1 Draw the graphs of the following linear equations:
(i) 2x + y + 3 = 0
Answers
PLZ MARK BEST AND THANK FOR THE HARD WORK.
HOPEIT HELPS.
2
x
+
y
=
3
⇒
2
(
0
)
+
y
=
3
⇒
0
+
y
=
3
⇒
y
=
3
So when
x
=
0
, we have
y
=
3
. Meaning the point
(
x
,
y
)
=
(
0
,
3
)
is a solution to
2
x
+
y
=
3
, and so our graph of the equation will pass through
(
0
,
3
)
.
We can get a few more points:
If
x
=
1
, then
2
x
+
y
=
3
⇒
2
(
1
)
+
y
=
3
⇒
2
+
y
=
3
⇒
y
=
1
So
(
x
,
y
)
=
(
1
,
1
)
is also on our graph.
Similarly, if
x
=
2
, then we get
y
=
–
1
, giving us the point
(
x
,
y
)
=
(
2
,
–
1
)
.
We then take the points we've computed,
(
0
,
3
)
,
(
1
,
1
)
,
(
2
,
–
1
)
, and plot them on a graph:
graph{((x)^2+(y-3)^2-.02)((x-1)^2+(y-1)^2-.02)((x-2)^2+(y+1)^
Step-by-step explanation:
solution:-
2x+y= 3
=> y= 3- 2x
so let us get all the values,
if x=1, y= ?
y= 3-2(1)
y= 3-2
y= 1
hence first coordinate is (1,1)
if x= 2, y= ?
y= 3-2(2)
y
= 3-4
y= -1
hence second coordinate is (2,-1)
if x= 0, y= ?
y= 3-2(0)
y= 3-0
y= 3
hence third coordinate is (0,3)
hey mate this is your answer.
please mark me as a brainlist.
