#### Question

Solve each of the following systems of equations by the method of cross-multiplication

5ax + 6by = 28

3ax + 4by = 18

#### Solution

The given system of equation is

5ax + 6by = 28

`=> 5ax + 6by - 28 = 0` .....(i)

and 3ax + 4by - 18 = 0

=> 3ax + 4by - 18 = 0 ....(ii)

Here

`a_1 = 5a , b_1 = 6b, c_1 = -28`

`a_2 = 3a, b_2 = 4b, c_2 = -18`

By cross multiplication we have

`= x/(6b xx (-18) - (-28) xx 4b) = (-4)/(5a xx (-18) - (-28) xx 3a) = 1/(5a xx 4b - 6b xx 3a)`

`=> x/(-108b + 112b) = (-y)/(-90a + 80a) = 1/(20ab - 18ab)`

`=> x/(4b) = (-y)/(-6a) = 1/(2ab)`

Now

`x/(4b) = (-y)/(-6a) = 1/(2ab)`

Now

`x/(4b) = 1/(2ab)`

`=> x = (5b - 2a)/10ab`

And

`(-y)/(-6a) = 1/(2ab)`

`=> y = (6a)/(2ab) = 3/b`

Hence `x = 2/a, y = 3/b` is the soluytion of the given system of equation.

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#### APPEARS IN

Solution Solve Each of the Following Systems of Equations by the Method of Cross-multiplication 5ax + 6by = 28 3ax + 4by = 18 Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method.