Solve the following system of equations using the cross multiplication method.
$7 x - y = 33 12 y - x = 19$

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Solution

Step 1:Applying & Analysing
The given system of equations may be written as
$7 x - y - 33 = 0, - x + 12 y - 19 = 0 \Rightarrow x - 12 y + 19 = 0$
The given system of equation is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
Step 2: Using cross multiplication method
$x = \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}, y = \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}} W h e r e, a_{1} = 7, b_{1} = - 1, c_{1} = - 33 a_{2} = 1, b_{2} = - 12, c_{2} = 19$
Substitute the given values in the above equation, We get
$x = \frac{- 1 x 19 - (- 12 x) - 33}{7 x - 12 - 1 x - 1} = \frac{- 191396}{- 84 + 1} = \frac{- 415}{- 83} = 5 y = \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}} = \frac{- 33 x 1 - 19 x 7}{7 x - 12 - 1 x - 1} = \frac{- 33 - 133}{- 84 + 1} = \frac{- 166}{- 83} = 2 T h e r e f o r e x = 5 & y = 2$
Step 3:Result
Hence x=5 & y=2 is the required solution

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