Given, $- 3 x + 4 y = 20$ and $6 x + 3 y = 15$
If $(x, y)$ is the solution to the system of equations above, what is the value of $x$ ?

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Solution

Let $- 3 x + 4 y = 20$ ....(1)
$6 x + 3 y = 15$ ....(2)
Multiplying equation (1) with $2$ , we get
$- 6 x + 8 y = 40$ ....(3)
Adding equations $1$ and $3$ , we get $11 y = 55$
$\Rightarrow y = 5$
Substituting in equation (1), we get
$- 3 x + 4 (5) = 20$
$\Rightarrow - 3 x + 20 = 20$
$\Rightarrow - 3 x = 0$
$\Rightarrow x = 0$

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