Question 153545
You're correct for both. 
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Now that you have the y intercept and the slope you can plot the graph.
The y-intercept for the first line is 5 or (0,5).
Plot it. 

{{{drawing( 300, 300, -20, 20, -20, 20,grid( 1 ),circle( 0, 5, .35 ))}}}
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You can find another point using the slope of -1.
That means for a positive step in x, you take an equal step in the negative direction in y.
From (0,5) moving 1 in x, -1 in y would give you (1,4).
If you don't want to use slope or it's cumbersome, 
another useful point to plot is the x-intercept.
Set y=0 and solve for y.
{{{y=-x+5}}}
{{{0=-x+5}}}
{{{x=5}}}
(5,0) Plot that too.
{{{drawing( 300, 300, -20, 20, -20, 20,grid( 1 ),circle(5,0,.4),circle( 0, 5, .4 ))}}}
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Then draw the line that connects those two points.
{{{drawing( 300, 300, -20, 20, -20,20,grid( 1 ),circle(5,0,.4),circle( 0, 5, .4 ),green(line( -20, 25, 20, -15)))}}}
Line 1 is done. 
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y=3x-11


Line 2- y intercept (0,-11)
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{{{drawing( 300, 300, -20, 20, -20,20,grid( 1 ),circle(5,0,.4),circle( 0, 5, .4 ),circle(0,-11,.4),green(line( -20, 25, 20, -15)))}}}
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The slope is 3, so for every step in x, you take 3 steps in y.
(0,-11)
(1,-8)
(2,-5)
(3,-2)
(4,1)
(5,4)
(6,7)
(7,10)
We have many to choose from.
(7,10)
{{{drawing( 300, 300, -20, 20, -20,20,grid( 1 ),circle(5,0,.4),circle( 0, 5, .4 ),circle(7,10,.4),circle(0,-11,.4),green(line( -20, 25, 20, -15)))}}}
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{{{drawing( 300, 300, -20, 20, -20,20,grid( 1 ),circle(5,0,.4),circle( 0, 5, .4 ),circle(7,10,.4),circle(0,-11,.4),blue(line( -20, -71, 20, 49)),green(line( -20, 25, 20, -15)))}}}
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Looks like an intersection at x=4
Line 1 : {{{y=-x+5=-4+5=1}}}
(4,1)
Line 2 : {{{y=3x-11=3(4)-11=1}}}
(4,1)
They share a common intersection point at (4,1).
That is the solution to system of equations. 
{{{system(x+y=5,
3x-y=11)}}}