solve the following pairs of linear equation by method of substitution 3x+y+1=0,2x-3y+8=0
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Hiiii friend,
3X + Y +1 = 0
3X + Y = -1------(1)
And,
2X - 3Y +8 = 0
2X - 3Y = -8......(2)
From equation (1) we get,
3X + Y = -1
X = -1 - Y/3.........(3)
Putting the value of Y in equation (2)
2X - 3Y = -8
2 × (-1-Y/3) - 3Y = -8
-2 - 2Y/3 - 3Y = -8
-2-2Y -9Y = -8×3
-11Y = -24+2
-11Y = -22
Y = -22/-11 => 2
Putting the value of X in equation (3)
X = -1-Y/3 => -1 - (2)/3 = -3/3 = -1
Hence,
X = -1 and Y = 2
HOPE IT WILL HELP YOU....... :-)
3X + Y +1 = 0
3X + Y = -1------(1)
And,
2X - 3Y +8 = 0
2X - 3Y = -8......(2)
From equation (1) we get,
3X + Y = -1
X = -1 - Y/3.........(3)
Putting the value of Y in equation (2)
2X - 3Y = -8
2 × (-1-Y/3) - 3Y = -8
-2 - 2Y/3 - 3Y = -8
-2-2Y -9Y = -8×3
-11Y = -24+2
-11Y = -22
Y = -22/-11 => 2
Putting the value of X in equation (3)
X = -1-Y/3 => -1 - (2)/3 = -3/3 = -1
Hence,
X = -1 and Y = 2
HOPE IT WILL HELP YOU....... :-)
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