Draw the graph of the linear equation $3x + 4y = 6$. At what points does the graph cut the $x$-axis and the $y$-axis?

(N/A) To draw the graph of the linear equation $3x + 4y = 6$,we first find some coordinate points that satisfy the equation.
$1$. To find the $x$-intercept,set $y = 0$:
$3x + 4(0) = 6 \implies 3x = 6 \implies x = 2$. So,the point is $(2, 0)$.
$2$. To find the $y$-intercept,set $x = 0$:
$3(0) + 4y = 6 \implies 4y = 6 \implies y = 1.5$. So,the point is $(0, 1.5)$.
$3$. To find another point,set $x = -2$:
$3(-2) + 4y = 6 \implies -6 + 4y = 6 \implies 4y = 12 \implies y = 3$. So,the point is $(-2, 3)$.
Summary table:

$x$	$2$	$0$	$-2$
$y$	$0$	$1.5$	$3$

Plot these points $(2, 0)$,$(0, 1.5)$,and $(-2, 3)$ on a Cartesian plane and join them to obtain a straight line.
The graph cuts the $x$-axis at $(2, 0)$ and the $y$-axis at $(0, 1.5)$.