Please try the following system first by elimination or substitution method and then use the x
and y intercepts and graph them to see if they intersect at the same point you found by the
method of eliminations or substitutions.
#1.
x+ y = 5 (1)
3x –y = 3 (2)
Adding both equations we have:
4x=8 then x=2
Using (1) 2+y=5
y=3
Solution: (2,3)
To graph x+y = 5 we use (0,5) and (5,0)
To graph 3x-y=3 we use (0,-3) and (1,0)
#2 Yes, lines intercept at (2,3)
2x + 3y = 13 (1)
3x - 5y = -9 (2)
10x+15y=65
9x-15y=-27
(multiply the first eqn by 5)
(multiply the second eqn by 3)
Then add them side-by-side
19x = 38
X=38/19= 2
Using (1)
2(2)+3y=13
4+3y=13
3y=9
Y=3
Solution : (2,3)
To graph 2x + 3y = 13
To graph 3x - 5y = -9
we use : (0,13/3) and (13/2,0)
we use: (0,9/5) and (-3,0)
Yes, lines intercept at (2,3)
x + y = 12
(1)
3x - 2y = 6
(2)
For the graph use the points: (12,
0) and (0, 12)
For the graph use the points: (2, 0) and (0, -3)
Multiply the first equation by -3:
-3x-3y=-36
3x-2y=6
Adding both equations:
-5y=-30
Y=6
Using (1)
-3x-3(6)=-36
-3x-18=-36
-3x=-18
X=6
Solution: (6,6)
Yes, lines intercept at (3,3)
Question for all: When using the method of elimination, after how many
steps, you would know for sure that the lines are going to be parallel and
there is no solution to the system of equations? Answer the same question
for infinitely many solutions.
We know that after the elimination of a variable (on both cases)
For example
2x+3y=8 (1)
4x+6y=11 (2)
Multiply (1) by -2
-4x-6y=-16
4x+6y=11
Adding:
0x+0y=-5 (no solution)