Systems of equations, help please?

Mathematics

1 answer:

lord [1]3 years ago

3 0

4x + 3y = 54 (1)
3x + 9y = 108 (2)

Multiply (1) by (-3)
-12x - 9y = -162
3x + 9y = 108
---------------------add
-9x = -54
x = 6

plug x = 6 into (1) to find y
4(6) + 3y = 54
24 + 3y = 54
3y = 30
y = 10

Answer
(6 , 10)

Hope it helps.

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Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
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