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_01. Which of the choices below constitutes a simultaneous solution to these equations? (2 pts.) (1) 3X + 2Y = 36 and (2) 5X

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Answer #1

Answer

The correct answer is "option a"
a. X=8, Y=6

Let
3X+2Y=36 be equation i
5X+4Y=64 be equation ii

multiplying
equation i by 5
equation ii by 3

3X+2Y=36 * 5
5X+4Y=64 * 3

we get

15X+10y= 180 let this be equation iii
15X+12y= 192 let this be equation iv

subtracting equation iv from equation iii

+15X+10y= 180
-15X-12y= -192
-----------------
-2Y=-12
Y=6

Substituting value of Y in equation i
3X+2Y=36
3X+2(6)=36
3X+12=36
3X=36-12
3X=24
X=8

So the solution is (8,6)


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