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# Plato Classroom Question #16

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Consider this system of equations:

$$^2/_3x+\text{ }^3/_5y=12$$ (equation A)

$$^5/_2y-3x=6$$(equation B)

The expression that gives the value of x is $$\boxed{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow}$$.
The solution for the system of equations is $$\boxed{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow}$$.

$$\boxed{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow}\\ \boxed{^{5a}/_3+\text{ }^{2b}/_5\text{ }\text{ }\text{ }\text{ }\text{ }\\\ ^{3a}/_5-\text{ }^{5b}/_2\\ ^{5a}/_2+\text{ }^{3b}/_5\\ ^{5a}/_3-\text{ }^{2b}/_5\\ ^{3a}/_2+\text{ }^b/_5}$$ $$\boxed{\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\downarrow}\\ \boxed{(^{1118}/_{47},\text{ }^{396}/_{47})\text{ }\text{ }\text{ }\\ (^{33}/_{13},\text{ }^{50}/_{13})\\ (^{99}/_{13},\text{ }^{150}/_{13})\\ (8,12)}$$

SpaceModo  Jan 24, 2018
edited by SpaceModo  Jan 24, 2018
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(2/3)x  + (3/5)y  =  12

-3x   + (5/2)y  =  6

Mutiply  the first equation through by the common denominator of 3 and 5 = 15

Multiply the second equation through by 2

So we have

10x  +  9y   =  180    multipy through by 6  =   60x + 54y  = 1080    (1)

-6x  +  5y      =  12  multiply through   by 10  =   -60x  + 50y  =   120    (2)

104y  =  1200     divide by  104

y =   1200/104 =   600/52  =  150 / 13

Obviously, only one answer has a "y" answer that  =  150/13.....so that must be the correct one

CPhill  Jan 24, 2018

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