Question 632161
You have 2 straight lines. The best way to get an
idea what they look like is to convert the equations
from the standard form ( as they now are ) into
the slope-intercept form
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{{{ 2x + y = 2 }}}
(1) {{{ y = -2x + 2 }}}
and
{{{ 6x + 4 y = 12 }}}
{{{ 3x + 2y = 6 }}}
(2) {{{ 2y = -3x + 6 }}}
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(1) and (2) are now in the form {{{ y = m*x + b }}}
where {{{ m }}} = slope and {{{ b }}} = y - intercept
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In (1), {{{ m = -2 }}}
In (2), {{{ m = -3 }}}
Both slope are negative, which means they go from
upper left to lower right on the graph, opposite to
what a positive slope does
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{{{ b }}} gives you a point for free on each line. It
is the y-intercept point which is (0,b), or
(0, 2) for equation (1)
(0,6) for equation (2)
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Now you just need 1 other point for both lines, and
since 2 points determine a line, you can draw
each line. Where the lines intersect is the solution
to the system of lines
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To find the other point, just plug in any value for
{{{ x }}}, and read off the value for {{{y}}}
For example:
(1) {{{ y = -2x + 2 }}}
I'll pick {{{ x = 1 }}}
(1) {{{ y = -2*1 + 2 }}}
(1) {{{ y = 0 }}}
So, I have the point (1,0)
and
(2) {{{ 2y = -3x + 6 }}}
I'll pick {{{ x = 2 }}}
(2) {{{ 2y = -3*2 + 6 }}}
(2) {{{ 2y = 0 }}}
(2) {{{ y = 0 }}}
So, I have the point (2,0)
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Now you have this information about the lines:
Line (1):
(0,2)
(1,0)
{{{ m = -2 }}}
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Line (2):
(0,6)
(2,0)
{{{ m = -3 }}}
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And that's all you need