[tex] \left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right. [/tex]
Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
[tex]\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right. [/tex]
[tex]\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right. [/tex]
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[tex]5y = 12[/tex]
[tex]\boxed{\boxed{y = \frac{12}{5}}}\end{array}}\qquad\quad\checkmark[/tex]
Let's replace the value found in the first equation:
[tex]4x-3y=16\:(I)[/tex]
[tex]4x-3*( \frac{12}{5}) =16[/tex]
[tex]4x - \frac{36}{5} = 16[/tex]
least common multiple (5)
[tex] \frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5} [/tex]
[tex]20x - 36 = 80[/tex]
[tex]20x = 80 + 36[/tex]
[tex]20x = 116[/tex]
[tex]x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark[/tex]
