]
[tex]\mathrm{Add\:}\frac{765}{2}\mathrm{\:to\:both\:sides}[/tex]
[tex]-15x-\frac{765}{2}+\frac{765}{2}=45+\frac{765}{2}[/tex]
[tex]-15x=\frac{855}{2}[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}-15[/tex]
[tex]\frac{-15x}{-15}=\frac{\frac{855}{2}}{-15}[/tex]
[tex]x=-\frac{57}{2}[/tex]
Therefore, the solution of the system of equations be:
[tex]x=-\frac{57}{2},\:y=85[/tex]
