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Sagot :
2x + 5y = 5 ----equation 1
-3x + 5y = 5 ----equation 2
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Subtract equation 2 from equation 1
2x + 5y = 5
- -3x + 5y = 5
5x + 0 = 0
5x = 0
x = 0
-----------------
Substitute x=0 to equation 1
2x + 5y = 5
2(0) + 5y = 5
5y = 5
y = 1
-----------------
Therefore the two lines intersect at (0,1)
--------------------
For the graph, let's consider the x and y intercepts of the two lines
LINE 1: 2x + 5y = 5
x-intercept, y=0
2x + 5(0) = 5
2x = 5
x = 5/2
x-intercept (5/2,0)
y-intercept, x=0
2(0) + 5y = 5
5y = 5
y = 1
y-intercept (0,1)
---------------------
LINE 2: -3x + 5y = 5
x-intercept, y=0
-3x + 5(0) = 5
-3x = 5
x = -5/3
x-intercept (-5/3,0)
y-intercept, x=0
-3(0) + 5y = 5
5y = 5
y = 1
y-intercept (0,1)
--------------------------
see attachment for the graph
-3x + 5y = 5 ----equation 2
------------------
Subtract equation 2 from equation 1
2x + 5y = 5
- -3x + 5y = 5
5x + 0 = 0
5x = 0
x = 0
-----------------
Substitute x=0 to equation 1
2x + 5y = 5
2(0) + 5y = 5
5y = 5
y = 1
-----------------
Therefore the two lines intersect at (0,1)
--------------------
For the graph, let's consider the x and y intercepts of the two lines
LINE 1: 2x + 5y = 5
x-intercept, y=0
2x + 5(0) = 5
2x = 5
x = 5/2
x-intercept (5/2,0)
y-intercept, x=0
2(0) + 5y = 5
5y = 5
y = 1
y-intercept (0,1)
---------------------
LINE 2: -3x + 5y = 5
x-intercept, y=0
-3x + 5(0) = 5
-3x = 5
x = -5/3
x-intercept (-5/3,0)
y-intercept, x=0
-3(0) + 5y = 5
5y = 5
y = 1
y-intercept (0,1)
--------------------------
see attachment for the graph
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