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Sagot :
First line y=(4+10x)/2=5x+2
Second line : y=5x+2
We see that the two have the same leading coefficient (5) hence they are parallel.
Second line : y=5x+2
We see that the two have the same leading coefficient (5) hence they are parallel.
Well,
If the slope of the lines are the same, then the lines are parralel.
We need to manipulate 2y - 10x = 4 into y = mx + b form.
Add 10x to both sides
2y = 10x + 4
Divide both sides by 2
y = 5x + 2
Do the same thing with the other equation.
Add 2 to both sides
y = 5x + 2
y = 5x + 2
It appears that, not only are they parallel, but they lie on exactly the same line! If this was a System of Simultaneous Linear Equations, then there would be an infinite number of solutions!
If the slope of the lines are the same, then the lines are parralel.
We need to manipulate 2y - 10x = 4 into y = mx + b form.
Add 10x to both sides
2y = 10x + 4
Divide both sides by 2
y = 5x + 2
Do the same thing with the other equation.
Add 2 to both sides
y = 5x + 2
y = 5x + 2
It appears that, not only are they parallel, but they lie on exactly the same line! If this was a System of Simultaneous Linear Equations, then there would be an infinite number of solutions!
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