Posted by Jessica on Tuesday, August 17, 2010 at 7:47pm.
(1.) first, graph the equations,, there are various ways how to graph them,, for linear equations, i usually get the x- and y- intercept (*x- and y- intercepts are points*):
*note: to get x-intercept, set y=0 and solve for x,, to get y-intercept, set x=0 and solve for y,,
>>in the 1st equation, 2x + 5=0, since there is no variable y, just solve for x and you'll get x=-5/2. the graph of x=-5/2 is a vertical line passing through -5/2 (or -2.5)
*another note: this equation has no y-intercept since there is no value of y in which x will be zero (that is, the graph of x=-5/2 will never pass the y-axis)
>>in the 2nd equation, 2x + y=8, to get x-intercept, set y=0, so:
2x + (0) = 8 *solve for x
x=4
therefore, x-int: (4,0)
for the y-intercept, set x=0, so:
2(0) + y = 8 *solve for y
y=8
therefore, y-int: (0,8)
plot (4,0) and (0,8) on the same cartesian plane and connect, and extend the line,, this is now the graph of the 2nd equation,,
>>now, locate the point of intersection~
*to check if it's really the point of intersection, do substitution,, to do this, choose one of the equations, and express one variable in terms of the other variable,, in this case we choose the 1st equation since it readily gives the value of x which is -5/2,,
not substitute this to the 2nd equation:
2(-5/2) + y = 8
y=13
>>thus, the point of intersection is at (-5/2, 13)
(2.) graph y=|x|
*to do this, first, graph y=x (the version which does not contain the absolute value),, now, since y=|x| means y is restricted only to positive values, look for the area in the graph in which y is negative (it's in 3rd quadrant, isn't it),, then make a "mirror image" of this in the 2nd quadrant (2nd quadrant, because y values are positive in there),, thus its graph should be V-shaped,,
so there,, =)
sorry for long explanation..
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