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Question
Solve the following pair of linear equations by the substitution and cross-multiplication methods
8x + 5y = 9
3x + 2y = 4
Solution
8x +5y = 9 ... (i)
3x +2y = 4 ... (ii)
From equation (ii), we get
`x = (4-2y)/3 ... (iii)`
Putting this value in equation (i), we get
`8((4-2y)/3) + 5y = 9`
32 - 16y +15y = 27
-y = -5
y = 5 ... (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, y = 5
By cross multiplication again, we get
8x + 5y -9 = 0
3x + 2y - 4 = 0
`x/(-20-(-18)) = y/(-27-(-32)) = 1/(16-15)`
`x/-2 = y/5 = 1/1`
x/-2 = 1 and y/5 = 1
x = -2 and y = 5
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