Answer:
x=3
y=5
Step-by-step explanation:
Method 1 ; Substitution Method
[tex]3x + 2y =19\\2x+2y=16\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:3x+2y=19:\\\quad x=\frac{19-2y}{3}\\\\\mathrm{Subsititute\:}x=\frac{19-2y}{3}\\\begin{bmatrix}2\times \frac{19-2y}{3}+2y=16\end{bmatrix}\\\\Simplify\\\begin{bmatrix}\frac{38+2y}{3}=16\end{bmatrix}\\\mathrm{Isolate}\:y\:\mathrm{for}\:\frac{38+2y}{3}=16:\quad y=5\\\\\mathrm{For\:}x=\frac{19-2y}{3}\\\\Substitute\:y\:= \:5\\x=\frac{19-2\times\:5}{3}\\\\\frac{19-2\times\:5}{3}=3\\\\x=3,\:y=5[/tex]
Method 2 : Elimination method
[tex]\begin{bmatrix}3x+2y=19\\ 2x+2y=16\end{bmatrix}\\\\\mathrm{Multiply\:}3x+2y=19\mathrm{\:by\:}2\:\\\mathrm{:}\:\quad \:6x+4y=38\\\\\mathrm{Multiply\:}2x+2y=16\mathrm{\:by\:}3\:\\\mathrm{:}\:\quad \:6x+6y=48\\\\\begin{bmatrix}6x+4y=38\\ 6x+6y=48\end{bmatrix}\\\\6x+6y=48\\-\\\underline{6x+4y=38}\\2y=10\\\\\begin{bmatrix}6x+4y=38\\ 2y=10\end{bmatrix}\\\\\mathrm{Solve}\:2y=10\:\mathrm{for}\:y:\quad y=5\\\mathrm{For\:}6x+4y=38\mathrm{\:plug\:in\:}y=5[/tex]
[tex]\mathrm{Solve}\:6x+4\times\:5=38\:\mathrm{for}\:x:\quad x=3\\\\x=3,\:y=5[/tex]
