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Solve the following pair of linear equations by the substitution and cross-multiplication methods

8x + 5y = 9

3x + 2y = 4

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#### 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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