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Nady [450]
3 years ago
8

15x - 12y =13 30x +9y =4 whats the value of x

Mathematics
2 answers:
makkiz [27]3 years ago
7 0
So assuming that x and y are the same value in both equations we can make the first equation negative (multiply the whole thing by -1) and get

-15x+12y=-13
we multiply it by 2 and get
-30x+24y=-26
we add this to the second equation
(30x+9y=4)+(-30x+24y=-26)
30x-30x+24y+9y=-22
or 33y=-22
divide both sides by 11
3y=-2
divide both sides by 3
y=-2/3

we subtitute it into the first equation
15x-12(-2/3)=13
15x-(-24/3)=13
15x+8=13
subtract 8 from both sides
15x=5
divide both sides by 5
3x=1
divide both sides by 3
x=1/3

so x=1/3
y=-2/3
monitta3 years ago
4 0
<u />\left \{ {{15x-12y=13\ \ | multiply\ by\ -2} \atop {30x+9y=4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \\\\ \left \{ {{-30x+24y=-26} \atop {30x+9y=4}} \right.\\addition\ method\\\\33y=-22\ \ \ \ | divide\ by\ 33\\\\&#10;y=-\frac{22}{33}\\\\&#10;15x=13+12y\\\\&#10;x=\frac{13+12y}{15}=\frac{13+12*(-\frac{22}{33})}{15}=\frac{13-\frac{264}{33}}{15}=\frac{13-8}{15}=\frac{1}{3}\\\\&#10;Solution \ is\ x=\frac{1}{3}.&#10;
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The average (arithmetic mean) of five different positive integers is 30. What is the greatest possible value of one of these int
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Step-by-step explanation:

Given:  The average (arithmetic mean) of five different positive integers is 30.

To find:  What is the greatest possible value of one of these integers?

Explanation: we are given that the arithmetic mean of five  positive integer

                   is 30.

Let the five positive integers are A,B,C,D,E

The arithmetic mean :

                   \frac{A+B+C+D+E}{5} =30.

On multiplying both side by 5.

                       A+B+C+D+E =150.

The least values of A ,B,C and D can be 1 ,2,3,4.

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