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LUCKY_DIMON [66]
3 years ago
10

Solve for x and y. 4x-3y=16 -2x+4y=-2 DUE 2/7/17!

Mathematics
1 answer:
Trava [24]3 years ago
4 0
\left \{ {{4x-3y=16\:(I)} \atop {-2x+4y=-2\:(II)}} \right.

<span>Multiply the 2nd equation by (-4). This will eliminate the x's when you add the two new equations together.
</span>\left \{ {{4x-3y=16\:\:\:\:\:\:\:\:\:\:} \atop {-2x+4y=-2\:*(-4)}} \right.
\left \{ {{\diagup\!\!\!\!4x-3y=16} \atop {-\diagup\!\!\!\!4x+8y=-4}} \right.
--------------------
5y = 12
\boxed{\boxed{y =  \frac{12}{5}}}\end{array}}\qquad\quad\checkmark

Let's replace the value found in the first equation:
4x-3y=16\:(I)
4x-3*( \frac{12}{5}) =16
4x -  \frac{36}{5} = 16
least common multiple (5)
\frac{20x}{\diagup\!\!\!\!5} - \frac{36}{\diagup\!\!\!\!5} = \frac{80}{\diagup\!\!\!\!5}
20x - 36 = 80
20x = 80 + 36
20x = 116
x = \frac{116}{20} \frac{\div4}{\div4} \to\: \boxed{\boxed{x = \frac{29}{5} }}\end{array}}\qquad\quad\checkmark



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The relationship between two quantities could be additive (In this case, one quantity is a result of adding a value to the other quantity) or multiplicative (In this case, one quantity is the result of multiplying the other quantity by a value).

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