Easy
Find the points where the graph of the equation $3x + 4y = 12$ cuts the $x$-axis and the $y$-axis.
(N/A) The graph of the linear equation $3x + 4y = 12$ cuts the $x$-axis at the point where $y = 0$.
Putting $y = 0$ in the equation,we get $3x + 4(0) = 12$,which simplifies to $3x = 12$,so $x = 4$.
Thus,the point on the $x$-axis is $(4, 0)$.
The graph of the linear equation $3x + 4y = 12$ cuts the $y$-axis at the point where $x = 0$.
Putting $x = 0$ in the equation,we get $3(0) + 4y = 12$,which simplifies to $4y = 12$,so $y = 3$.
Thus,the point on the $y$-axis is $(0, 3)$.
Putting $y = 0$ in the equation,we get $3x + 4(0) = 12$,which simplifies to $3x = 12$,so $x = 4$.
Thus,the point on the $x$-axis is $(4, 0)$.
The graph of the linear equation $3x + 4y = 12$ cuts the $y$-axis at the point where $x = 0$.
Putting $x = 0$ in the equation,we get $3(0) + 4y = 12$,which simplifies to $4y = 12$,so $y = 3$.
Thus,the point on the $y$-axis is $(0, 3)$.
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