Q4 of 71 Page 68

Solve the following pair of equations by cross multiplication method:





Putting these value in given equations, we get


4b + 7a = 16… (i), 10b + 3a = 11 ….. (ii)


So, the given equations transforms into linear equation in two variables.


Multiply equation (i) by 5 and (ii) by 2,


20b + 35a = 80 … (iii)


20b + 6a = 22….. (iv)


Subtract (iv) from (iii),


20b + 35a – 20b – 6a = 80 – 22


⇒ 29a = 58


⇒ a = 2


Putting above value in (ii),


10b + 3× 2 = 11


⇒ 10b = 11 – 6




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