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If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arith
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02 Feb 2019, 02:33
02 Feb 2019, 02:33
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Question Stats:
89% (01:25) correct
10% (01:23) wrong based on 64 sessions
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If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arithmetic mean) of x and y?
Show: :: OA
\(\frac{18}{10}\)
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Joined: 27 Nov 2018
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Re: If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arith
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06 Feb 2019, 01:58
06 Feb 2019, 01:58
2
18/10.
7x + 3y = 17 +
3x + 7y = 19
=
10x+10y=36
x+y=36/10
average of x,y = x+y/2 = 36/20=18/10
7x + 3y = 17 +
3x + 7y = 19
=
10x+10y=36
x+y=36/10
average of x,y = x+y/2 = 36/20=18/10
Re: If 7x + 3y = 17 and 3x + 7y = 19, what is the average (arith
[#permalink]
28 Sep 2019, 07:15
28 Sep 2019, 07:15
1
Lets solve these two equations first
EQ1 : 7x+3y = 17
EQ2: 3x+7y = 19
Let’s multiply eq1 by 3 to get : 21x+9y = 51
Let’s multiply eq2 by 7 to get : 21x+49y = 133
Now, lets subtract EQ1-EQ2 to get : -40y = -82
So, y = 82/40=41/20
Now, lets substitute y in EQ1 to get : 7x+ 3(41/20)=17
7x + 123/20=17
7x = 17 – 123/20
7x = (340-123)/20
7x = 217/20
X = 31/20
So, X+Y = 31/20+41/20= 72/20=36/10
Average of x & y is (36/10)/2= 18/10
EQ1 : 7x+3y = 17
EQ2: 3x+7y = 19
Let’s multiply eq1 by 3 to get : 21x+9y = 51
Let’s multiply eq2 by 7 to get : 21x+49y = 133
Now, lets subtract EQ1-EQ2 to get : -40y = -82
So, y = 82/40=41/20
Now, lets substitute y in EQ1 to get : 7x+ 3(41/20)=17
7x + 123/20=17
7x = 17 – 123/20
7x = (340-123)/20
7x = 217/20
X = 31/20
So, X+Y = 31/20+41/20= 72/20=36/10
Average of x & y is (36/10)/2= 18/10
